find the value of c that makes each trinomial a perfect … square root of a negative number has no...

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Find the value of c that makes each trinomial a perfect square. 1. x 2 18x + c SOLUTION: In this trinomial, b = 18. So, c must be 81 to make the trinomial a perfect square. 2. x 2 + 22x + c SOLUTION: In this trinomial, b = 22. So, c must be 121 to make the trinomial a perfect square. 3. x 2 + 9x + c SOLUTION: In this trinomial, b = 9. So, c must be to make the trinomial a perfect square. 2 eSolutions Manual - Powered by Cognero Page 1 9 - 4 Solving Quadratic Equations by Completing the Square

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Find the value of c that makes each trinomial a perfect square.

1. x2 − 18x + c

SOLUTION:  In this trinomial, b = –18.

So, c must be 81 to make the trinomial a perfect square.

2. x2 + 22x + c

SOLUTION:  In this trinomial, b = 22.

So, c must be 121 to make the trinomial a perfect square.

3. x2 + 9x + c

SOLUTION:  In this trinomial, b = 9.

 

So, c must be  to make the trinomial a perfect square.

4. x2 − 7x + c

SOLUTION:  In this trinomial, b = –7.

 

So, c must be  to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

5. x2 + 4x = 6

SOLUTION:  

The solutions are about –5.2 and 1.2.

6. x2 − 8x = −9

SOLUTION:  

The solutions are about 1.4 and 6.6.

7. 4x2 + 9x − 1 = 0

SOLUTION:  

The solutions are about –2.4 and 0.1.

8. −2x2 + 10x + 22 = 4

SOLUTION:  

The solutions are about –1.4 and 6.4.

9. CCSS MODELING Collin is building a deck on the back of his family’s house. He has enough lumber for the deck to be 144 square feet. The length should be 10 feet more than its width. What should the dimensions of the deck be?

SOLUTION:  Let x = the width of the deck and x + 10 = the length of the deck.

The dimensions of the deck can not be negative. So, x = –5 + 13 or 8. The width of the deck is 8 feet and the length of the deck is 8 + 10 or 18 feet.

Find the value of c that makes each trinomial a perfect square.

10. x2 + 26x + c

SOLUTION:  In this trinomial, b = 26.

So, c must be 169 to make the trinomial a perfect square.

11. x2 − 24x + c

SOLUTION:  In this trinomial, b = –24.

So, c must be 144 to make the trinomial a perfect square.

12. x2 − 19x + c

SOLUTION:  In this trinomial, b = –19.

So, c must be  to make the trinomial a perfect square.

13. x2 + 17x + c

SOLUTION:  In this trinomial, b = 17.

So, c must be  to make the trinomial a perfect square.

14. x2 + 5x + c

SOLUTION:  In this trinomial, b = 5.

So, c must be  to make the trinomial a perfect square.

15. x2 − 13x + c

SOLUTION:  In this trinomial, b = –13.

So, c must be  to make the trinomial a perfect square.

16. x2 − 22x + c

SOLUTION:  In this trinomial, b = –22.

So, c must be 121 to make the trinomial a perfect square.

17. x2 − 15x + c

SOLUTION:  In this trinomial, b = –15.

So, c must be  to make the trinomial a perfect square.

18. x2 + 24x + c

SOLUTION:  In this trinomial, b = 24.

So, c must be 144 to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

19. x2 + 6x − 16 = 0

SOLUTION:  

  The solutions are –8 and 2.

20. x2 − 2x − 14 = 0

SOLUTION:  

  The solutions are about –2.9 and 4.9.

21. x2 − 8x − 1 = 8

SOLUTION:  

  The solutions are –1 and 9.

22. x2 + 3x + 21 = 22

SOLUTION:  

  The solutions are about –3.3 and 0.3.

23. x2 − 11x + 3 = 5

SOLUTION:  

  The solutions are about –0.2 and 11.2.

24. 5x2 − 10x = 23

SOLUTION:  

  The solutions are about –1.4 and 3.4.

25. 2x2 − 2x + 7 = 5

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

26. 3x2 + 12x + 81 = 15

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

27. 4x2 + 6x = 12

SOLUTION:  

  The solutions are about –2.6 and 1.1.

28. 4x2 + 5 = 10x

SOLUTION:  

  The solutions are about 0.7 and 1.8.

29. −2x2 + 10x = −14

SOLUTION:  

  The solutions are about –1.1 and 6.1.

30. −3x2 − 12 = 14x

SOLUTION:  

  The solutions are about –3.5 and –1.1.

31. STOCK The price p in dollars for a particular stock can be modeled by the quadratic equation p = 3.5t − 0.05t2, wh

the number of days after the stock is purchased. When is the stock worth $60?

SOLUTION:  Let p = 60.

  The stock is worth $60 on the 30th and 40th days after purchase.

GEOMETRY  Find the value of x for each figure. Round to the nearest tenth if necessary.

32. area = 45 in2

SOLUTION:  

  The height cannot be negative. So, x ≈ 6.3.

33. area = 110 ft2

SOLUTION:  

  The dimensions of the rectangle cannot be negative. So, x ≈ 5.3.

34. NUMBER THEORY The product of two consecutive even integers is 224. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers are 14 and 16 or –16 and –14.

35. CCSS PRECISION The product of two consecutive negative odd integers is 483. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers must be negative. So, they are –23 and –21.

36. GEOMETRY Find the area of the triangle.

SOLUTION:  

  The dimensions of the triangle cannot be negative. So, x = 18. The base of the triangle is 18 meters and the height is 18 + 6 or 24 meters.

The area of the triangle is 216 square meters.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

37. 0.2x2 − 0.2x − 0.4 = 0

SOLUTION:  

  The solutions are –1 and 2.

38. 0.5x2 = 2x − 0.3

SOLUTION:  

  The solutions are about 0.2 and 3.8.

39. 2x2 −

SOLUTION:  

  The solutions are about 0.2 and 0.9.

40. 

SOLUTION:  

  The solutions are –0.5 and 2.5.

41. 

SOLUTION:  

  The solutions are about –8.2 and 0.2.

42. 

SOLUTION:  

  The solutions are about –5.1 and 0.1.

43. ASTRONOMY The height of an object t seconds after it is dropped is given by the equation ,

where h0 is the initial height and g is the acceleration due to gravity. The acceleration due to gravity near the surface

of Mars is 3.73 m/s2, while on Earth it is 9.8 m/s

2. Suppose an object is dropped from an initial height of 120 meters

above the surface of each planet. a. On which planet would the object reach the ground first? b. How long would it take the object to reach the ground on each planet? Round each answer to the nearest tenth. c. Do the times that it takes the object to reach the ground seem reasonable? Explain your reasoning.

SOLUTION:  a. The object on Earth will reach the ground first because it is falling at a faster rate.   b. Mars:

So, t ≈ 8.0 seconds. Earth:

So, t ≈ 4.9 seconds.   c. Sample answer: Yes; the acceleration due to gravity is much greater on Earth than on Mars, so the time to reach the ground should be much less.

44. Find all values of c that make x2 + cx + 100 a perfect square trinomial.

SOLUTION:  

So, c can be –20 or 20 to make the trinomial a perfect square.

45. Find all values of c that make x2 + cx + 225 a perfect square trinomial.

SOLUTION:  

So, c can be –30 or 30 to make the trinomial a perfect square.

46. PAINTING Before she begins painting a picture, Donna stretches her canvas over a wood frame. The frame has a length of 60 inches and a width of 4 inches. She has enough canvas to cover 480 square inches. Donna decides to increase the dimensions of the frame. If the increase in the length is 10 times the increase in the width, what will the dimensions of the frame be?

SOLUTION:  Let x = the increase in the width and let 10x = the increase in the length. So, x + 4 = the new width and 10x + 60 = the new length.  

  The dimensions cannot be negative, so x = 2. The width is 2 + 4 or 6 inches and the length is 10(2) + 60 or 80 inches.

47. MULTIPLE REPRESENTATIONS In this problem, you will investigate a property of quadratic equations. a. TABULAR  Copy the table shown and complete the second column.

b. ALGEBRAIC Set each trinomial equal to zero, and solve the equation by completing the square. Complete the last column of the table with the number of roots of each equation.

c. VERBAL Compare the number of roots of each equation to the result in the b2 − 4ac column. Is there a

relationship between these values? If so, describe it.

d. ANALYTICAL Predict how many solutions 2x2 − 9x + 15 = 0 will have.  Verify your prediction by solving the 

equation.

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3    

3x2 + 6x + 9    

x2 − 2x + 7    

x2 + 10x + 25    

x2 + 3x − 12    

SOLUTION:  a.

b.

The trinomial x2 − 8x + 16 has 1 root.

 

The trinomial 2x

2 − 11x + 3 has 2 roots.

 

The square root of a negative number has no real roots. So, the trinomial 3x2 + 6x + 9 has 0 roots.

 

The square root of a negative number has no real roots. So, the trinomial x2 − 2x + 7.

 

The trinomial x2 + 10x + 25 has 1 root.

 

The trinomial x2 + 3x − 12 has 2 roots.

 

c. If b2 − 4ac is negative, the equation has no real solutions. If b

2 − 4ac is zero, the equation has one solution. If b2

− 4ac is positive, the equation has 2 solutions. d.

The equation 2x2 − 9x + 15 = 0 has 0 real solutions because b

2 − 4ac is negative.

 

The equation cannot be solved because the square root of a negative number has no real roots.

Trinomial b2 − 4ac Number

of Roots

x2 − 8x + 16 (–8)

2 – 4(1)(16) = 0 1

2x2 − 11x + 3 (–11)

2 – 4(2)(3) = 97  

3x2 + 6x + 9 (6)

2 – 4(3)(9) = −72  

x2 − 2x + 7 (–2)

2 – 4(1)(7) = −24  

x2 + 10x + 25 (10)

2 – 4(1)(25) = 0  

x2 + 3x − 12 (3)

2 – 4(1)(–12) = 57  

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3 97 2

3x2 + 6x + 9 −72 0

x2 − 2x + 7 −24 0

x2 + 10x + 25 0 1

x2 + 3x − 12 57 2

48. CCSS PERSEVERANCE Given y = ax2 + bx + c with a ≠ 0, derive the equation for the axis of symmetry by 

completing the square and rewriting the equation in the form y = a(x − h)2 + k .

SOLUTION:  

Let    and   , then the equation becomes y = (x - h)2 + k .

For an equation in the form y = ax2 + bx + c,  the equation for the axis of symmetry is   .

So, for an equation in the form  y = (x - h)2 + k, the equation for the axis of symmetry becomes x = h.

49. REASONING Determine the number of solutions x2 + bx = c has if . Explain.

SOLUTION:  

None; Sample answer: If you add to each side of the equation and each side of the inequality, you get

. Since the left side of the last equation is a perfect square, it cannot

equal the negative number . So, there are no real solutions.

50. WHICH ONE DOESN’T BELONG? Identify the expression that does not belong with the other three. Explain your reasoning.

 

SOLUTION:  The first 3 trinomials are perfect squares.

The trinomial  is not a perfect square. So, it does not belong.

51. OPEN ENDED Write a quadratic equation for which the only solution is 4.

SOLUTION:  Find a quadratic that has two roots of 4.

Verify that the solution is 4.

So, the quadratic equation x2 − 8x + 16 = 0 has a solution of 4.

52. WRITING IN MATH Compare and contrast the following strategies for solving x2 − 5x − 7 = 0: completing the

square, graphing, and factoring.

SOLUTION:  Sample answer: Because the leading coefficient is 1, solving the equation by completing the square is simpler and yields exact answers for the solutions.   

 

To solve the equation by graphing, use a graphing calculator to graph the related function y = x2 - 5x - 7. Select the

zero option from the 2nd [CALC] menu to determine the roots. 

   

  The roots are given as decimal approximations. So, for this equation the solutions would have to be given as estimations.   There are no factors of –7 that have a sum of –5, so solving by factoring is not possible.

53. The length of a rectangle is 3 times its width. The area of the rectangle is 75 square feet. Find the length of the rectangle in feet. A 25 B 15 C 10 D 5

SOLUTION:  Let x = the width of the rectangle and let 3x = the length of the rectangle.

The length of the rectangle is 3(5) or 15 feet. Choice B is the correct answer.

54. PROBABILITY At a festival, winners of a game draw a token for a prize. There is one token for each prize. The prizes include 9 movie passes, 8 stuffed animals, 5 hats, 10 jump ropes, and 4 glow necklaces. What is the probabilitythat the first person to draw a token will win a movie pass?

F  

G

H

J  

SOLUTION:  There are 9 + 8 + 5 + 10 + 4 or 36 possible outcomes.

P(movie pass) =  or  . Choice J is the correct answer.

55. GRIDDED RESPONSE The population of a town can be modeled by P = 22,000 + 125t, where P represents the population and t represents the number of years from 2000. How many years after 2000 will the population be 26,000?

SOLUTION:  

In 32 years the population of the town will be 26,000.

56. Percy delivers pizzas for Pizza King. He is paid $6 an hour plus $2.50 for each pizza he delivers. Percy earned $280 last week. If he worked a total of 30 hours, how many pizzas did he deliver? A 250 pizzas B 184 pizzas C 40 pizzas D 34 pizzas

SOLUTION:  Let p = the number of pizzas Percy delivered.

Percy delivered 40 pizzas. Choice C is the correct answer.

Describe how the graph of each function is related to the graph of f (x) = x2.

57. g(x) = −12 + x2

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –12, the translation is

down. So, the graph is shifted down 12 units from the parent function. 

58. h(x) = (x + 2)2

SOLUTION:  

The graph of f (x) = (x – c)2 represents a translation left or right from the parent graph. Since c = –2, the translation

is to the left by 2 units. 

 

59. g(x) = 2x2 + 5

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 2 and c = 5. Since 2 > 0 and  > 1, the graph of y = 2x

2 + 5

is the graph of y = x2 vertically stretched and shifted up 5 units.

60. 

SOLUTION:  

The function can be written f (x) = a(x – b)2, where a = and b = 6. Since > 0 and  < 1, the graph of 

is the graph of y = x2 vertically compressed and shifted right 6 units.

61. g(x) = 6 + x2

SOLUTION:  

The function can be written f (x) = ax2 + c, where a =  and c = 6. Since  > 0 and   > 1, the graph of y = 6 +

x2 is the graph of y = x

2 vertically stretched and shifted up 6 units.

62. h(x) = − 1 −   x2

SOLUTION:  

The function can be written f (x) = –ax2 + c, where a =  and c = –1. Since  < 0 and   > 1, the graph of y

= –1 – x2 is the graph of y = x

2 vertically stretched, shifted down 1 unit and reflected across the x-axis.

63. RIDES A popular amusement park ride whisks riders to the top of a 250-foot tower and drops them. A function for

the height of a rider is h = −16t2 + 250, where h is the height and t is the time in seconds. The ride stops the descent

of the rider 40 feet above the ground. Write an equation that models the drop of the rider. How long does it take to fall from 250 feet to 40 feet?

SOLUTION:  

The ride stops the descent of the rider at 40 feet above ground, so h = 40. Then, the equation 40 = −16t2 + 250

models the drop of the rider.

Time cannot be negative. So, it takes about 3.6 seconds to complete the ride.

Simplify. Assume that no denominator is equal to zero.

64. 

SOLUTION:  

65. 

SOLUTION:  

66. 

SOLUTION:  

67. 

SOLUTION:  

68. 

SOLUTION:  

69. b3(m

−3)(b

−6)

SOLUTION:  

Solve each open sentence.

70. |y − 2| > 7

SOLUTION:  

The solution set is {y |y > 9 or y < −5}.

Case 1 y – 2 is positive.

and Case 2 y – 2 is negative.

 

71. |z + 5| < 3

SOLUTION:  

The solution set is {z |−8 < z < −2}.

Case 1 z +5 is positive.

and Case 2 z +5 is negative.

 

72. |2b + 7| ≤ −6

SOLUTION:  

cannot be negative. So,  cannot be less than or equal to –6. The solution set is empty.

73. |3 − 2y | ≥ 8

SOLUTION:  

The solution set is {y |y  ≥ 5.5 or y  ≤ −2.5}.

Case 1 3 – 2y is positive.

and Case 2 3 – 2y is negative.

 

74. |9 − 4m| < −1

SOLUTION:  

cannot be negative. So, cannot be less than or equal to –1. The solution set is empty.

75. |5c − 2| ≤ 13

SOLUTION:  

The solution set is {c|−2.2 ≤ c ≤ 3}.

Case 1 5c – 2 is positive.

and Case 2 5c – 2 is negative.

 

Evaluate  for each set of values. Round to the nearest tenth if necessary.

76. a = 2, b = −5, c = 2

SOLUTION:  

77. a = 1, b = 12, c = 11

SOLUTION:  

78. a = −9, b = 10, c = −1

SOLUTION:  

79. a = 1, b = 7, c = −3

SOLUTION:  

80. a = 2, b = −4, c = −6

SOLUTION:  

81. a = 3, b = 1, c = 2

SOLUTION:  

This value is not a real number.

eSolutions Manual - Powered by Cognero Page 1

9-4 Solving Quadratic Equations by Completing the Square

Find the value of c that makes each trinomial a perfect square.

1. x2 − 18x + c

SOLUTION:  In this trinomial, b = –18.

So, c must be 81 to make the trinomial a perfect square.

2. x2 + 22x + c

SOLUTION:  In this trinomial, b = 22.

So, c must be 121 to make the trinomial a perfect square.

3. x2 + 9x + c

SOLUTION:  In this trinomial, b = 9.

 

So, c must be  to make the trinomial a perfect square.

4. x2 − 7x + c

SOLUTION:  In this trinomial, b = –7.

 

So, c must be  to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

5. x2 + 4x = 6

SOLUTION:  

The solutions are about –5.2 and 1.2.

6. x2 − 8x = −9

SOLUTION:  

The solutions are about 1.4 and 6.6.

7. 4x2 + 9x − 1 = 0

SOLUTION:  

The solutions are about –2.4 and 0.1.

8. −2x2 + 10x + 22 = 4

SOLUTION:  

The solutions are about –1.4 and 6.4.

9. CCSS MODELING Collin is building a deck on the back of his family’s house. He has enough lumber for the deck to be 144 square feet. The length should be 10 feet more than its width. What should the dimensions of the deck be?

SOLUTION:  Let x = the width of the deck and x + 10 = the length of the deck.

The dimensions of the deck can not be negative. So, x = –5 + 13 or 8. The width of the deck is 8 feet and the length of the deck is 8 + 10 or 18 feet.

Find the value of c that makes each trinomial a perfect square.

10. x2 + 26x + c

SOLUTION:  In this trinomial, b = 26.

So, c must be 169 to make the trinomial a perfect square.

11. x2 − 24x + c

SOLUTION:  In this trinomial, b = –24.

So, c must be 144 to make the trinomial a perfect square.

12. x2 − 19x + c

SOLUTION:  In this trinomial, b = –19.

So, c must be  to make the trinomial a perfect square.

13. x2 + 17x + c

SOLUTION:  In this trinomial, b = 17.

So, c must be  to make the trinomial a perfect square.

14. x2 + 5x + c

SOLUTION:  In this trinomial, b = 5.

So, c must be  to make the trinomial a perfect square.

15. x2 − 13x + c

SOLUTION:  In this trinomial, b = –13.

So, c must be  to make the trinomial a perfect square.

16. x2 − 22x + c

SOLUTION:  In this trinomial, b = –22.

So, c must be 121 to make the trinomial a perfect square.

17. x2 − 15x + c

SOLUTION:  In this trinomial, b = –15.

So, c must be  to make the trinomial a perfect square.

18. x2 + 24x + c

SOLUTION:  In this trinomial, b = 24.

So, c must be 144 to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

19. x2 + 6x − 16 = 0

SOLUTION:  

  The solutions are –8 and 2.

20. x2 − 2x − 14 = 0

SOLUTION:  

  The solutions are about –2.9 and 4.9.

21. x2 − 8x − 1 = 8

SOLUTION:  

  The solutions are –1 and 9.

22. x2 + 3x + 21 = 22

SOLUTION:  

  The solutions are about –3.3 and 0.3.

23. x2 − 11x + 3 = 5

SOLUTION:  

  The solutions are about –0.2 and 11.2.

24. 5x2 − 10x = 23

SOLUTION:  

  The solutions are about –1.4 and 3.4.

25. 2x2 − 2x + 7 = 5

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

26. 3x2 + 12x + 81 = 15

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

27. 4x2 + 6x = 12

SOLUTION:  

  The solutions are about –2.6 and 1.1.

28. 4x2 + 5 = 10x

SOLUTION:  

  The solutions are about 0.7 and 1.8.

29. −2x2 + 10x = −14

SOLUTION:  

  The solutions are about –1.1 and 6.1.

30. −3x2 − 12 = 14x

SOLUTION:  

  The solutions are about –3.5 and –1.1.

31. STOCK The price p in dollars for a particular stock can be modeled by the quadratic equation p = 3.5t − 0.05t2, wh

the number of days after the stock is purchased. When is the stock worth $60?

SOLUTION:  Let p = 60.

  The stock is worth $60 on the 30th and 40th days after purchase.

GEOMETRY  Find the value of x for each figure. Round to the nearest tenth if necessary.

32. area = 45 in2

SOLUTION:  

  The height cannot be negative. So, x ≈ 6.3.

33. area = 110 ft2

SOLUTION:  

  The dimensions of the rectangle cannot be negative. So, x ≈ 5.3.

34. NUMBER THEORY The product of two consecutive even integers is 224. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers are 14 and 16 or –16 and –14.

35. CCSS PRECISION The product of two consecutive negative odd integers is 483. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers must be negative. So, they are –23 and –21.

36. GEOMETRY Find the area of the triangle.

SOLUTION:  

  The dimensions of the triangle cannot be negative. So, x = 18. The base of the triangle is 18 meters and the height is 18 + 6 or 24 meters.

The area of the triangle is 216 square meters.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

37. 0.2x2 − 0.2x − 0.4 = 0

SOLUTION:  

  The solutions are –1 and 2.

38. 0.5x2 = 2x − 0.3

SOLUTION:  

  The solutions are about 0.2 and 3.8.

39. 2x2 −

SOLUTION:  

  The solutions are about 0.2 and 0.9.

40. 

SOLUTION:  

  The solutions are –0.5 and 2.5.

41. 

SOLUTION:  

  The solutions are about –8.2 and 0.2.

42. 

SOLUTION:  

  The solutions are about –5.1 and 0.1.

43. ASTRONOMY The height of an object t seconds after it is dropped is given by the equation ,

where h0 is the initial height and g is the acceleration due to gravity. The acceleration due to gravity near the surface

of Mars is 3.73 m/s2, while on Earth it is 9.8 m/s

2. Suppose an object is dropped from an initial height of 120 meters

above the surface of each planet. a. On which planet would the object reach the ground first? b. How long would it take the object to reach the ground on each planet? Round each answer to the nearest tenth. c. Do the times that it takes the object to reach the ground seem reasonable? Explain your reasoning.

SOLUTION:  a. The object on Earth will reach the ground first because it is falling at a faster rate.   b. Mars:

So, t ≈ 8.0 seconds. Earth:

So, t ≈ 4.9 seconds.   c. Sample answer: Yes; the acceleration due to gravity is much greater on Earth than on Mars, so the time to reach the ground should be much less.

44. Find all values of c that make x2 + cx + 100 a perfect square trinomial.

SOLUTION:  

So, c can be –20 or 20 to make the trinomial a perfect square.

45. Find all values of c that make x2 + cx + 225 a perfect square trinomial.

SOLUTION:  

So, c can be –30 or 30 to make the trinomial a perfect square.

46. PAINTING Before she begins painting a picture, Donna stretches her canvas over a wood frame. The frame has a length of 60 inches and a width of 4 inches. She has enough canvas to cover 480 square inches. Donna decides to increase the dimensions of the frame. If the increase in the length is 10 times the increase in the width, what will the dimensions of the frame be?

SOLUTION:  Let x = the increase in the width and let 10x = the increase in the length. So, x + 4 = the new width and 10x + 60 = the new length.  

  The dimensions cannot be negative, so x = 2. The width is 2 + 4 or 6 inches and the length is 10(2) + 60 or 80 inches.

47. MULTIPLE REPRESENTATIONS In this problem, you will investigate a property of quadratic equations. a. TABULAR  Copy the table shown and complete the second column.

b. ALGEBRAIC Set each trinomial equal to zero, and solve the equation by completing the square. Complete the last column of the table with the number of roots of each equation.

c. VERBAL Compare the number of roots of each equation to the result in the b2 − 4ac column. Is there a

relationship between these values? If so, describe it.

d. ANALYTICAL Predict how many solutions 2x2 − 9x + 15 = 0 will have.  Verify your prediction by solving the 

equation.

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3    

3x2 + 6x + 9    

x2 − 2x + 7    

x2 + 10x + 25    

x2 + 3x − 12    

SOLUTION:  a.

b.

The trinomial x2 − 8x + 16 has 1 root.

 

The trinomial 2x

2 − 11x + 3 has 2 roots.

 

The square root of a negative number has no real roots. So, the trinomial 3x2 + 6x + 9 has 0 roots.

 

The square root of a negative number has no real roots. So, the trinomial x2 − 2x + 7.

 

The trinomial x2 + 10x + 25 has 1 root.

 

The trinomial x2 + 3x − 12 has 2 roots.

 

c. If b2 − 4ac is negative, the equation has no real solutions. If b

2 − 4ac is zero, the equation has one solution. If b2

− 4ac is positive, the equation has 2 solutions. d.

The equation 2x2 − 9x + 15 = 0 has 0 real solutions because b

2 − 4ac is negative.

 

The equation cannot be solved because the square root of a negative number has no real roots.

Trinomial b2 − 4ac Number

of Roots

x2 − 8x + 16 (–8)

2 – 4(1)(16) = 0 1

2x2 − 11x + 3 (–11)

2 – 4(2)(3) = 97  

3x2 + 6x + 9 (6)

2 – 4(3)(9) = −72  

x2 − 2x + 7 (–2)

2 – 4(1)(7) = −24  

x2 + 10x + 25 (10)

2 – 4(1)(25) = 0  

x2 + 3x − 12 (3)

2 – 4(1)(–12) = 57  

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3 97 2

3x2 + 6x + 9 −72 0

x2 − 2x + 7 −24 0

x2 + 10x + 25 0 1

x2 + 3x − 12 57 2

48. CCSS PERSEVERANCE Given y = ax2 + bx + c with a ≠ 0, derive the equation for the axis of symmetry by 

completing the square and rewriting the equation in the form y = a(x − h)2 + k .

SOLUTION:  

Let    and   , then the equation becomes y = (x - h)2 + k .

For an equation in the form y = ax2 + bx + c,  the equation for the axis of symmetry is   .

So, for an equation in the form  y = (x - h)2 + k, the equation for the axis of symmetry becomes x = h.

49. REASONING Determine the number of solutions x2 + bx = c has if . Explain.

SOLUTION:  

None; Sample answer: If you add to each side of the equation and each side of the inequality, you get

. Since the left side of the last equation is a perfect square, it cannot

equal the negative number . So, there are no real solutions.

50. WHICH ONE DOESN’T BELONG? Identify the expression that does not belong with the other three. Explain your reasoning.

 

SOLUTION:  The first 3 trinomials are perfect squares.

The trinomial  is not a perfect square. So, it does not belong.

51. OPEN ENDED Write a quadratic equation for which the only solution is 4.

SOLUTION:  Find a quadratic that has two roots of 4.

Verify that the solution is 4.

So, the quadratic equation x2 − 8x + 16 = 0 has a solution of 4.

52. WRITING IN MATH Compare and contrast the following strategies for solving x2 − 5x − 7 = 0: completing the

square, graphing, and factoring.

SOLUTION:  Sample answer: Because the leading coefficient is 1, solving the equation by completing the square is simpler and yields exact answers for the solutions.   

 

To solve the equation by graphing, use a graphing calculator to graph the related function y = x2 - 5x - 7. Select the

zero option from the 2nd [CALC] menu to determine the roots. 

   

  The roots are given as decimal approximations. So, for this equation the solutions would have to be given as estimations.   There are no factors of –7 that have a sum of –5, so solving by factoring is not possible.

53. The length of a rectangle is 3 times its width. The area of the rectangle is 75 square feet. Find the length of the rectangle in feet. A 25 B 15 C 10 D 5

SOLUTION:  Let x = the width of the rectangle and let 3x = the length of the rectangle.

The length of the rectangle is 3(5) or 15 feet. Choice B is the correct answer.

54. PROBABILITY At a festival, winners of a game draw a token for a prize. There is one token for each prize. The prizes include 9 movie passes, 8 stuffed animals, 5 hats, 10 jump ropes, and 4 glow necklaces. What is the probabilitythat the first person to draw a token will win a movie pass?

F  

G

H

J  

SOLUTION:  There are 9 + 8 + 5 + 10 + 4 or 36 possible outcomes.

P(movie pass) =  or  . Choice J is the correct answer.

55. GRIDDED RESPONSE The population of a town can be modeled by P = 22,000 + 125t, where P represents the population and t represents the number of years from 2000. How many years after 2000 will the population be 26,000?

SOLUTION:  

In 32 years the population of the town will be 26,000.

56. Percy delivers pizzas for Pizza King. He is paid $6 an hour plus $2.50 for each pizza he delivers. Percy earned $280 last week. If he worked a total of 30 hours, how many pizzas did he deliver? A 250 pizzas B 184 pizzas C 40 pizzas D 34 pizzas

SOLUTION:  Let p = the number of pizzas Percy delivered.

Percy delivered 40 pizzas. Choice C is the correct answer.

Describe how the graph of each function is related to the graph of f (x) = x2.

57. g(x) = −12 + x2

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –12, the translation is

down. So, the graph is shifted down 12 units from the parent function. 

58. h(x) = (x + 2)2

SOLUTION:  

The graph of f (x) = (x – c)2 represents a translation left or right from the parent graph. Since c = –2, the translation

is to the left by 2 units. 

 

59. g(x) = 2x2 + 5

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 2 and c = 5. Since 2 > 0 and  > 1, the graph of y = 2x

2 + 5

is the graph of y = x2 vertically stretched and shifted up 5 units.

60. 

SOLUTION:  

The function can be written f (x) = a(x – b)2, where a = and b = 6. Since > 0 and  < 1, the graph of 

is the graph of y = x2 vertically compressed and shifted right 6 units.

61. g(x) = 6 + x2

SOLUTION:  

The function can be written f (x) = ax2 + c, where a =  and c = 6. Since  > 0 and   > 1, the graph of y = 6 +

x2 is the graph of y = x

2 vertically stretched and shifted up 6 units.

62. h(x) = − 1 −   x2

SOLUTION:  

The function can be written f (x) = –ax2 + c, where a =  and c = –1. Since  < 0 and   > 1, the graph of y

= –1 – x2 is the graph of y = x

2 vertically stretched, shifted down 1 unit and reflected across the x-axis.

63. RIDES A popular amusement park ride whisks riders to the top of a 250-foot tower and drops them. A function for

the height of a rider is h = −16t2 + 250, where h is the height and t is the time in seconds. The ride stops the descent

of the rider 40 feet above the ground. Write an equation that models the drop of the rider. How long does it take to fall from 250 feet to 40 feet?

SOLUTION:  

The ride stops the descent of the rider at 40 feet above ground, so h = 40. Then, the equation 40 = −16t2 + 250

models the drop of the rider.

Time cannot be negative. So, it takes about 3.6 seconds to complete the ride.

Simplify. Assume that no denominator is equal to zero.

64. 

SOLUTION:  

65. 

SOLUTION:  

66. 

SOLUTION:  

67. 

SOLUTION:  

68. 

SOLUTION:  

69. b3(m

−3)(b

−6)

SOLUTION:  

Solve each open sentence.

70. |y − 2| > 7

SOLUTION:  

The solution set is {y |y > 9 or y < −5}.

Case 1 y – 2 is positive.

and Case 2 y – 2 is negative.

 

71. |z + 5| < 3

SOLUTION:  

The solution set is {z |−8 < z < −2}.

Case 1 z +5 is positive.

and Case 2 z +5 is negative.

 

72. |2b + 7| ≤ −6

SOLUTION:  

cannot be negative. So,  cannot be less than or equal to –6. The solution set is empty.

73. |3 − 2y | ≥ 8

SOLUTION:  

The solution set is {y |y  ≥ 5.5 or y  ≤ −2.5}.

Case 1 3 – 2y is positive.

and Case 2 3 – 2y is negative.

 

74. |9 − 4m| < −1

SOLUTION:  

cannot be negative. So, cannot be less than or equal to –1. The solution set is empty.

75. |5c − 2| ≤ 13

SOLUTION:  

The solution set is {c|−2.2 ≤ c ≤ 3}.

Case 1 5c – 2 is positive.

and Case 2 5c – 2 is negative.

 

Evaluate  for each set of values. Round to the nearest tenth if necessary.

76. a = 2, b = −5, c = 2

SOLUTION:  

77. a = 1, b = 12, c = 11

SOLUTION:  

78. a = −9, b = 10, c = −1

SOLUTION:  

79. a = 1, b = 7, c = −3

SOLUTION:  

80. a = 2, b = −4, c = −6

SOLUTION:  

81. a = 3, b = 1, c = 2

SOLUTION:  

This value is not a real number.

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9-4 Solving Quadratic Equations by Completing the Square

Find the value of c that makes each trinomial a perfect square.

1. x2 − 18x + c

SOLUTION:  In this trinomial, b = –18.

So, c must be 81 to make the trinomial a perfect square.

2. x2 + 22x + c

SOLUTION:  In this trinomial, b = 22.

So, c must be 121 to make the trinomial a perfect square.

3. x2 + 9x + c

SOLUTION:  In this trinomial, b = 9.

 

So, c must be  to make the trinomial a perfect square.

4. x2 − 7x + c

SOLUTION:  In this trinomial, b = –7.

 

So, c must be  to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

5. x2 + 4x = 6

SOLUTION:  

The solutions are about –5.2 and 1.2.

6. x2 − 8x = −9

SOLUTION:  

The solutions are about 1.4 and 6.6.

7. 4x2 + 9x − 1 = 0

SOLUTION:  

The solutions are about –2.4 and 0.1.

8. −2x2 + 10x + 22 = 4

SOLUTION:  

The solutions are about –1.4 and 6.4.

9. CCSS MODELING Collin is building a deck on the back of his family’s house. He has enough lumber for the deck to be 144 square feet. The length should be 10 feet more than its width. What should the dimensions of the deck be?

SOLUTION:  Let x = the width of the deck and x + 10 = the length of the deck.

The dimensions of the deck can not be negative. So, x = –5 + 13 or 8. The width of the deck is 8 feet and the length of the deck is 8 + 10 or 18 feet.

Find the value of c that makes each trinomial a perfect square.

10. x2 + 26x + c

SOLUTION:  In this trinomial, b = 26.

So, c must be 169 to make the trinomial a perfect square.

11. x2 − 24x + c

SOLUTION:  In this trinomial, b = –24.

So, c must be 144 to make the trinomial a perfect square.

12. x2 − 19x + c

SOLUTION:  In this trinomial, b = –19.

So, c must be  to make the trinomial a perfect square.

13. x2 + 17x + c

SOLUTION:  In this trinomial, b = 17.

So, c must be  to make the trinomial a perfect square.

14. x2 + 5x + c

SOLUTION:  In this trinomial, b = 5.

So, c must be  to make the trinomial a perfect square.

15. x2 − 13x + c

SOLUTION:  In this trinomial, b = –13.

So, c must be  to make the trinomial a perfect square.

16. x2 − 22x + c

SOLUTION:  In this trinomial, b = –22.

So, c must be 121 to make the trinomial a perfect square.

17. x2 − 15x + c

SOLUTION:  In this trinomial, b = –15.

So, c must be  to make the trinomial a perfect square.

18. x2 + 24x + c

SOLUTION:  In this trinomial, b = 24.

So, c must be 144 to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

19. x2 + 6x − 16 = 0

SOLUTION:  

  The solutions are –8 and 2.

20. x2 − 2x − 14 = 0

SOLUTION:  

  The solutions are about –2.9 and 4.9.

21. x2 − 8x − 1 = 8

SOLUTION:  

  The solutions are –1 and 9.

22. x2 + 3x + 21 = 22

SOLUTION:  

  The solutions are about –3.3 and 0.3.

23. x2 − 11x + 3 = 5

SOLUTION:  

  The solutions are about –0.2 and 11.2.

24. 5x2 − 10x = 23

SOLUTION:  

  The solutions are about –1.4 and 3.4.

25. 2x2 − 2x + 7 = 5

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

26. 3x2 + 12x + 81 = 15

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

27. 4x2 + 6x = 12

SOLUTION:  

  The solutions are about –2.6 and 1.1.

28. 4x2 + 5 = 10x

SOLUTION:  

  The solutions are about 0.7 and 1.8.

29. −2x2 + 10x = −14

SOLUTION:  

  The solutions are about –1.1 and 6.1.

30. −3x2 − 12 = 14x

SOLUTION:  

  The solutions are about –3.5 and –1.1.

31. STOCK The price p in dollars for a particular stock can be modeled by the quadratic equation p = 3.5t − 0.05t2, wh

the number of days after the stock is purchased. When is the stock worth $60?

SOLUTION:  Let p = 60.

  The stock is worth $60 on the 30th and 40th days after purchase.

GEOMETRY  Find the value of x for each figure. Round to the nearest tenth if necessary.

32. area = 45 in2

SOLUTION:  

  The height cannot be negative. So, x ≈ 6.3.

33. area = 110 ft2

SOLUTION:  

  The dimensions of the rectangle cannot be negative. So, x ≈ 5.3.

34. NUMBER THEORY The product of two consecutive even integers is 224. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers are 14 and 16 or –16 and –14.

35. CCSS PRECISION The product of two consecutive negative odd integers is 483. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers must be negative. So, they are –23 and –21.

36. GEOMETRY Find the area of the triangle.

SOLUTION:  

  The dimensions of the triangle cannot be negative. So, x = 18. The base of the triangle is 18 meters and the height is 18 + 6 or 24 meters.

The area of the triangle is 216 square meters.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

37. 0.2x2 − 0.2x − 0.4 = 0

SOLUTION:  

  The solutions are –1 and 2.

38. 0.5x2 = 2x − 0.3

SOLUTION:  

  The solutions are about 0.2 and 3.8.

39. 2x2 −

SOLUTION:  

  The solutions are about 0.2 and 0.9.

40. 

SOLUTION:  

  The solutions are –0.5 and 2.5.

41. 

SOLUTION:  

  The solutions are about –8.2 and 0.2.

42. 

SOLUTION:  

  The solutions are about –5.1 and 0.1.

43. ASTRONOMY The height of an object t seconds after it is dropped is given by the equation ,

where h0 is the initial height and g is the acceleration due to gravity. The acceleration due to gravity near the surface

of Mars is 3.73 m/s2, while on Earth it is 9.8 m/s

2. Suppose an object is dropped from an initial height of 120 meters

above the surface of each planet. a. On which planet would the object reach the ground first? b. How long would it take the object to reach the ground on each planet? Round each answer to the nearest tenth. c. Do the times that it takes the object to reach the ground seem reasonable? Explain your reasoning.

SOLUTION:  a. The object on Earth will reach the ground first because it is falling at a faster rate.   b. Mars:

So, t ≈ 8.0 seconds. Earth:

So, t ≈ 4.9 seconds.   c. Sample answer: Yes; the acceleration due to gravity is much greater on Earth than on Mars, so the time to reach the ground should be much less.

44. Find all values of c that make x2 + cx + 100 a perfect square trinomial.

SOLUTION:  

So, c can be –20 or 20 to make the trinomial a perfect square.

45. Find all values of c that make x2 + cx + 225 a perfect square trinomial.

SOLUTION:  

So, c can be –30 or 30 to make the trinomial a perfect square.

46. PAINTING Before she begins painting a picture, Donna stretches her canvas over a wood frame. The frame has a length of 60 inches and a width of 4 inches. She has enough canvas to cover 480 square inches. Donna decides to increase the dimensions of the frame. If the increase in the length is 10 times the increase in the width, what will the dimensions of the frame be?

SOLUTION:  Let x = the increase in the width and let 10x = the increase in the length. So, x + 4 = the new width and 10x + 60 = the new length.  

  The dimensions cannot be negative, so x = 2. The width is 2 + 4 or 6 inches and the length is 10(2) + 60 or 80 inches.

47. MULTIPLE REPRESENTATIONS In this problem, you will investigate a property of quadratic equations. a. TABULAR  Copy the table shown and complete the second column.

b. ALGEBRAIC Set each trinomial equal to zero, and solve the equation by completing the square. Complete the last column of the table with the number of roots of each equation.

c. VERBAL Compare the number of roots of each equation to the result in the b2 − 4ac column. Is there a

relationship between these values? If so, describe it.

d. ANALYTICAL Predict how many solutions 2x2 − 9x + 15 = 0 will have.  Verify your prediction by solving the 

equation.

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3    

3x2 + 6x + 9    

x2 − 2x + 7    

x2 + 10x + 25    

x2 + 3x − 12    

SOLUTION:  a.

b.

The trinomial x2 − 8x + 16 has 1 root.

 

The trinomial 2x

2 − 11x + 3 has 2 roots.

 

The square root of a negative number has no real roots. So, the trinomial 3x2 + 6x + 9 has 0 roots.

 

The square root of a negative number has no real roots. So, the trinomial x2 − 2x + 7.

 

The trinomial x2 + 10x + 25 has 1 root.

 

The trinomial x2 + 3x − 12 has 2 roots.

 

c. If b2 − 4ac is negative, the equation has no real solutions. If b

2 − 4ac is zero, the equation has one solution. If b2

− 4ac is positive, the equation has 2 solutions. d.

The equation 2x2 − 9x + 15 = 0 has 0 real solutions because b

2 − 4ac is negative.

 

The equation cannot be solved because the square root of a negative number has no real roots.

Trinomial b2 − 4ac Number

of Roots

x2 − 8x + 16 (–8)

2 – 4(1)(16) = 0 1

2x2 − 11x + 3 (–11)

2 – 4(2)(3) = 97  

3x2 + 6x + 9 (6)

2 – 4(3)(9) = −72  

x2 − 2x + 7 (–2)

2 – 4(1)(7) = −24  

x2 + 10x + 25 (10)

2 – 4(1)(25) = 0  

x2 + 3x − 12 (3)

2 – 4(1)(–12) = 57  

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3 97 2

3x2 + 6x + 9 −72 0

x2 − 2x + 7 −24 0

x2 + 10x + 25 0 1

x2 + 3x − 12 57 2

48. CCSS PERSEVERANCE Given y = ax2 + bx + c with a ≠ 0, derive the equation for the axis of symmetry by 

completing the square and rewriting the equation in the form y = a(x − h)2 + k .

SOLUTION:  

Let    and   , then the equation becomes y = (x - h)2 + k .

For an equation in the form y = ax2 + bx + c,  the equation for the axis of symmetry is   .

So, for an equation in the form  y = (x - h)2 + k, the equation for the axis of symmetry becomes x = h.

49. REASONING Determine the number of solutions x2 + bx = c has if . Explain.

SOLUTION:  

None; Sample answer: If you add to each side of the equation and each side of the inequality, you get

. Since the left side of the last equation is a perfect square, it cannot

equal the negative number . So, there are no real solutions.

50. WHICH ONE DOESN’T BELONG? Identify the expression that does not belong with the other three. Explain your reasoning.

 

SOLUTION:  The first 3 trinomials are perfect squares.

The trinomial  is not a perfect square. So, it does not belong.

51. OPEN ENDED Write a quadratic equation for which the only solution is 4.

SOLUTION:  Find a quadratic that has two roots of 4.

Verify that the solution is 4.

So, the quadratic equation x2 − 8x + 16 = 0 has a solution of 4.

52. WRITING IN MATH Compare and contrast the following strategies for solving x2 − 5x − 7 = 0: completing the

square, graphing, and factoring.

SOLUTION:  Sample answer: Because the leading coefficient is 1, solving the equation by completing the square is simpler and yields exact answers for the solutions.   

 

To solve the equation by graphing, use a graphing calculator to graph the related function y = x2 - 5x - 7. Select the

zero option from the 2nd [CALC] menu to determine the roots. 

   

  The roots are given as decimal approximations. So, for this equation the solutions would have to be given as estimations.   There are no factors of –7 that have a sum of –5, so solving by factoring is not possible.

53. The length of a rectangle is 3 times its width. The area of the rectangle is 75 square feet. Find the length of the rectangle in feet. A 25 B 15 C 10 D 5

SOLUTION:  Let x = the width of the rectangle and let 3x = the length of the rectangle.

The length of the rectangle is 3(5) or 15 feet. Choice B is the correct answer.

54. PROBABILITY At a festival, winners of a game draw a token for a prize. There is one token for each prize. The prizes include 9 movie passes, 8 stuffed animals, 5 hats, 10 jump ropes, and 4 glow necklaces. What is the probabilitythat the first person to draw a token will win a movie pass?

F  

G

H

J  

SOLUTION:  There are 9 + 8 + 5 + 10 + 4 or 36 possible outcomes.

P(movie pass) =  or  . Choice J is the correct answer.

55. GRIDDED RESPONSE The population of a town can be modeled by P = 22,000 + 125t, where P represents the population and t represents the number of years from 2000. How many years after 2000 will the population be 26,000?

SOLUTION:  

In 32 years the population of the town will be 26,000.

56. Percy delivers pizzas for Pizza King. He is paid $6 an hour plus $2.50 for each pizza he delivers. Percy earned $280 last week. If he worked a total of 30 hours, how many pizzas did he deliver? A 250 pizzas B 184 pizzas C 40 pizzas D 34 pizzas

SOLUTION:  Let p = the number of pizzas Percy delivered.

Percy delivered 40 pizzas. Choice C is the correct answer.

Describe how the graph of each function is related to the graph of f (x) = x2.

57. g(x) = −12 + x2

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –12, the translation is

down. So, the graph is shifted down 12 units from the parent function. 

58. h(x) = (x + 2)2

SOLUTION:  

The graph of f (x) = (x – c)2 represents a translation left or right from the parent graph. Since c = –2, the translation

is to the left by 2 units. 

 

59. g(x) = 2x2 + 5

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 2 and c = 5. Since 2 > 0 and  > 1, the graph of y = 2x

2 + 5

is the graph of y = x2 vertically stretched and shifted up 5 units.

60. 

SOLUTION:  

The function can be written f (x) = a(x – b)2, where a = and b = 6. Since > 0 and  < 1, the graph of 

is the graph of y = x2 vertically compressed and shifted right 6 units.

61. g(x) = 6 + x2

SOLUTION:  

The function can be written f (x) = ax2 + c, where a =  and c = 6. Since  > 0 and   > 1, the graph of y = 6 +

x2 is the graph of y = x

2 vertically stretched and shifted up 6 units.

62. h(x) = − 1 −   x2

SOLUTION:  

The function can be written f (x) = –ax2 + c, where a =  and c = –1. Since  < 0 and   > 1, the graph of y

= –1 – x2 is the graph of y = x

2 vertically stretched, shifted down 1 unit and reflected across the x-axis.

63. RIDES A popular amusement park ride whisks riders to the top of a 250-foot tower and drops them. A function for

the height of a rider is h = −16t2 + 250, where h is the height and t is the time in seconds. The ride stops the descent

of the rider 40 feet above the ground. Write an equation that models the drop of the rider. How long does it take to fall from 250 feet to 40 feet?

SOLUTION:  

The ride stops the descent of the rider at 40 feet above ground, so h = 40. Then, the equation 40 = −16t2 + 250

models the drop of the rider.

Time cannot be negative. So, it takes about 3.6 seconds to complete the ride.

Simplify. Assume that no denominator is equal to zero.

64. 

SOLUTION:  

65. 

SOLUTION:  

66. 

SOLUTION:  

67. 

SOLUTION:  

68. 

SOLUTION:  

69. b3(m

−3)(b

−6)

SOLUTION:  

Solve each open sentence.

70. |y − 2| > 7

SOLUTION:  

The solution set is {y |y > 9 or y < −5}.

Case 1 y – 2 is positive.

and Case 2 y – 2 is negative.

 

71. |z + 5| < 3

SOLUTION:  

The solution set is {z |−8 < z < −2}.

Case 1 z +5 is positive.

and Case 2 z +5 is negative.

 

72. |2b + 7| ≤ −6

SOLUTION:  

cannot be negative. So,  cannot be less than or equal to –6. The solution set is empty.

73. |3 − 2y | ≥ 8

SOLUTION:  

The solution set is {y |y  ≥ 5.5 or y  ≤ −2.5}.

Case 1 3 – 2y is positive.

and Case 2 3 – 2y is negative.

 

74. |9 − 4m| < −1

SOLUTION:  

cannot be negative. So, cannot be less than or equal to –1. The solution set is empty.

75. |5c − 2| ≤ 13

SOLUTION:  

The solution set is {c|−2.2 ≤ c ≤ 3}.

Case 1 5c – 2 is positive.

and Case 2 5c – 2 is negative.

 

Evaluate  for each set of values. Round to the nearest tenth if necessary.

76. a = 2, b = −5, c = 2

SOLUTION:  

77. a = 1, b = 12, c = 11

SOLUTION:  

78. a = −9, b = 10, c = −1

SOLUTION:  

79. a = 1, b = 7, c = −3

SOLUTION:  

80. a = 2, b = −4, c = −6

SOLUTION:  

81. a = 3, b = 1, c = 2

SOLUTION:  

This value is not a real number.

eSolutions Manual - Powered by Cognero Page 3

9-4 Solving Quadratic Equations by Completing the Square

Find the value of c that makes each trinomial a perfect square.

1. x2 − 18x + c

SOLUTION:  In this trinomial, b = –18.

So, c must be 81 to make the trinomial a perfect square.

2. x2 + 22x + c

SOLUTION:  In this trinomial, b = 22.

So, c must be 121 to make the trinomial a perfect square.

3. x2 + 9x + c

SOLUTION:  In this trinomial, b = 9.

 

So, c must be  to make the trinomial a perfect square.

4. x2 − 7x + c

SOLUTION:  In this trinomial, b = –7.

 

So, c must be  to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

5. x2 + 4x = 6

SOLUTION:  

The solutions are about –5.2 and 1.2.

6. x2 − 8x = −9

SOLUTION:  

The solutions are about 1.4 and 6.6.

7. 4x2 + 9x − 1 = 0

SOLUTION:  

The solutions are about –2.4 and 0.1.

8. −2x2 + 10x + 22 = 4

SOLUTION:  

The solutions are about –1.4 and 6.4.

9. CCSS MODELING Collin is building a deck on the back of his family’s house. He has enough lumber for the deck to be 144 square feet. The length should be 10 feet more than its width. What should the dimensions of the deck be?

SOLUTION:  Let x = the width of the deck and x + 10 = the length of the deck.

The dimensions of the deck can not be negative. So, x = –5 + 13 or 8. The width of the deck is 8 feet and the length of the deck is 8 + 10 or 18 feet.

Find the value of c that makes each trinomial a perfect square.

10. x2 + 26x + c

SOLUTION:  In this trinomial, b = 26.

So, c must be 169 to make the trinomial a perfect square.

11. x2 − 24x + c

SOLUTION:  In this trinomial, b = –24.

So, c must be 144 to make the trinomial a perfect square.

12. x2 − 19x + c

SOLUTION:  In this trinomial, b = –19.

So, c must be  to make the trinomial a perfect square.

13. x2 + 17x + c

SOLUTION:  In this trinomial, b = 17.

So, c must be  to make the trinomial a perfect square.

14. x2 + 5x + c

SOLUTION:  In this trinomial, b = 5.

So, c must be  to make the trinomial a perfect square.

15. x2 − 13x + c

SOLUTION:  In this trinomial, b = –13.

So, c must be  to make the trinomial a perfect square.

16. x2 − 22x + c

SOLUTION:  In this trinomial, b = –22.

So, c must be 121 to make the trinomial a perfect square.

17. x2 − 15x + c

SOLUTION:  In this trinomial, b = –15.

So, c must be  to make the trinomial a perfect square.

18. x2 + 24x + c

SOLUTION:  In this trinomial, b = 24.

So, c must be 144 to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

19. x2 + 6x − 16 = 0

SOLUTION:  

  The solutions are –8 and 2.

20. x2 − 2x − 14 = 0

SOLUTION:  

  The solutions are about –2.9 and 4.9.

21. x2 − 8x − 1 = 8

SOLUTION:  

  The solutions are –1 and 9.

22. x2 + 3x + 21 = 22

SOLUTION:  

  The solutions are about –3.3 and 0.3.

23. x2 − 11x + 3 = 5

SOLUTION:  

  The solutions are about –0.2 and 11.2.

24. 5x2 − 10x = 23

SOLUTION:  

  The solutions are about –1.4 and 3.4.

25. 2x2 − 2x + 7 = 5

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

26. 3x2 + 12x + 81 = 15

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

27. 4x2 + 6x = 12

SOLUTION:  

  The solutions are about –2.6 and 1.1.

28. 4x2 + 5 = 10x

SOLUTION:  

  The solutions are about 0.7 and 1.8.

29. −2x2 + 10x = −14

SOLUTION:  

  The solutions are about –1.1 and 6.1.

30. −3x2 − 12 = 14x

SOLUTION:  

  The solutions are about –3.5 and –1.1.

31. STOCK The price p in dollars for a particular stock can be modeled by the quadratic equation p = 3.5t − 0.05t2, wh

the number of days after the stock is purchased. When is the stock worth $60?

SOLUTION:  Let p = 60.

  The stock is worth $60 on the 30th and 40th days after purchase.

GEOMETRY  Find the value of x for each figure. Round to the nearest tenth if necessary.

32. area = 45 in2

SOLUTION:  

  The height cannot be negative. So, x ≈ 6.3.

33. area = 110 ft2

SOLUTION:  

  The dimensions of the rectangle cannot be negative. So, x ≈ 5.3.

34. NUMBER THEORY The product of two consecutive even integers is 224. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers are 14 and 16 or –16 and –14.

35. CCSS PRECISION The product of two consecutive negative odd integers is 483. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers must be negative. So, they are –23 and –21.

36. GEOMETRY Find the area of the triangle.

SOLUTION:  

  The dimensions of the triangle cannot be negative. So, x = 18. The base of the triangle is 18 meters and the height is 18 + 6 or 24 meters.

The area of the triangle is 216 square meters.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

37. 0.2x2 − 0.2x − 0.4 = 0

SOLUTION:  

  The solutions are –1 and 2.

38. 0.5x2 = 2x − 0.3

SOLUTION:  

  The solutions are about 0.2 and 3.8.

39. 2x2 −

SOLUTION:  

  The solutions are about 0.2 and 0.9.

40. 

SOLUTION:  

  The solutions are –0.5 and 2.5.

41. 

SOLUTION:  

  The solutions are about –8.2 and 0.2.

42. 

SOLUTION:  

  The solutions are about –5.1 and 0.1.

43. ASTRONOMY The height of an object t seconds after it is dropped is given by the equation ,

where h0 is the initial height and g is the acceleration due to gravity. The acceleration due to gravity near the surface

of Mars is 3.73 m/s2, while on Earth it is 9.8 m/s

2. Suppose an object is dropped from an initial height of 120 meters

above the surface of each planet. a. On which planet would the object reach the ground first? b. How long would it take the object to reach the ground on each planet? Round each answer to the nearest tenth. c. Do the times that it takes the object to reach the ground seem reasonable? Explain your reasoning.

SOLUTION:  a. The object on Earth will reach the ground first because it is falling at a faster rate.   b. Mars:

So, t ≈ 8.0 seconds. Earth:

So, t ≈ 4.9 seconds.   c. Sample answer: Yes; the acceleration due to gravity is much greater on Earth than on Mars, so the time to reach the ground should be much less.

44. Find all values of c that make x2 + cx + 100 a perfect square trinomial.

SOLUTION:  

So, c can be –20 or 20 to make the trinomial a perfect square.

45. Find all values of c that make x2 + cx + 225 a perfect square trinomial.

SOLUTION:  

So, c can be –30 or 30 to make the trinomial a perfect square.

46. PAINTING Before she begins painting a picture, Donna stretches her canvas over a wood frame. The frame has a length of 60 inches and a width of 4 inches. She has enough canvas to cover 480 square inches. Donna decides to increase the dimensions of the frame. If the increase in the length is 10 times the increase in the width, what will the dimensions of the frame be?

SOLUTION:  Let x = the increase in the width and let 10x = the increase in the length. So, x + 4 = the new width and 10x + 60 = the new length.  

  The dimensions cannot be negative, so x = 2. The width is 2 + 4 or 6 inches and the length is 10(2) + 60 or 80 inches.

47. MULTIPLE REPRESENTATIONS In this problem, you will investigate a property of quadratic equations. a. TABULAR  Copy the table shown and complete the second column.

b. ALGEBRAIC Set each trinomial equal to zero, and solve the equation by completing the square. Complete the last column of the table with the number of roots of each equation.

c. VERBAL Compare the number of roots of each equation to the result in the b2 − 4ac column. Is there a

relationship between these values? If so, describe it.

d. ANALYTICAL Predict how many solutions 2x2 − 9x + 15 = 0 will have.  Verify your prediction by solving the 

equation.

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3    

3x2 + 6x + 9    

x2 − 2x + 7    

x2 + 10x + 25    

x2 + 3x − 12    

SOLUTION:  a.

b.

The trinomial x2 − 8x + 16 has 1 root.

 

The trinomial 2x

2 − 11x + 3 has 2 roots.

 

The square root of a negative number has no real roots. So, the trinomial 3x2 + 6x + 9 has 0 roots.

 

The square root of a negative number has no real roots. So, the trinomial x2 − 2x + 7.

 

The trinomial x2 + 10x + 25 has 1 root.

 

The trinomial x2 + 3x − 12 has 2 roots.

 

c. If b2 − 4ac is negative, the equation has no real solutions. If b

2 − 4ac is zero, the equation has one solution. If b2

− 4ac is positive, the equation has 2 solutions. d.

The equation 2x2 − 9x + 15 = 0 has 0 real solutions because b

2 − 4ac is negative.

 

The equation cannot be solved because the square root of a negative number has no real roots.

Trinomial b2 − 4ac Number

of Roots

x2 − 8x + 16 (–8)

2 – 4(1)(16) = 0 1

2x2 − 11x + 3 (–11)

2 – 4(2)(3) = 97  

3x2 + 6x + 9 (6)

2 – 4(3)(9) = −72  

x2 − 2x + 7 (–2)

2 – 4(1)(7) = −24  

x2 + 10x + 25 (10)

2 – 4(1)(25) = 0  

x2 + 3x − 12 (3)

2 – 4(1)(–12) = 57  

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3 97 2

3x2 + 6x + 9 −72 0

x2 − 2x + 7 −24 0

x2 + 10x + 25 0 1

x2 + 3x − 12 57 2

48. CCSS PERSEVERANCE Given y = ax2 + bx + c with a ≠ 0, derive the equation for the axis of symmetry by 

completing the square and rewriting the equation in the form y = a(x − h)2 + k .

SOLUTION:  

Let    and   , then the equation becomes y = (x - h)2 + k .

For an equation in the form y = ax2 + bx + c,  the equation for the axis of symmetry is   .

So, for an equation in the form  y = (x - h)2 + k, the equation for the axis of symmetry becomes x = h.

49. REASONING Determine the number of solutions x2 + bx = c has if . Explain.

SOLUTION:  

None; Sample answer: If you add to each side of the equation and each side of the inequality, you get

. Since the left side of the last equation is a perfect square, it cannot

equal the negative number . So, there are no real solutions.

50. WHICH ONE DOESN’T BELONG? Identify the expression that does not belong with the other three. Explain your reasoning.

 

SOLUTION:  The first 3 trinomials are perfect squares.

The trinomial  is not a perfect square. So, it does not belong.

51. OPEN ENDED Write a quadratic equation for which the only solution is 4.

SOLUTION:  Find a quadratic that has two roots of 4.

Verify that the solution is 4.

So, the quadratic equation x2 − 8x + 16 = 0 has a solution of 4.

52. WRITING IN MATH Compare and contrast the following strategies for solving x2 − 5x − 7 = 0: completing the

square, graphing, and factoring.

SOLUTION:  Sample answer: Because the leading coefficient is 1, solving the equation by completing the square is simpler and yields exact answers for the solutions.   

 

To solve the equation by graphing, use a graphing calculator to graph the related function y = x2 - 5x - 7. Select the

zero option from the 2nd [CALC] menu to determine the roots. 

   

  The roots are given as decimal approximations. So, for this equation the solutions would have to be given as estimations.   There are no factors of –7 that have a sum of –5, so solving by factoring is not possible.

53. The length of a rectangle is 3 times its width. The area of the rectangle is 75 square feet. Find the length of the rectangle in feet. A 25 B 15 C 10 D 5

SOLUTION:  Let x = the width of the rectangle and let 3x = the length of the rectangle.

The length of the rectangle is 3(5) or 15 feet. Choice B is the correct answer.

54. PROBABILITY At a festival, winners of a game draw a token for a prize. There is one token for each prize. The prizes include 9 movie passes, 8 stuffed animals, 5 hats, 10 jump ropes, and 4 glow necklaces. What is the probabilitythat the first person to draw a token will win a movie pass?

F  

G

H

J  

SOLUTION:  There are 9 + 8 + 5 + 10 + 4 or 36 possible outcomes.

P(movie pass) =  or  . Choice J is the correct answer.

55. GRIDDED RESPONSE The population of a town can be modeled by P = 22,000 + 125t, where P represents the population and t represents the number of years from 2000. How many years after 2000 will the population be 26,000?

SOLUTION:  

In 32 years the population of the town will be 26,000.

56. Percy delivers pizzas for Pizza King. He is paid $6 an hour plus $2.50 for each pizza he delivers. Percy earned $280 last week. If he worked a total of 30 hours, how many pizzas did he deliver? A 250 pizzas B 184 pizzas C 40 pizzas D 34 pizzas

SOLUTION:  Let p = the number of pizzas Percy delivered.

Percy delivered 40 pizzas. Choice C is the correct answer.

Describe how the graph of each function is related to the graph of f (x) = x2.

57. g(x) = −12 + x2

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –12, the translation is

down. So, the graph is shifted down 12 units from the parent function. 

58. h(x) = (x + 2)2

SOLUTION:  

The graph of f (x) = (x – c)2 represents a translation left or right from the parent graph. Since c = –2, the translation

is to the left by 2 units. 

 

59. g(x) = 2x2 + 5

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 2 and c = 5. Since 2 > 0 and  > 1, the graph of y = 2x

2 + 5

is the graph of y = x2 vertically stretched and shifted up 5 units.

60. 

SOLUTION:  

The function can be written f (x) = a(x – b)2, where a = and b = 6. Since > 0 and  < 1, the graph of 

is the graph of y = x2 vertically compressed and shifted right 6 units.

61. g(x) = 6 + x2

SOLUTION:  

The function can be written f (x) = ax2 + c, where a =  and c = 6. Since  > 0 and   > 1, the graph of y = 6 +

x2 is the graph of y = x

2 vertically stretched and shifted up 6 units.

62. h(x) = − 1 −   x2

SOLUTION:  

The function can be written f (x) = –ax2 + c, where a =  and c = –1. Since  < 0 and   > 1, the graph of y

= –1 – x2 is the graph of y = x

2 vertically stretched, shifted down 1 unit and reflected across the x-axis.

63. RIDES A popular amusement park ride whisks riders to the top of a 250-foot tower and drops them. A function for

the height of a rider is h = −16t2 + 250, where h is the height and t is the time in seconds. The ride stops the descent

of the rider 40 feet above the ground. Write an equation that models the drop of the rider. How long does it take to fall from 250 feet to 40 feet?

SOLUTION:  

The ride stops the descent of the rider at 40 feet above ground, so h = 40. Then, the equation 40 = −16t2 + 250

models the drop of the rider.

Time cannot be negative. So, it takes about 3.6 seconds to complete the ride.

Simplify. Assume that no denominator is equal to zero.

64. 

SOLUTION:  

65. 

SOLUTION:  

66. 

SOLUTION:  

67. 

SOLUTION:  

68. 

SOLUTION:  

69. b3(m

−3)(b

−6)

SOLUTION:  

Solve each open sentence.

70. |y − 2| > 7

SOLUTION:  

The solution set is {y |y > 9 or y < −5}.

Case 1 y – 2 is positive.

and Case 2 y – 2 is negative.

 

71. |z + 5| < 3

SOLUTION:  

The solution set is {z |−8 < z < −2}.

Case 1 z +5 is positive.

and Case 2 z +5 is negative.

 

72. |2b + 7| ≤ −6

SOLUTION:  

cannot be negative. So,  cannot be less than or equal to –6. The solution set is empty.

73. |3 − 2y | ≥ 8

SOLUTION:  

The solution set is {y |y  ≥ 5.5 or y  ≤ −2.5}.

Case 1 3 – 2y is positive.

and Case 2 3 – 2y is negative.

 

74. |9 − 4m| < −1

SOLUTION:  

cannot be negative. So, cannot be less than or equal to –1. The solution set is empty.

75. |5c − 2| ≤ 13

SOLUTION:  

The solution set is {c|−2.2 ≤ c ≤ 3}.

Case 1 5c – 2 is positive.

and Case 2 5c – 2 is negative.

 

Evaluate  for each set of values. Round to the nearest tenth if necessary.

76. a = 2, b = −5, c = 2

SOLUTION:  

77. a = 1, b = 12, c = 11

SOLUTION:  

78. a = −9, b = 10, c = −1

SOLUTION:  

79. a = 1, b = 7, c = −3

SOLUTION:  

80. a = 2, b = −4, c = −6

SOLUTION:  

81. a = 3, b = 1, c = 2

SOLUTION:  

This value is not a real number.

eSolutions Manual - Powered by Cognero Page 4

9-4 Solving Quadratic Equations by Completing the Square

Find the value of c that makes each trinomial a perfect square.

1. x2 − 18x + c

SOLUTION:  In this trinomial, b = –18.

So, c must be 81 to make the trinomial a perfect square.

2. x2 + 22x + c

SOLUTION:  In this trinomial, b = 22.

So, c must be 121 to make the trinomial a perfect square.

3. x2 + 9x + c

SOLUTION:  In this trinomial, b = 9.

 

So, c must be  to make the trinomial a perfect square.

4. x2 − 7x + c

SOLUTION:  In this trinomial, b = –7.

 

So, c must be  to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

5. x2 + 4x = 6

SOLUTION:  

The solutions are about –5.2 and 1.2.

6. x2 − 8x = −9

SOLUTION:  

The solutions are about 1.4 and 6.6.

7. 4x2 + 9x − 1 = 0

SOLUTION:  

The solutions are about –2.4 and 0.1.

8. −2x2 + 10x + 22 = 4

SOLUTION:  

The solutions are about –1.4 and 6.4.

9. CCSS MODELING Collin is building a deck on the back of his family’s house. He has enough lumber for the deck to be 144 square feet. The length should be 10 feet more than its width. What should the dimensions of the deck be?

SOLUTION:  Let x = the width of the deck and x + 10 = the length of the deck.

The dimensions of the deck can not be negative. So, x = –5 + 13 or 8. The width of the deck is 8 feet and the length of the deck is 8 + 10 or 18 feet.

Find the value of c that makes each trinomial a perfect square.

10. x2 + 26x + c

SOLUTION:  In this trinomial, b = 26.

So, c must be 169 to make the trinomial a perfect square.

11. x2 − 24x + c

SOLUTION:  In this trinomial, b = –24.

So, c must be 144 to make the trinomial a perfect square.

12. x2 − 19x + c

SOLUTION:  In this trinomial, b = –19.

So, c must be  to make the trinomial a perfect square.

13. x2 + 17x + c

SOLUTION:  In this trinomial, b = 17.

So, c must be  to make the trinomial a perfect square.

14. x2 + 5x + c

SOLUTION:  In this trinomial, b = 5.

So, c must be  to make the trinomial a perfect square.

15. x2 − 13x + c

SOLUTION:  In this trinomial, b = –13.

So, c must be  to make the trinomial a perfect square.

16. x2 − 22x + c

SOLUTION:  In this trinomial, b = –22.

So, c must be 121 to make the trinomial a perfect square.

17. x2 − 15x + c

SOLUTION:  In this trinomial, b = –15.

So, c must be  to make the trinomial a perfect square.

18. x2 + 24x + c

SOLUTION:  In this trinomial, b = 24.

So, c must be 144 to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

19. x2 + 6x − 16 = 0

SOLUTION:  

  The solutions are –8 and 2.

20. x2 − 2x − 14 = 0

SOLUTION:  

  The solutions are about –2.9 and 4.9.

21. x2 − 8x − 1 = 8

SOLUTION:  

  The solutions are –1 and 9.

22. x2 + 3x + 21 = 22

SOLUTION:  

  The solutions are about –3.3 and 0.3.

23. x2 − 11x + 3 = 5

SOLUTION:  

  The solutions are about –0.2 and 11.2.

24. 5x2 − 10x = 23

SOLUTION:  

  The solutions are about –1.4 and 3.4.

25. 2x2 − 2x + 7 = 5

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

26. 3x2 + 12x + 81 = 15

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

27. 4x2 + 6x = 12

SOLUTION:  

  The solutions are about –2.6 and 1.1.

28. 4x2 + 5 = 10x

SOLUTION:  

  The solutions are about 0.7 and 1.8.

29. −2x2 + 10x = −14

SOLUTION:  

  The solutions are about –1.1 and 6.1.

30. −3x2 − 12 = 14x

SOLUTION:  

  The solutions are about –3.5 and –1.1.

31. STOCK The price p in dollars for a particular stock can be modeled by the quadratic equation p = 3.5t − 0.05t2, wh

the number of days after the stock is purchased. When is the stock worth $60?

SOLUTION:  Let p = 60.

  The stock is worth $60 on the 30th and 40th days after purchase.

GEOMETRY  Find the value of x for each figure. Round to the nearest tenth if necessary.

32. area = 45 in2

SOLUTION:  

  The height cannot be negative. So, x ≈ 6.3.

33. area = 110 ft2

SOLUTION:  

  The dimensions of the rectangle cannot be negative. So, x ≈ 5.3.

34. NUMBER THEORY The product of two consecutive even integers is 224. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers are 14 and 16 or –16 and –14.

35. CCSS PRECISION The product of two consecutive negative odd integers is 483. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers must be negative. So, they are –23 and –21.

36. GEOMETRY Find the area of the triangle.

SOLUTION:  

  The dimensions of the triangle cannot be negative. So, x = 18. The base of the triangle is 18 meters and the height is 18 + 6 or 24 meters.

The area of the triangle is 216 square meters.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

37. 0.2x2 − 0.2x − 0.4 = 0

SOLUTION:  

  The solutions are –1 and 2.

38. 0.5x2 = 2x − 0.3

SOLUTION:  

  The solutions are about 0.2 and 3.8.

39. 2x2 −

SOLUTION:  

  The solutions are about 0.2 and 0.9.

40. 

SOLUTION:  

  The solutions are –0.5 and 2.5.

41. 

SOLUTION:  

  The solutions are about –8.2 and 0.2.

42. 

SOLUTION:  

  The solutions are about –5.1 and 0.1.

43. ASTRONOMY The height of an object t seconds after it is dropped is given by the equation ,

where h0 is the initial height and g is the acceleration due to gravity. The acceleration due to gravity near the surface

of Mars is 3.73 m/s2, while on Earth it is 9.8 m/s

2. Suppose an object is dropped from an initial height of 120 meters

above the surface of each planet. a. On which planet would the object reach the ground first? b. How long would it take the object to reach the ground on each planet? Round each answer to the nearest tenth. c. Do the times that it takes the object to reach the ground seem reasonable? Explain your reasoning.

SOLUTION:  a. The object on Earth will reach the ground first because it is falling at a faster rate.   b. Mars:

So, t ≈ 8.0 seconds. Earth:

So, t ≈ 4.9 seconds.   c. Sample answer: Yes; the acceleration due to gravity is much greater on Earth than on Mars, so the time to reach the ground should be much less.

44. Find all values of c that make x2 + cx + 100 a perfect square trinomial.

SOLUTION:  

So, c can be –20 or 20 to make the trinomial a perfect square.

45. Find all values of c that make x2 + cx + 225 a perfect square trinomial.

SOLUTION:  

So, c can be –30 or 30 to make the trinomial a perfect square.

46. PAINTING Before she begins painting a picture, Donna stretches her canvas over a wood frame. The frame has a length of 60 inches and a width of 4 inches. She has enough canvas to cover 480 square inches. Donna decides to increase the dimensions of the frame. If the increase in the length is 10 times the increase in the width, what will the dimensions of the frame be?

SOLUTION:  Let x = the increase in the width and let 10x = the increase in the length. So, x + 4 = the new width and 10x + 60 = the new length.  

  The dimensions cannot be negative, so x = 2. The width is 2 + 4 or 6 inches and the length is 10(2) + 60 or 80 inches.

47. MULTIPLE REPRESENTATIONS In this problem, you will investigate a property of quadratic equations. a. TABULAR  Copy the table shown and complete the second column.

b. ALGEBRAIC Set each trinomial equal to zero, and solve the equation by completing the square. Complete the last column of the table with the number of roots of each equation.

c. VERBAL Compare the number of roots of each equation to the result in the b2 − 4ac column. Is there a

relationship between these values? If so, describe it.

d. ANALYTICAL Predict how many solutions 2x2 − 9x + 15 = 0 will have.  Verify your prediction by solving the 

equation.

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3    

3x2 + 6x + 9    

x2 − 2x + 7    

x2 + 10x + 25    

x2 + 3x − 12    

SOLUTION:  a.

b.

The trinomial x2 − 8x + 16 has 1 root.

 

The trinomial 2x

2 − 11x + 3 has 2 roots.

 

The square root of a negative number has no real roots. So, the trinomial 3x2 + 6x + 9 has 0 roots.

 

The square root of a negative number has no real roots. So, the trinomial x2 − 2x + 7.

 

The trinomial x2 + 10x + 25 has 1 root.

 

The trinomial x2 + 3x − 12 has 2 roots.

 

c. If b2 − 4ac is negative, the equation has no real solutions. If b

2 − 4ac is zero, the equation has one solution. If b2

− 4ac is positive, the equation has 2 solutions. d.

The equation 2x2 − 9x + 15 = 0 has 0 real solutions because b

2 − 4ac is negative.

 

The equation cannot be solved because the square root of a negative number has no real roots.

Trinomial b2 − 4ac Number

of Roots

x2 − 8x + 16 (–8)

2 – 4(1)(16) = 0 1

2x2 − 11x + 3 (–11)

2 – 4(2)(3) = 97  

3x2 + 6x + 9 (6)

2 – 4(3)(9) = −72  

x2 − 2x + 7 (–2)

2 – 4(1)(7) = −24  

x2 + 10x + 25 (10)

2 – 4(1)(25) = 0  

x2 + 3x − 12 (3)

2 – 4(1)(–12) = 57  

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3 97 2

3x2 + 6x + 9 −72 0

x2 − 2x + 7 −24 0

x2 + 10x + 25 0 1

x2 + 3x − 12 57 2

48. CCSS PERSEVERANCE Given y = ax2 + bx + c with a ≠ 0, derive the equation for the axis of symmetry by 

completing the square and rewriting the equation in the form y = a(x − h)2 + k .

SOLUTION:  

Let    and   , then the equation becomes y = (x - h)2 + k .

For an equation in the form y = ax2 + bx + c,  the equation for the axis of symmetry is   .

So, for an equation in the form  y = (x - h)2 + k, the equation for the axis of symmetry becomes x = h.

49. REASONING Determine the number of solutions x2 + bx = c has if . Explain.

SOLUTION:  

None; Sample answer: If you add to each side of the equation and each side of the inequality, you get

. Since the left side of the last equation is a perfect square, it cannot

equal the negative number . So, there are no real solutions.

50. WHICH ONE DOESN’T BELONG? Identify the expression that does not belong with the other three. Explain your reasoning.

 

SOLUTION:  The first 3 trinomials are perfect squares.

The trinomial  is not a perfect square. So, it does not belong.

51. OPEN ENDED Write a quadratic equation for which the only solution is 4.

SOLUTION:  Find a quadratic that has two roots of 4.

Verify that the solution is 4.

So, the quadratic equation x2 − 8x + 16 = 0 has a solution of 4.

52. WRITING IN MATH Compare and contrast the following strategies for solving x2 − 5x − 7 = 0: completing the

square, graphing, and factoring.

SOLUTION:  Sample answer: Because the leading coefficient is 1, solving the equation by completing the square is simpler and yields exact answers for the solutions.   

 

To solve the equation by graphing, use a graphing calculator to graph the related function y = x2 - 5x - 7. Select the

zero option from the 2nd [CALC] menu to determine the roots. 

   

  The roots are given as decimal approximations. So, for this equation the solutions would have to be given as estimations.   There are no factors of –7 that have a sum of –5, so solving by factoring is not possible.

53. The length of a rectangle is 3 times its width. The area of the rectangle is 75 square feet. Find the length of the rectangle in feet. A 25 B 15 C 10 D 5

SOLUTION:  Let x = the width of the rectangle and let 3x = the length of the rectangle.

The length of the rectangle is 3(5) or 15 feet. Choice B is the correct answer.

54. PROBABILITY At a festival, winners of a game draw a token for a prize. There is one token for each prize. The prizes include 9 movie passes, 8 stuffed animals, 5 hats, 10 jump ropes, and 4 glow necklaces. What is the probabilitythat the first person to draw a token will win a movie pass?

F  

G

H

J  

SOLUTION:  There are 9 + 8 + 5 + 10 + 4 or 36 possible outcomes.

P(movie pass) =  or  . Choice J is the correct answer.

55. GRIDDED RESPONSE The population of a town can be modeled by P = 22,000 + 125t, where P represents the population and t represents the number of years from 2000. How many years after 2000 will the population be 26,000?

SOLUTION:  

In 32 years the population of the town will be 26,000.

56. Percy delivers pizzas for Pizza King. He is paid $6 an hour plus $2.50 for each pizza he delivers. Percy earned $280 last week. If he worked a total of 30 hours, how many pizzas did he deliver? A 250 pizzas B 184 pizzas C 40 pizzas D 34 pizzas

SOLUTION:  Let p = the number of pizzas Percy delivered.

Percy delivered 40 pizzas. Choice C is the correct answer.

Describe how the graph of each function is related to the graph of f (x) = x2.

57. g(x) = −12 + x2

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –12, the translation is

down. So, the graph is shifted down 12 units from the parent function. 

58. h(x) = (x + 2)2

SOLUTION:  

The graph of f (x) = (x – c)2 represents a translation left or right from the parent graph. Since c = –2, the translation

is to the left by 2 units. 

 

59. g(x) = 2x2 + 5

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 2 and c = 5. Since 2 > 0 and  > 1, the graph of y = 2x

2 + 5

is the graph of y = x2 vertically stretched and shifted up 5 units.

60. 

SOLUTION:  

The function can be written f (x) = a(x – b)2, where a = and b = 6. Since > 0 and  < 1, the graph of 

is the graph of y = x2 vertically compressed and shifted right 6 units.

61. g(x) = 6 + x2

SOLUTION:  

The function can be written f (x) = ax2 + c, where a =  and c = 6. Since  > 0 and   > 1, the graph of y = 6 +

x2 is the graph of y = x

2 vertically stretched and shifted up 6 units.

62. h(x) = − 1 −   x2

SOLUTION:  

The function can be written f (x) = –ax2 + c, where a =  and c = –1. Since  < 0 and   > 1, the graph of y

= –1 – x2 is the graph of y = x

2 vertically stretched, shifted down 1 unit and reflected across the x-axis.

63. RIDES A popular amusement park ride whisks riders to the top of a 250-foot tower and drops them. A function for

the height of a rider is h = −16t2 + 250, where h is the height and t is the time in seconds. The ride stops the descent

of the rider 40 feet above the ground. Write an equation that models the drop of the rider. How long does it take to fall from 250 feet to 40 feet?

SOLUTION:  

The ride stops the descent of the rider at 40 feet above ground, so h = 40. Then, the equation 40 = −16t2 + 250

models the drop of the rider.

Time cannot be negative. So, it takes about 3.6 seconds to complete the ride.

Simplify. Assume that no denominator is equal to zero.

64. 

SOLUTION:  

65. 

SOLUTION:  

66. 

SOLUTION:  

67. 

SOLUTION:  

68. 

SOLUTION:  

69. b3(m

−3)(b

−6)

SOLUTION:  

Solve each open sentence.

70. |y − 2| > 7

SOLUTION:  

The solution set is {y |y > 9 or y < −5}.

Case 1 y – 2 is positive.

and Case 2 y – 2 is negative.

 

71. |z + 5| < 3

SOLUTION:  

The solution set is {z |−8 < z < −2}.

Case 1 z +5 is positive.

and Case 2 z +5 is negative.

 

72. |2b + 7| ≤ −6

SOLUTION:  

cannot be negative. So,  cannot be less than or equal to –6. The solution set is empty.

73. |3 − 2y | ≥ 8

SOLUTION:  

The solution set is {y |y  ≥ 5.5 or y  ≤ −2.5}.

Case 1 3 – 2y is positive.

and Case 2 3 – 2y is negative.

 

74. |9 − 4m| < −1

SOLUTION:  

cannot be negative. So, cannot be less than or equal to –1. The solution set is empty.

75. |5c − 2| ≤ 13

SOLUTION:  

The solution set is {c|−2.2 ≤ c ≤ 3}.

Case 1 5c – 2 is positive.

and Case 2 5c – 2 is negative.

 

Evaluate  for each set of values. Round to the nearest tenth if necessary.

76. a = 2, b = −5, c = 2

SOLUTION:  

77. a = 1, b = 12, c = 11

SOLUTION:  

78. a = −9, b = 10, c = −1

SOLUTION:  

79. a = 1, b = 7, c = −3

SOLUTION:  

80. a = 2, b = −4, c = −6

SOLUTION:  

81. a = 3, b = 1, c = 2

SOLUTION:  

This value is not a real number.

eSolutions Manual - Powered by Cognero Page 5

9-4 Solving Quadratic Equations by Completing the Square

Find the value of c that makes each trinomial a perfect square.

1. x2 − 18x + c

SOLUTION:  In this trinomial, b = –18.

So, c must be 81 to make the trinomial a perfect square.

2. x2 + 22x + c

SOLUTION:  In this trinomial, b = 22.

So, c must be 121 to make the trinomial a perfect square.

3. x2 + 9x + c

SOLUTION:  In this trinomial, b = 9.

 

So, c must be  to make the trinomial a perfect square.

4. x2 − 7x + c

SOLUTION:  In this trinomial, b = –7.

 

So, c must be  to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

5. x2 + 4x = 6

SOLUTION:  

The solutions are about –5.2 and 1.2.

6. x2 − 8x = −9

SOLUTION:  

The solutions are about 1.4 and 6.6.

7. 4x2 + 9x − 1 = 0

SOLUTION:  

The solutions are about –2.4 and 0.1.

8. −2x2 + 10x + 22 = 4

SOLUTION:  

The solutions are about –1.4 and 6.4.

9. CCSS MODELING Collin is building a deck on the back of his family’s house. He has enough lumber for the deck to be 144 square feet. The length should be 10 feet more than its width. What should the dimensions of the deck be?

SOLUTION:  Let x = the width of the deck and x + 10 = the length of the deck.

The dimensions of the deck can not be negative. So, x = –5 + 13 or 8. The width of the deck is 8 feet and the length of the deck is 8 + 10 or 18 feet.

Find the value of c that makes each trinomial a perfect square.

10. x2 + 26x + c

SOLUTION:  In this trinomial, b = 26.

So, c must be 169 to make the trinomial a perfect square.

11. x2 − 24x + c

SOLUTION:  In this trinomial, b = –24.

So, c must be 144 to make the trinomial a perfect square.

12. x2 − 19x + c

SOLUTION:  In this trinomial, b = –19.

So, c must be  to make the trinomial a perfect square.

13. x2 + 17x + c

SOLUTION:  In this trinomial, b = 17.

So, c must be  to make the trinomial a perfect square.

14. x2 + 5x + c

SOLUTION:  In this trinomial, b = 5.

So, c must be  to make the trinomial a perfect square.

15. x2 − 13x + c

SOLUTION:  In this trinomial, b = –13.

So, c must be  to make the trinomial a perfect square.

16. x2 − 22x + c

SOLUTION:  In this trinomial, b = –22.

So, c must be 121 to make the trinomial a perfect square.

17. x2 − 15x + c

SOLUTION:  In this trinomial, b = –15.

So, c must be  to make the trinomial a perfect square.

18. x2 + 24x + c

SOLUTION:  In this trinomial, b = 24.

So, c must be 144 to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

19. x2 + 6x − 16 = 0

SOLUTION:  

  The solutions are –8 and 2.

20. x2 − 2x − 14 = 0

SOLUTION:  

  The solutions are about –2.9 and 4.9.

21. x2 − 8x − 1 = 8

SOLUTION:  

  The solutions are –1 and 9.

22. x2 + 3x + 21 = 22

SOLUTION:  

  The solutions are about –3.3 and 0.3.

23. x2 − 11x + 3 = 5

SOLUTION:  

  The solutions are about –0.2 and 11.2.

24. 5x2 − 10x = 23

SOLUTION:  

  The solutions are about –1.4 and 3.4.

25. 2x2 − 2x + 7 = 5

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

26. 3x2 + 12x + 81 = 15

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

27. 4x2 + 6x = 12

SOLUTION:  

  The solutions are about –2.6 and 1.1.

28. 4x2 + 5 = 10x

SOLUTION:  

  The solutions are about 0.7 and 1.8.

29. −2x2 + 10x = −14

SOLUTION:  

  The solutions are about –1.1 and 6.1.

30. −3x2 − 12 = 14x

SOLUTION:  

  The solutions are about –3.5 and –1.1.

31. STOCK The price p in dollars for a particular stock can be modeled by the quadratic equation p = 3.5t − 0.05t2, wh

the number of days after the stock is purchased. When is the stock worth $60?

SOLUTION:  Let p = 60.

  The stock is worth $60 on the 30th and 40th days after purchase.

GEOMETRY  Find the value of x for each figure. Round to the nearest tenth if necessary.

32. area = 45 in2

SOLUTION:  

  The height cannot be negative. So, x ≈ 6.3.

33. area = 110 ft2

SOLUTION:  

  The dimensions of the rectangle cannot be negative. So, x ≈ 5.3.

34. NUMBER THEORY The product of two consecutive even integers is 224. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers are 14 and 16 or –16 and –14.

35. CCSS PRECISION The product of two consecutive negative odd integers is 483. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers must be negative. So, they are –23 and –21.

36. GEOMETRY Find the area of the triangle.

SOLUTION:  

  The dimensions of the triangle cannot be negative. So, x = 18. The base of the triangle is 18 meters and the height is 18 + 6 or 24 meters.

The area of the triangle is 216 square meters.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

37. 0.2x2 − 0.2x − 0.4 = 0

SOLUTION:  

  The solutions are –1 and 2.

38. 0.5x2 = 2x − 0.3

SOLUTION:  

  The solutions are about 0.2 and 3.8.

39. 2x2 −

SOLUTION:  

  The solutions are about 0.2 and 0.9.

40. 

SOLUTION:  

  The solutions are –0.5 and 2.5.

41. 

SOLUTION:  

  The solutions are about –8.2 and 0.2.

42. 

SOLUTION:  

  The solutions are about –5.1 and 0.1.

43. ASTRONOMY The height of an object t seconds after it is dropped is given by the equation ,

where h0 is the initial height and g is the acceleration due to gravity. The acceleration due to gravity near the surface

of Mars is 3.73 m/s2, while on Earth it is 9.8 m/s

2. Suppose an object is dropped from an initial height of 120 meters

above the surface of each planet. a. On which planet would the object reach the ground first? b. How long would it take the object to reach the ground on each planet? Round each answer to the nearest tenth. c. Do the times that it takes the object to reach the ground seem reasonable? Explain your reasoning.

SOLUTION:  a. The object on Earth will reach the ground first because it is falling at a faster rate.   b. Mars:

So, t ≈ 8.0 seconds. Earth:

So, t ≈ 4.9 seconds.   c. Sample answer: Yes; the acceleration due to gravity is much greater on Earth than on Mars, so the time to reach the ground should be much less.

44. Find all values of c that make x2 + cx + 100 a perfect square trinomial.

SOLUTION:  

So, c can be –20 or 20 to make the trinomial a perfect square.

45. Find all values of c that make x2 + cx + 225 a perfect square trinomial.

SOLUTION:  

So, c can be –30 or 30 to make the trinomial a perfect square.

46. PAINTING Before she begins painting a picture, Donna stretches her canvas over a wood frame. The frame has a length of 60 inches and a width of 4 inches. She has enough canvas to cover 480 square inches. Donna decides to increase the dimensions of the frame. If the increase in the length is 10 times the increase in the width, what will the dimensions of the frame be?

SOLUTION:  Let x = the increase in the width and let 10x = the increase in the length. So, x + 4 = the new width and 10x + 60 = the new length.  

  The dimensions cannot be negative, so x = 2. The width is 2 + 4 or 6 inches and the length is 10(2) + 60 or 80 inches.

47. MULTIPLE REPRESENTATIONS In this problem, you will investigate a property of quadratic equations. a. TABULAR  Copy the table shown and complete the second column.

b. ALGEBRAIC Set each trinomial equal to zero, and solve the equation by completing the square. Complete the last column of the table with the number of roots of each equation.

c. VERBAL Compare the number of roots of each equation to the result in the b2 − 4ac column. Is there a

relationship between these values? If so, describe it.

d. ANALYTICAL Predict how many solutions 2x2 − 9x + 15 = 0 will have.  Verify your prediction by solving the 

equation.

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3    

3x2 + 6x + 9    

x2 − 2x + 7    

x2 + 10x + 25    

x2 + 3x − 12    

SOLUTION:  a.

b.

The trinomial x2 − 8x + 16 has 1 root.

 

The trinomial 2x

2 − 11x + 3 has 2 roots.

 

The square root of a negative number has no real roots. So, the trinomial 3x2 + 6x + 9 has 0 roots.

 

The square root of a negative number has no real roots. So, the trinomial x2 − 2x + 7.

 

The trinomial x2 + 10x + 25 has 1 root.

 

The trinomial x2 + 3x − 12 has 2 roots.

 

c. If b2 − 4ac is negative, the equation has no real solutions. If b

2 − 4ac is zero, the equation has one solution. If b2

− 4ac is positive, the equation has 2 solutions. d.

The equation 2x2 − 9x + 15 = 0 has 0 real solutions because b

2 − 4ac is negative.

 

The equation cannot be solved because the square root of a negative number has no real roots.

Trinomial b2 − 4ac Number

of Roots

x2 − 8x + 16 (–8)

2 – 4(1)(16) = 0 1

2x2 − 11x + 3 (–11)

2 – 4(2)(3) = 97  

3x2 + 6x + 9 (6)

2 – 4(3)(9) = −72  

x2 − 2x + 7 (–2)

2 – 4(1)(7) = −24  

x2 + 10x + 25 (10)

2 – 4(1)(25) = 0  

x2 + 3x − 12 (3)

2 – 4(1)(–12) = 57  

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3 97 2

3x2 + 6x + 9 −72 0

x2 − 2x + 7 −24 0

x2 + 10x + 25 0 1

x2 + 3x − 12 57 2

48. CCSS PERSEVERANCE Given y = ax2 + bx + c with a ≠ 0, derive the equation for the axis of symmetry by 

completing the square and rewriting the equation in the form y = a(x − h)2 + k .

SOLUTION:  

Let    and   , then the equation becomes y = (x - h)2 + k .

For an equation in the form y = ax2 + bx + c,  the equation for the axis of symmetry is   .

So, for an equation in the form  y = (x - h)2 + k, the equation for the axis of symmetry becomes x = h.

49. REASONING Determine the number of solutions x2 + bx = c has if . Explain.

SOLUTION:  

None; Sample answer: If you add to each side of the equation and each side of the inequality, you get

. Since the left side of the last equation is a perfect square, it cannot

equal the negative number . So, there are no real solutions.

50. WHICH ONE DOESN’T BELONG? Identify the expression that does not belong with the other three. Explain your reasoning.

 

SOLUTION:  The first 3 trinomials are perfect squares.

The trinomial  is not a perfect square. So, it does not belong.

51. OPEN ENDED Write a quadratic equation for which the only solution is 4.

SOLUTION:  Find a quadratic that has two roots of 4.

Verify that the solution is 4.

So, the quadratic equation x2 − 8x + 16 = 0 has a solution of 4.

52. WRITING IN MATH Compare and contrast the following strategies for solving x2 − 5x − 7 = 0: completing the

square, graphing, and factoring.

SOLUTION:  Sample answer: Because the leading coefficient is 1, solving the equation by completing the square is simpler and yields exact answers for the solutions.   

 

To solve the equation by graphing, use a graphing calculator to graph the related function y = x2 - 5x - 7. Select the

zero option from the 2nd [CALC] menu to determine the roots. 

   

  The roots are given as decimal approximations. So, for this equation the solutions would have to be given as estimations.   There are no factors of –7 that have a sum of –5, so solving by factoring is not possible.

53. The length of a rectangle is 3 times its width. The area of the rectangle is 75 square feet. Find the length of the rectangle in feet. A 25 B 15 C 10 D 5

SOLUTION:  Let x = the width of the rectangle and let 3x = the length of the rectangle.

The length of the rectangle is 3(5) or 15 feet. Choice B is the correct answer.

54. PROBABILITY At a festival, winners of a game draw a token for a prize. There is one token for each prize. The prizes include 9 movie passes, 8 stuffed animals, 5 hats, 10 jump ropes, and 4 glow necklaces. What is the probabilitythat the first person to draw a token will win a movie pass?

F  

G

H

J  

SOLUTION:  There are 9 + 8 + 5 + 10 + 4 or 36 possible outcomes.

P(movie pass) =  or  . Choice J is the correct answer.

55. GRIDDED RESPONSE The population of a town can be modeled by P = 22,000 + 125t, where P represents the population and t represents the number of years from 2000. How many years after 2000 will the population be 26,000?

SOLUTION:  

In 32 years the population of the town will be 26,000.

56. Percy delivers pizzas for Pizza King. He is paid $6 an hour plus $2.50 for each pizza he delivers. Percy earned $280 last week. If he worked a total of 30 hours, how many pizzas did he deliver? A 250 pizzas B 184 pizzas C 40 pizzas D 34 pizzas

SOLUTION:  Let p = the number of pizzas Percy delivered.

Percy delivered 40 pizzas. Choice C is the correct answer.

Describe how the graph of each function is related to the graph of f (x) = x2.

57. g(x) = −12 + x2

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –12, the translation is

down. So, the graph is shifted down 12 units from the parent function. 

58. h(x) = (x + 2)2

SOLUTION:  

The graph of f (x) = (x – c)2 represents a translation left or right from the parent graph. Since c = –2, the translation

is to the left by 2 units. 

 

59. g(x) = 2x2 + 5

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 2 and c = 5. Since 2 > 0 and  > 1, the graph of y = 2x

2 + 5

is the graph of y = x2 vertically stretched and shifted up 5 units.

60. 

SOLUTION:  

The function can be written f (x) = a(x – b)2, where a = and b = 6. Since > 0 and  < 1, the graph of 

is the graph of y = x2 vertically compressed and shifted right 6 units.

61. g(x) = 6 + x2

SOLUTION:  

The function can be written f (x) = ax2 + c, where a =  and c = 6. Since  > 0 and   > 1, the graph of y = 6 +

x2 is the graph of y = x

2 vertically stretched and shifted up 6 units.

62. h(x) = − 1 −   x2

SOLUTION:  

The function can be written f (x) = –ax2 + c, where a =  and c = –1. Since  < 0 and   > 1, the graph of y

= –1 – x2 is the graph of y = x

2 vertically stretched, shifted down 1 unit and reflected across the x-axis.

63. RIDES A popular amusement park ride whisks riders to the top of a 250-foot tower and drops them. A function for

the height of a rider is h = −16t2 + 250, where h is the height and t is the time in seconds. The ride stops the descent

of the rider 40 feet above the ground. Write an equation that models the drop of the rider. How long does it take to fall from 250 feet to 40 feet?

SOLUTION:  

The ride stops the descent of the rider at 40 feet above ground, so h = 40. Then, the equation 40 = −16t2 + 250

models the drop of the rider.

Time cannot be negative. So, it takes about 3.6 seconds to complete the ride.

Simplify. Assume that no denominator is equal to zero.

64. 

SOLUTION:  

65. 

SOLUTION:  

66. 

SOLUTION:  

67. 

SOLUTION:  

68. 

SOLUTION:  

69. b3(m

−3)(b

−6)

SOLUTION:  

Solve each open sentence.

70. |y − 2| > 7

SOLUTION:  

The solution set is {y |y > 9 or y < −5}.

Case 1 y – 2 is positive.

and Case 2 y – 2 is negative.

 

71. |z + 5| < 3

SOLUTION:  

The solution set is {z |−8 < z < −2}.

Case 1 z +5 is positive.

and Case 2 z +5 is negative.

 

72. |2b + 7| ≤ −6

SOLUTION:  

cannot be negative. So,  cannot be less than or equal to –6. The solution set is empty.

73. |3 − 2y | ≥ 8

SOLUTION:  

The solution set is {y |y  ≥ 5.5 or y  ≤ −2.5}.

Case 1 3 – 2y is positive.

and Case 2 3 – 2y is negative.

 

74. |9 − 4m| < −1

SOLUTION:  

cannot be negative. So, cannot be less than or equal to –1. The solution set is empty.

75. |5c − 2| ≤ 13

SOLUTION:  

The solution set is {c|−2.2 ≤ c ≤ 3}.

Case 1 5c – 2 is positive.

and Case 2 5c – 2 is negative.

 

Evaluate  for each set of values. Round to the nearest tenth if necessary.

76. a = 2, b = −5, c = 2

SOLUTION:  

77. a = 1, b = 12, c = 11

SOLUTION:  

78. a = −9, b = 10, c = −1

SOLUTION:  

79. a = 1, b = 7, c = −3

SOLUTION:  

80. a = 2, b = −4, c = −6

SOLUTION:  

81. a = 3, b = 1, c = 2

SOLUTION:  

This value is not a real number.

eSolutions Manual - Powered by Cognero Page 6

9-4 Solving Quadratic Equations by Completing the Square

Find the value of c that makes each trinomial a perfect square.

1. x2 − 18x + c

SOLUTION:  In this trinomial, b = –18.

So, c must be 81 to make the trinomial a perfect square.

2. x2 + 22x + c

SOLUTION:  In this trinomial, b = 22.

So, c must be 121 to make the trinomial a perfect square.

3. x2 + 9x + c

SOLUTION:  In this trinomial, b = 9.

 

So, c must be  to make the trinomial a perfect square.

4. x2 − 7x + c

SOLUTION:  In this trinomial, b = –7.

 

So, c must be  to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

5. x2 + 4x = 6

SOLUTION:  

The solutions are about –5.2 and 1.2.

6. x2 − 8x = −9

SOLUTION:  

The solutions are about 1.4 and 6.6.

7. 4x2 + 9x − 1 = 0

SOLUTION:  

The solutions are about –2.4 and 0.1.

8. −2x2 + 10x + 22 = 4

SOLUTION:  

The solutions are about –1.4 and 6.4.

9. CCSS MODELING Collin is building a deck on the back of his family’s house. He has enough lumber for the deck to be 144 square feet. The length should be 10 feet more than its width. What should the dimensions of the deck be?

SOLUTION:  Let x = the width of the deck and x + 10 = the length of the deck.

The dimensions of the deck can not be negative. So, x = –5 + 13 or 8. The width of the deck is 8 feet and the length of the deck is 8 + 10 or 18 feet.

Find the value of c that makes each trinomial a perfect square.

10. x2 + 26x + c

SOLUTION:  In this trinomial, b = 26.

So, c must be 169 to make the trinomial a perfect square.

11. x2 − 24x + c

SOLUTION:  In this trinomial, b = –24.

So, c must be 144 to make the trinomial a perfect square.

12. x2 − 19x + c

SOLUTION:  In this trinomial, b = –19.

So, c must be  to make the trinomial a perfect square.

13. x2 + 17x + c

SOLUTION:  In this trinomial, b = 17.

So, c must be  to make the trinomial a perfect square.

14. x2 + 5x + c

SOLUTION:  In this trinomial, b = 5.

So, c must be  to make the trinomial a perfect square.

15. x2 − 13x + c

SOLUTION:  In this trinomial, b = –13.

So, c must be  to make the trinomial a perfect square.

16. x2 − 22x + c

SOLUTION:  In this trinomial, b = –22.

So, c must be 121 to make the trinomial a perfect square.

17. x2 − 15x + c

SOLUTION:  In this trinomial, b = –15.

So, c must be  to make the trinomial a perfect square.

18. x2 + 24x + c

SOLUTION:  In this trinomial, b = 24.

So, c must be 144 to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

19. x2 + 6x − 16 = 0

SOLUTION:  

  The solutions are –8 and 2.

20. x2 − 2x − 14 = 0

SOLUTION:  

  The solutions are about –2.9 and 4.9.

21. x2 − 8x − 1 = 8

SOLUTION:  

  The solutions are –1 and 9.

22. x2 + 3x + 21 = 22

SOLUTION:  

  The solutions are about –3.3 and 0.3.

23. x2 − 11x + 3 = 5

SOLUTION:  

  The solutions are about –0.2 and 11.2.

24. 5x2 − 10x = 23

SOLUTION:  

  The solutions are about –1.4 and 3.4.

25. 2x2 − 2x + 7 = 5

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

26. 3x2 + 12x + 81 = 15

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

27. 4x2 + 6x = 12

SOLUTION:  

  The solutions are about –2.6 and 1.1.

28. 4x2 + 5 = 10x

SOLUTION:  

  The solutions are about 0.7 and 1.8.

29. −2x2 + 10x = −14

SOLUTION:  

  The solutions are about –1.1 and 6.1.

30. −3x2 − 12 = 14x

SOLUTION:  

  The solutions are about –3.5 and –1.1.

31. STOCK The price p in dollars for a particular stock can be modeled by the quadratic equation p = 3.5t − 0.05t2, wh

the number of days after the stock is purchased. When is the stock worth $60?

SOLUTION:  Let p = 60.

  The stock is worth $60 on the 30th and 40th days after purchase.

GEOMETRY  Find the value of x for each figure. Round to the nearest tenth if necessary.

32. area = 45 in2

SOLUTION:  

  The height cannot be negative. So, x ≈ 6.3.

33. area = 110 ft2

SOLUTION:  

  The dimensions of the rectangle cannot be negative. So, x ≈ 5.3.

34. NUMBER THEORY The product of two consecutive even integers is 224. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers are 14 and 16 or –16 and –14.

35. CCSS PRECISION The product of two consecutive negative odd integers is 483. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers must be negative. So, they are –23 and –21.

36. GEOMETRY Find the area of the triangle.

SOLUTION:  

  The dimensions of the triangle cannot be negative. So, x = 18. The base of the triangle is 18 meters and the height is 18 + 6 or 24 meters.

The area of the triangle is 216 square meters.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

37. 0.2x2 − 0.2x − 0.4 = 0

SOLUTION:  

  The solutions are –1 and 2.

38. 0.5x2 = 2x − 0.3

SOLUTION:  

  The solutions are about 0.2 and 3.8.

39. 2x2 −

SOLUTION:  

  The solutions are about 0.2 and 0.9.

40. 

SOLUTION:  

  The solutions are –0.5 and 2.5.

41. 

SOLUTION:  

  The solutions are about –8.2 and 0.2.

42. 

SOLUTION:  

  The solutions are about –5.1 and 0.1.

43. ASTRONOMY The height of an object t seconds after it is dropped is given by the equation ,

where h0 is the initial height and g is the acceleration due to gravity. The acceleration due to gravity near the surface

of Mars is 3.73 m/s2, while on Earth it is 9.8 m/s

2. Suppose an object is dropped from an initial height of 120 meters

above the surface of each planet. a. On which planet would the object reach the ground first? b. How long would it take the object to reach the ground on each planet? Round each answer to the nearest tenth. c. Do the times that it takes the object to reach the ground seem reasonable? Explain your reasoning.

SOLUTION:  a. The object on Earth will reach the ground first because it is falling at a faster rate.   b. Mars:

So, t ≈ 8.0 seconds. Earth:

So, t ≈ 4.9 seconds.   c. Sample answer: Yes; the acceleration due to gravity is much greater on Earth than on Mars, so the time to reach the ground should be much less.

44. Find all values of c that make x2 + cx + 100 a perfect square trinomial.

SOLUTION:  

So, c can be –20 or 20 to make the trinomial a perfect square.

45. Find all values of c that make x2 + cx + 225 a perfect square trinomial.

SOLUTION:  

So, c can be –30 or 30 to make the trinomial a perfect square.

46. PAINTING Before she begins painting a picture, Donna stretches her canvas over a wood frame. The frame has a length of 60 inches and a width of 4 inches. She has enough canvas to cover 480 square inches. Donna decides to increase the dimensions of the frame. If the increase in the length is 10 times the increase in the width, what will the dimensions of the frame be?

SOLUTION:  Let x = the increase in the width and let 10x = the increase in the length. So, x + 4 = the new width and 10x + 60 = the new length.  

  The dimensions cannot be negative, so x = 2. The width is 2 + 4 or 6 inches and the length is 10(2) + 60 or 80 inches.

47. MULTIPLE REPRESENTATIONS In this problem, you will investigate a property of quadratic equations. a. TABULAR  Copy the table shown and complete the second column.

b. ALGEBRAIC Set each trinomial equal to zero, and solve the equation by completing the square. Complete the last column of the table with the number of roots of each equation.

c. VERBAL Compare the number of roots of each equation to the result in the b2 − 4ac column. Is there a

relationship between these values? If so, describe it.

d. ANALYTICAL Predict how many solutions 2x2 − 9x + 15 = 0 will have.  Verify your prediction by solving the 

equation.

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3    

3x2 + 6x + 9    

x2 − 2x + 7    

x2 + 10x + 25    

x2 + 3x − 12    

SOLUTION:  a.

b.

The trinomial x2 − 8x + 16 has 1 root.

 

The trinomial 2x

2 − 11x + 3 has 2 roots.

 

The square root of a negative number has no real roots. So, the trinomial 3x2 + 6x + 9 has 0 roots.

 

The square root of a negative number has no real roots. So, the trinomial x2 − 2x + 7.

 

The trinomial x2 + 10x + 25 has 1 root.

 

The trinomial x2 + 3x − 12 has 2 roots.

 

c. If b2 − 4ac is negative, the equation has no real solutions. If b

2 − 4ac is zero, the equation has one solution. If b2

− 4ac is positive, the equation has 2 solutions. d.

The equation 2x2 − 9x + 15 = 0 has 0 real solutions because b

2 − 4ac is negative.

 

The equation cannot be solved because the square root of a negative number has no real roots.

Trinomial b2 − 4ac Number

of Roots

x2 − 8x + 16 (–8)

2 – 4(1)(16) = 0 1

2x2 − 11x + 3 (–11)

2 – 4(2)(3) = 97  

3x2 + 6x + 9 (6)

2 – 4(3)(9) = −72  

x2 − 2x + 7 (–2)

2 – 4(1)(7) = −24  

x2 + 10x + 25 (10)

2 – 4(1)(25) = 0  

x2 + 3x − 12 (3)

2 – 4(1)(–12) = 57  

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3 97 2

3x2 + 6x + 9 −72 0

x2 − 2x + 7 −24 0

x2 + 10x + 25 0 1

x2 + 3x − 12 57 2

48. CCSS PERSEVERANCE Given y = ax2 + bx + c with a ≠ 0, derive the equation for the axis of symmetry by 

completing the square and rewriting the equation in the form y = a(x − h)2 + k .

SOLUTION:  

Let    and   , then the equation becomes y = (x - h)2 + k .

For an equation in the form y = ax2 + bx + c,  the equation for the axis of symmetry is   .

So, for an equation in the form  y = (x - h)2 + k, the equation for the axis of symmetry becomes x = h.

49. REASONING Determine the number of solutions x2 + bx = c has if . Explain.

SOLUTION:  

None; Sample answer: If you add to each side of the equation and each side of the inequality, you get

. Since the left side of the last equation is a perfect square, it cannot

equal the negative number . So, there are no real solutions.

50. WHICH ONE DOESN’T BELONG? Identify the expression that does not belong with the other three. Explain your reasoning.

 

SOLUTION:  The first 3 trinomials are perfect squares.

The trinomial  is not a perfect square. So, it does not belong.

51. OPEN ENDED Write a quadratic equation for which the only solution is 4.

SOLUTION:  Find a quadratic that has two roots of 4.

Verify that the solution is 4.

So, the quadratic equation x2 − 8x + 16 = 0 has a solution of 4.

52. WRITING IN MATH Compare and contrast the following strategies for solving x2 − 5x − 7 = 0: completing the

square, graphing, and factoring.

SOLUTION:  Sample answer: Because the leading coefficient is 1, solving the equation by completing the square is simpler and yields exact answers for the solutions.   

 

To solve the equation by graphing, use a graphing calculator to graph the related function y = x2 - 5x - 7. Select the

zero option from the 2nd [CALC] menu to determine the roots. 

   

  The roots are given as decimal approximations. So, for this equation the solutions would have to be given as estimations.   There are no factors of –7 that have a sum of –5, so solving by factoring is not possible.

53. The length of a rectangle is 3 times its width. The area of the rectangle is 75 square feet. Find the length of the rectangle in feet. A 25 B 15 C 10 D 5

SOLUTION:  Let x = the width of the rectangle and let 3x = the length of the rectangle.

The length of the rectangle is 3(5) or 15 feet. Choice B is the correct answer.

54. PROBABILITY At a festival, winners of a game draw a token for a prize. There is one token for each prize. The prizes include 9 movie passes, 8 stuffed animals, 5 hats, 10 jump ropes, and 4 glow necklaces. What is the probabilitythat the first person to draw a token will win a movie pass?

F  

G

H

J  

SOLUTION:  There are 9 + 8 + 5 + 10 + 4 or 36 possible outcomes.

P(movie pass) =  or  . Choice J is the correct answer.

55. GRIDDED RESPONSE The population of a town can be modeled by P = 22,000 + 125t, where P represents the population and t represents the number of years from 2000. How many years after 2000 will the population be 26,000?

SOLUTION:  

In 32 years the population of the town will be 26,000.

56. Percy delivers pizzas for Pizza King. He is paid $6 an hour plus $2.50 for each pizza he delivers. Percy earned $280 last week. If he worked a total of 30 hours, how many pizzas did he deliver? A 250 pizzas B 184 pizzas C 40 pizzas D 34 pizzas

SOLUTION:  Let p = the number of pizzas Percy delivered.

Percy delivered 40 pizzas. Choice C is the correct answer.

Describe how the graph of each function is related to the graph of f (x) = x2.

57. g(x) = −12 + x2

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –12, the translation is

down. So, the graph is shifted down 12 units from the parent function. 

58. h(x) = (x + 2)2

SOLUTION:  

The graph of f (x) = (x – c)2 represents a translation left or right from the parent graph. Since c = –2, the translation

is to the left by 2 units. 

 

59. g(x) = 2x2 + 5

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 2 and c = 5. Since 2 > 0 and  > 1, the graph of y = 2x

2 + 5

is the graph of y = x2 vertically stretched and shifted up 5 units.

60. 

SOLUTION:  

The function can be written f (x) = a(x – b)2, where a = and b = 6. Since > 0 and  < 1, the graph of 

is the graph of y = x2 vertically compressed and shifted right 6 units.

61. g(x) = 6 + x2

SOLUTION:  

The function can be written f (x) = ax2 + c, where a =  and c = 6. Since  > 0 and   > 1, the graph of y = 6 +

x2 is the graph of y = x

2 vertically stretched and shifted up 6 units.

62. h(x) = − 1 −   x2

SOLUTION:  

The function can be written f (x) = –ax2 + c, where a =  and c = –1. Since  < 0 and   > 1, the graph of y

= –1 – x2 is the graph of y = x

2 vertically stretched, shifted down 1 unit and reflected across the x-axis.

63. RIDES A popular amusement park ride whisks riders to the top of a 250-foot tower and drops them. A function for

the height of a rider is h = −16t2 + 250, where h is the height and t is the time in seconds. The ride stops the descent

of the rider 40 feet above the ground. Write an equation that models the drop of the rider. How long does it take to fall from 250 feet to 40 feet?

SOLUTION:  

The ride stops the descent of the rider at 40 feet above ground, so h = 40. Then, the equation 40 = −16t2 + 250

models the drop of the rider.

Time cannot be negative. So, it takes about 3.6 seconds to complete the ride.

Simplify. Assume that no denominator is equal to zero.

64. 

SOLUTION:  

65. 

SOLUTION:  

66. 

SOLUTION:  

67. 

SOLUTION:  

68. 

SOLUTION:  

69. b3(m

−3)(b

−6)

SOLUTION:  

Solve each open sentence.

70. |y − 2| > 7

SOLUTION:  

The solution set is {y |y > 9 or y < −5}.

Case 1 y – 2 is positive.

and Case 2 y – 2 is negative.

 

71. |z + 5| < 3

SOLUTION:  

The solution set is {z |−8 < z < −2}.

Case 1 z +5 is positive.

and Case 2 z +5 is negative.

 

72. |2b + 7| ≤ −6

SOLUTION:  

cannot be negative. So,  cannot be less than or equal to –6. The solution set is empty.

73. |3 − 2y | ≥ 8

SOLUTION:  

The solution set is {y |y  ≥ 5.5 or y  ≤ −2.5}.

Case 1 3 – 2y is positive.

and Case 2 3 – 2y is negative.

 

74. |9 − 4m| < −1

SOLUTION:  

cannot be negative. So, cannot be less than or equal to –1. The solution set is empty.

75. |5c − 2| ≤ 13

SOLUTION:  

The solution set is {c|−2.2 ≤ c ≤ 3}.

Case 1 5c – 2 is positive.

and Case 2 5c – 2 is negative.

 

Evaluate  for each set of values. Round to the nearest tenth if necessary.

76. a = 2, b = −5, c = 2

SOLUTION:  

77. a = 1, b = 12, c = 11

SOLUTION:  

78. a = −9, b = 10, c = −1

SOLUTION:  

79. a = 1, b = 7, c = −3

SOLUTION:  

80. a = 2, b = −4, c = −6

SOLUTION:  

81. a = 3, b = 1, c = 2

SOLUTION:  

This value is not a real number.

eSolutions Manual - Powered by Cognero Page 7

9-4 Solving Quadratic Equations by Completing the Square

Find the value of c that makes each trinomial a perfect square.

1. x2 − 18x + c

SOLUTION:  In this trinomial, b = –18.

So, c must be 81 to make the trinomial a perfect square.

2. x2 + 22x + c

SOLUTION:  In this trinomial, b = 22.

So, c must be 121 to make the trinomial a perfect square.

3. x2 + 9x + c

SOLUTION:  In this trinomial, b = 9.

 

So, c must be  to make the trinomial a perfect square.

4. x2 − 7x + c

SOLUTION:  In this trinomial, b = –7.

 

So, c must be  to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

5. x2 + 4x = 6

SOLUTION:  

The solutions are about –5.2 and 1.2.

6. x2 − 8x = −9

SOLUTION:  

The solutions are about 1.4 and 6.6.

7. 4x2 + 9x − 1 = 0

SOLUTION:  

The solutions are about –2.4 and 0.1.

8. −2x2 + 10x + 22 = 4

SOLUTION:  

The solutions are about –1.4 and 6.4.

9. CCSS MODELING Collin is building a deck on the back of his family’s house. He has enough lumber for the deck to be 144 square feet. The length should be 10 feet more than its width. What should the dimensions of the deck be?

SOLUTION:  Let x = the width of the deck and x + 10 = the length of the deck.

The dimensions of the deck can not be negative. So, x = –5 + 13 or 8. The width of the deck is 8 feet and the length of the deck is 8 + 10 or 18 feet.

Find the value of c that makes each trinomial a perfect square.

10. x2 + 26x + c

SOLUTION:  In this trinomial, b = 26.

So, c must be 169 to make the trinomial a perfect square.

11. x2 − 24x + c

SOLUTION:  In this trinomial, b = –24.

So, c must be 144 to make the trinomial a perfect square.

12. x2 − 19x + c

SOLUTION:  In this trinomial, b = –19.

So, c must be  to make the trinomial a perfect square.

13. x2 + 17x + c

SOLUTION:  In this trinomial, b = 17.

So, c must be  to make the trinomial a perfect square.

14. x2 + 5x + c

SOLUTION:  In this trinomial, b = 5.

So, c must be  to make the trinomial a perfect square.

15. x2 − 13x + c

SOLUTION:  In this trinomial, b = –13.

So, c must be  to make the trinomial a perfect square.

16. x2 − 22x + c

SOLUTION:  In this trinomial, b = –22.

So, c must be 121 to make the trinomial a perfect square.

17. x2 − 15x + c

SOLUTION:  In this trinomial, b = –15.

So, c must be  to make the trinomial a perfect square.

18. x2 + 24x + c

SOLUTION:  In this trinomial, b = 24.

So, c must be 144 to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

19. x2 + 6x − 16 = 0

SOLUTION:  

  The solutions are –8 and 2.

20. x2 − 2x − 14 = 0

SOLUTION:  

  The solutions are about –2.9 and 4.9.

21. x2 − 8x − 1 = 8

SOLUTION:  

  The solutions are –1 and 9.

22. x2 + 3x + 21 = 22

SOLUTION:  

  The solutions are about –3.3 and 0.3.

23. x2 − 11x + 3 = 5

SOLUTION:  

  The solutions are about –0.2 and 11.2.

24. 5x2 − 10x = 23

SOLUTION:  

  The solutions are about –1.4 and 3.4.

25. 2x2 − 2x + 7 = 5

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

26. 3x2 + 12x + 81 = 15

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

27. 4x2 + 6x = 12

SOLUTION:  

  The solutions are about –2.6 and 1.1.

28. 4x2 + 5 = 10x

SOLUTION:  

  The solutions are about 0.7 and 1.8.

29. −2x2 + 10x = −14

SOLUTION:  

  The solutions are about –1.1 and 6.1.

30. −3x2 − 12 = 14x

SOLUTION:  

  The solutions are about –3.5 and –1.1.

31. STOCK The price p in dollars for a particular stock can be modeled by the quadratic equation p = 3.5t − 0.05t2, wh

the number of days after the stock is purchased. When is the stock worth $60?

SOLUTION:  Let p = 60.

  The stock is worth $60 on the 30th and 40th days after purchase.

GEOMETRY  Find the value of x for each figure. Round to the nearest tenth if necessary.

32. area = 45 in2

SOLUTION:  

  The height cannot be negative. So, x ≈ 6.3.

33. area = 110 ft2

SOLUTION:  

  The dimensions of the rectangle cannot be negative. So, x ≈ 5.3.

34. NUMBER THEORY The product of two consecutive even integers is 224. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers are 14 and 16 or –16 and –14.

35. CCSS PRECISION The product of two consecutive negative odd integers is 483. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers must be negative. So, they are –23 and –21.

36. GEOMETRY Find the area of the triangle.

SOLUTION:  

  The dimensions of the triangle cannot be negative. So, x = 18. The base of the triangle is 18 meters and the height is 18 + 6 or 24 meters.

The area of the triangle is 216 square meters.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

37. 0.2x2 − 0.2x − 0.4 = 0

SOLUTION:  

  The solutions are –1 and 2.

38. 0.5x2 = 2x − 0.3

SOLUTION:  

  The solutions are about 0.2 and 3.8.

39. 2x2 −

SOLUTION:  

  The solutions are about 0.2 and 0.9.

40. 

SOLUTION:  

  The solutions are –0.5 and 2.5.

41. 

SOLUTION:  

  The solutions are about –8.2 and 0.2.

42. 

SOLUTION:  

  The solutions are about –5.1 and 0.1.

43. ASTRONOMY The height of an object t seconds after it is dropped is given by the equation ,

where h0 is the initial height and g is the acceleration due to gravity. The acceleration due to gravity near the surface

of Mars is 3.73 m/s2, while on Earth it is 9.8 m/s

2. Suppose an object is dropped from an initial height of 120 meters

above the surface of each planet. a. On which planet would the object reach the ground first? b. How long would it take the object to reach the ground on each planet? Round each answer to the nearest tenth. c. Do the times that it takes the object to reach the ground seem reasonable? Explain your reasoning.

SOLUTION:  a. The object on Earth will reach the ground first because it is falling at a faster rate.   b. Mars:

So, t ≈ 8.0 seconds. Earth:

So, t ≈ 4.9 seconds.   c. Sample answer: Yes; the acceleration due to gravity is much greater on Earth than on Mars, so the time to reach the ground should be much less.

44. Find all values of c that make x2 + cx + 100 a perfect square trinomial.

SOLUTION:  

So, c can be –20 or 20 to make the trinomial a perfect square.

45. Find all values of c that make x2 + cx + 225 a perfect square trinomial.

SOLUTION:  

So, c can be –30 or 30 to make the trinomial a perfect square.

46. PAINTING Before she begins painting a picture, Donna stretches her canvas over a wood frame. The frame has a length of 60 inches and a width of 4 inches. She has enough canvas to cover 480 square inches. Donna decides to increase the dimensions of the frame. If the increase in the length is 10 times the increase in the width, what will the dimensions of the frame be?

SOLUTION:  Let x = the increase in the width and let 10x = the increase in the length. So, x + 4 = the new width and 10x + 60 = the new length.  

  The dimensions cannot be negative, so x = 2. The width is 2 + 4 or 6 inches and the length is 10(2) + 60 or 80 inches.

47. MULTIPLE REPRESENTATIONS In this problem, you will investigate a property of quadratic equations. a. TABULAR  Copy the table shown and complete the second column.

b. ALGEBRAIC Set each trinomial equal to zero, and solve the equation by completing the square. Complete the last column of the table with the number of roots of each equation.

c. VERBAL Compare the number of roots of each equation to the result in the b2 − 4ac column. Is there a

relationship between these values? If so, describe it.

d. ANALYTICAL Predict how many solutions 2x2 − 9x + 15 = 0 will have.  Verify your prediction by solving the 

equation.

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3    

3x2 + 6x + 9    

x2 − 2x + 7    

x2 + 10x + 25    

x2 + 3x − 12    

SOLUTION:  a.

b.

The trinomial x2 − 8x + 16 has 1 root.

 

The trinomial 2x

2 − 11x + 3 has 2 roots.

 

The square root of a negative number has no real roots. So, the trinomial 3x2 + 6x + 9 has 0 roots.

 

The square root of a negative number has no real roots. So, the trinomial x2 − 2x + 7.

 

The trinomial x2 + 10x + 25 has 1 root.

 

The trinomial x2 + 3x − 12 has 2 roots.

 

c. If b2 − 4ac is negative, the equation has no real solutions. If b

2 − 4ac is zero, the equation has one solution. If b2

− 4ac is positive, the equation has 2 solutions. d.

The equation 2x2 − 9x + 15 = 0 has 0 real solutions because b

2 − 4ac is negative.

 

The equation cannot be solved because the square root of a negative number has no real roots.

Trinomial b2 − 4ac Number

of Roots

x2 − 8x + 16 (–8)

2 – 4(1)(16) = 0 1

2x2 − 11x + 3 (–11)

2 – 4(2)(3) = 97  

3x2 + 6x + 9 (6)

2 – 4(3)(9) = −72  

x2 − 2x + 7 (–2)

2 – 4(1)(7) = −24  

x2 + 10x + 25 (10)

2 – 4(1)(25) = 0  

x2 + 3x − 12 (3)

2 – 4(1)(–12) = 57  

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3 97 2

3x2 + 6x + 9 −72 0

x2 − 2x + 7 −24 0

x2 + 10x + 25 0 1

x2 + 3x − 12 57 2

48. CCSS PERSEVERANCE Given y = ax2 + bx + c with a ≠ 0, derive the equation for the axis of symmetry by 

completing the square and rewriting the equation in the form y = a(x − h)2 + k .

SOLUTION:  

Let    and   , then the equation becomes y = (x - h)2 + k .

For an equation in the form y = ax2 + bx + c,  the equation for the axis of symmetry is   .

So, for an equation in the form  y = (x - h)2 + k, the equation for the axis of symmetry becomes x = h.

49. REASONING Determine the number of solutions x2 + bx = c has if . Explain.

SOLUTION:  

None; Sample answer: If you add to each side of the equation and each side of the inequality, you get

. Since the left side of the last equation is a perfect square, it cannot

equal the negative number . So, there are no real solutions.

50. WHICH ONE DOESN’T BELONG? Identify the expression that does not belong with the other three. Explain your reasoning.

 

SOLUTION:  The first 3 trinomials are perfect squares.

The trinomial  is not a perfect square. So, it does not belong.

51. OPEN ENDED Write a quadratic equation for which the only solution is 4.

SOLUTION:  Find a quadratic that has two roots of 4.

Verify that the solution is 4.

So, the quadratic equation x2 − 8x + 16 = 0 has a solution of 4.

52. WRITING IN MATH Compare and contrast the following strategies for solving x2 − 5x − 7 = 0: completing the

square, graphing, and factoring.

SOLUTION:  Sample answer: Because the leading coefficient is 1, solving the equation by completing the square is simpler and yields exact answers for the solutions.   

 

To solve the equation by graphing, use a graphing calculator to graph the related function y = x2 - 5x - 7. Select the

zero option from the 2nd [CALC] menu to determine the roots. 

   

  The roots are given as decimal approximations. So, for this equation the solutions would have to be given as estimations.   There are no factors of –7 that have a sum of –5, so solving by factoring is not possible.

53. The length of a rectangle is 3 times its width. The area of the rectangle is 75 square feet. Find the length of the rectangle in feet. A 25 B 15 C 10 D 5

SOLUTION:  Let x = the width of the rectangle and let 3x = the length of the rectangle.

The length of the rectangle is 3(5) or 15 feet. Choice B is the correct answer.

54. PROBABILITY At a festival, winners of a game draw a token for a prize. There is one token for each prize. The prizes include 9 movie passes, 8 stuffed animals, 5 hats, 10 jump ropes, and 4 glow necklaces. What is the probabilitythat the first person to draw a token will win a movie pass?

F  

G

H

J  

SOLUTION:  There are 9 + 8 + 5 + 10 + 4 or 36 possible outcomes.

P(movie pass) =  or  . Choice J is the correct answer.

55. GRIDDED RESPONSE The population of a town can be modeled by P = 22,000 + 125t, where P represents the population and t represents the number of years from 2000. How many years after 2000 will the population be 26,000?

SOLUTION:  

In 32 years the population of the town will be 26,000.

56. Percy delivers pizzas for Pizza King. He is paid $6 an hour plus $2.50 for each pizza he delivers. Percy earned $280 last week. If he worked a total of 30 hours, how many pizzas did he deliver? A 250 pizzas B 184 pizzas C 40 pizzas D 34 pizzas

SOLUTION:  Let p = the number of pizzas Percy delivered.

Percy delivered 40 pizzas. Choice C is the correct answer.

Describe how the graph of each function is related to the graph of f (x) = x2.

57. g(x) = −12 + x2

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –12, the translation is

down. So, the graph is shifted down 12 units from the parent function. 

58. h(x) = (x + 2)2

SOLUTION:  

The graph of f (x) = (x – c)2 represents a translation left or right from the parent graph. Since c = –2, the translation

is to the left by 2 units. 

 

59. g(x) = 2x2 + 5

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 2 and c = 5. Since 2 > 0 and  > 1, the graph of y = 2x

2 + 5

is the graph of y = x2 vertically stretched and shifted up 5 units.

60. 

SOLUTION:  

The function can be written f (x) = a(x – b)2, where a = and b = 6. Since > 0 and  < 1, the graph of 

is the graph of y = x2 vertically compressed and shifted right 6 units.

61. g(x) = 6 + x2

SOLUTION:  

The function can be written f (x) = ax2 + c, where a =  and c = 6. Since  > 0 and   > 1, the graph of y = 6 +

x2 is the graph of y = x

2 vertically stretched and shifted up 6 units.

62. h(x) = − 1 −   x2

SOLUTION:  

The function can be written f (x) = –ax2 + c, where a =  and c = –1. Since  < 0 and   > 1, the graph of y

= –1 – x2 is the graph of y = x

2 vertically stretched, shifted down 1 unit and reflected across the x-axis.

63. RIDES A popular amusement park ride whisks riders to the top of a 250-foot tower and drops them. A function for

the height of a rider is h = −16t2 + 250, where h is the height and t is the time in seconds. The ride stops the descent

of the rider 40 feet above the ground. Write an equation that models the drop of the rider. How long does it take to fall from 250 feet to 40 feet?

SOLUTION:  

The ride stops the descent of the rider at 40 feet above ground, so h = 40. Then, the equation 40 = −16t2 + 250

models the drop of the rider.

Time cannot be negative. So, it takes about 3.6 seconds to complete the ride.

Simplify. Assume that no denominator is equal to zero.

64. 

SOLUTION:  

65. 

SOLUTION:  

66. 

SOLUTION:  

67. 

SOLUTION:  

68. 

SOLUTION:  

69. b3(m

−3)(b

−6)

SOLUTION:  

Solve each open sentence.

70. |y − 2| > 7

SOLUTION:  

The solution set is {y |y > 9 or y < −5}.

Case 1 y – 2 is positive.

and Case 2 y – 2 is negative.

 

71. |z + 5| < 3

SOLUTION:  

The solution set is {z |−8 < z < −2}.

Case 1 z +5 is positive.

and Case 2 z +5 is negative.

 

72. |2b + 7| ≤ −6

SOLUTION:  

cannot be negative. So,  cannot be less than or equal to –6. The solution set is empty.

73. |3 − 2y | ≥ 8

SOLUTION:  

The solution set is {y |y  ≥ 5.5 or y  ≤ −2.5}.

Case 1 3 – 2y is positive.

and Case 2 3 – 2y is negative.

 

74. |9 − 4m| < −1

SOLUTION:  

cannot be negative. So, cannot be less than or equal to –1. The solution set is empty.

75. |5c − 2| ≤ 13

SOLUTION:  

The solution set is {c|−2.2 ≤ c ≤ 3}.

Case 1 5c – 2 is positive.

and Case 2 5c – 2 is negative.

 

Evaluate  for each set of values. Round to the nearest tenth if necessary.

76. a = 2, b = −5, c = 2

SOLUTION:  

77. a = 1, b = 12, c = 11

SOLUTION:  

78. a = −9, b = 10, c = −1

SOLUTION:  

79. a = 1, b = 7, c = −3

SOLUTION:  

80. a = 2, b = −4, c = −6

SOLUTION:  

81. a = 3, b = 1, c = 2

SOLUTION:  

This value is not a real number.

eSolutions Manual - Powered by Cognero Page 8

9-4 Solving Quadratic Equations by Completing the Square

Find the value of c that makes each trinomial a perfect square.

1. x2 − 18x + c

SOLUTION:  In this trinomial, b = –18.

So, c must be 81 to make the trinomial a perfect square.

2. x2 + 22x + c

SOLUTION:  In this trinomial, b = 22.

So, c must be 121 to make the trinomial a perfect square.

3. x2 + 9x + c

SOLUTION:  In this trinomial, b = 9.

 

So, c must be  to make the trinomial a perfect square.

4. x2 − 7x + c

SOLUTION:  In this trinomial, b = –7.

 

So, c must be  to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

5. x2 + 4x = 6

SOLUTION:  

The solutions are about –5.2 and 1.2.

6. x2 − 8x = −9

SOLUTION:  

The solutions are about 1.4 and 6.6.

7. 4x2 + 9x − 1 = 0

SOLUTION:  

The solutions are about –2.4 and 0.1.

8. −2x2 + 10x + 22 = 4

SOLUTION:  

The solutions are about –1.4 and 6.4.

9. CCSS MODELING Collin is building a deck on the back of his family’s house. He has enough lumber for the deck to be 144 square feet. The length should be 10 feet more than its width. What should the dimensions of the deck be?

SOLUTION:  Let x = the width of the deck and x + 10 = the length of the deck.

The dimensions of the deck can not be negative. So, x = –5 + 13 or 8. The width of the deck is 8 feet and the length of the deck is 8 + 10 or 18 feet.

Find the value of c that makes each trinomial a perfect square.

10. x2 + 26x + c

SOLUTION:  In this trinomial, b = 26.

So, c must be 169 to make the trinomial a perfect square.

11. x2 − 24x + c

SOLUTION:  In this trinomial, b = –24.

So, c must be 144 to make the trinomial a perfect square.

12. x2 − 19x + c

SOLUTION:  In this trinomial, b = –19.

So, c must be  to make the trinomial a perfect square.

13. x2 + 17x + c

SOLUTION:  In this trinomial, b = 17.

So, c must be  to make the trinomial a perfect square.

14. x2 + 5x + c

SOLUTION:  In this trinomial, b = 5.

So, c must be  to make the trinomial a perfect square.

15. x2 − 13x + c

SOLUTION:  In this trinomial, b = –13.

So, c must be  to make the trinomial a perfect square.

16. x2 − 22x + c

SOLUTION:  In this trinomial, b = –22.

So, c must be 121 to make the trinomial a perfect square.

17. x2 − 15x + c

SOLUTION:  In this trinomial, b = –15.

So, c must be  to make the trinomial a perfect square.

18. x2 + 24x + c

SOLUTION:  In this trinomial, b = 24.

So, c must be 144 to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

19. x2 + 6x − 16 = 0

SOLUTION:  

  The solutions are –8 and 2.

20. x2 − 2x − 14 = 0

SOLUTION:  

  The solutions are about –2.9 and 4.9.

21. x2 − 8x − 1 = 8

SOLUTION:  

  The solutions are –1 and 9.

22. x2 + 3x + 21 = 22

SOLUTION:  

  The solutions are about –3.3 and 0.3.

23. x2 − 11x + 3 = 5

SOLUTION:  

  The solutions are about –0.2 and 11.2.

24. 5x2 − 10x = 23

SOLUTION:  

  The solutions are about –1.4 and 3.4.

25. 2x2 − 2x + 7 = 5

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

26. 3x2 + 12x + 81 = 15

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

27. 4x2 + 6x = 12

SOLUTION:  

  The solutions are about –2.6 and 1.1.

28. 4x2 + 5 = 10x

SOLUTION:  

  The solutions are about 0.7 and 1.8.

29. −2x2 + 10x = −14

SOLUTION:  

  The solutions are about –1.1 and 6.1.

30. −3x2 − 12 = 14x

SOLUTION:  

  The solutions are about –3.5 and –1.1.

31. STOCK The price p in dollars for a particular stock can be modeled by the quadratic equation p = 3.5t − 0.05t2, wh

the number of days after the stock is purchased. When is the stock worth $60?

SOLUTION:  Let p = 60.

  The stock is worth $60 on the 30th and 40th days after purchase.

GEOMETRY  Find the value of x for each figure. Round to the nearest tenth if necessary.

32. area = 45 in2

SOLUTION:  

  The height cannot be negative. So, x ≈ 6.3.

33. area = 110 ft2

SOLUTION:  

  The dimensions of the rectangle cannot be negative. So, x ≈ 5.3.

34. NUMBER THEORY The product of two consecutive even integers is 224. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers are 14 and 16 or –16 and –14.

35. CCSS PRECISION The product of two consecutive negative odd integers is 483. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers must be negative. So, they are –23 and –21.

36. GEOMETRY Find the area of the triangle.

SOLUTION:  

  The dimensions of the triangle cannot be negative. So, x = 18. The base of the triangle is 18 meters and the height is 18 + 6 or 24 meters.

The area of the triangle is 216 square meters.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

37. 0.2x2 − 0.2x − 0.4 = 0

SOLUTION:  

  The solutions are –1 and 2.

38. 0.5x2 = 2x − 0.3

SOLUTION:  

  The solutions are about 0.2 and 3.8.

39. 2x2 −

SOLUTION:  

  The solutions are about 0.2 and 0.9.

40. 

SOLUTION:  

  The solutions are –0.5 and 2.5.

41. 

SOLUTION:  

  The solutions are about –8.2 and 0.2.

42. 

SOLUTION:  

  The solutions are about –5.1 and 0.1.

43. ASTRONOMY The height of an object t seconds after it is dropped is given by the equation ,

where h0 is the initial height and g is the acceleration due to gravity. The acceleration due to gravity near the surface

of Mars is 3.73 m/s2, while on Earth it is 9.8 m/s

2. Suppose an object is dropped from an initial height of 120 meters

above the surface of each planet. a. On which planet would the object reach the ground first? b. How long would it take the object to reach the ground on each planet? Round each answer to the nearest tenth. c. Do the times that it takes the object to reach the ground seem reasonable? Explain your reasoning.

SOLUTION:  a. The object on Earth will reach the ground first because it is falling at a faster rate.   b. Mars:

So, t ≈ 8.0 seconds. Earth:

So, t ≈ 4.9 seconds.   c. Sample answer: Yes; the acceleration due to gravity is much greater on Earth than on Mars, so the time to reach the ground should be much less.

44. Find all values of c that make x2 + cx + 100 a perfect square trinomial.

SOLUTION:  

So, c can be –20 or 20 to make the trinomial a perfect square.

45. Find all values of c that make x2 + cx + 225 a perfect square trinomial.

SOLUTION:  

So, c can be –30 or 30 to make the trinomial a perfect square.

46. PAINTING Before she begins painting a picture, Donna stretches her canvas over a wood frame. The frame has a length of 60 inches and a width of 4 inches. She has enough canvas to cover 480 square inches. Donna decides to increase the dimensions of the frame. If the increase in the length is 10 times the increase in the width, what will the dimensions of the frame be?

SOLUTION:  Let x = the increase in the width and let 10x = the increase in the length. So, x + 4 = the new width and 10x + 60 = the new length.  

  The dimensions cannot be negative, so x = 2. The width is 2 + 4 or 6 inches and the length is 10(2) + 60 or 80 inches.

47. MULTIPLE REPRESENTATIONS In this problem, you will investigate a property of quadratic equations. a. TABULAR  Copy the table shown and complete the second column.

b. ALGEBRAIC Set each trinomial equal to zero, and solve the equation by completing the square. Complete the last column of the table with the number of roots of each equation.

c. VERBAL Compare the number of roots of each equation to the result in the b2 − 4ac column. Is there a

relationship between these values? If so, describe it.

d. ANALYTICAL Predict how many solutions 2x2 − 9x + 15 = 0 will have.  Verify your prediction by solving the 

equation.

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3    

3x2 + 6x + 9    

x2 − 2x + 7    

x2 + 10x + 25    

x2 + 3x − 12    

SOLUTION:  a.

b.

The trinomial x2 − 8x + 16 has 1 root.

 

The trinomial 2x

2 − 11x + 3 has 2 roots.

 

The square root of a negative number has no real roots. So, the trinomial 3x2 + 6x + 9 has 0 roots.

 

The square root of a negative number has no real roots. So, the trinomial x2 − 2x + 7.

 

The trinomial x2 + 10x + 25 has 1 root.

 

The trinomial x2 + 3x − 12 has 2 roots.

 

c. If b2 − 4ac is negative, the equation has no real solutions. If b

2 − 4ac is zero, the equation has one solution. If b2

− 4ac is positive, the equation has 2 solutions. d.

The equation 2x2 − 9x + 15 = 0 has 0 real solutions because b

2 − 4ac is negative.

 

The equation cannot be solved because the square root of a negative number has no real roots.

Trinomial b2 − 4ac Number

of Roots

x2 − 8x + 16 (–8)

2 – 4(1)(16) = 0 1

2x2 − 11x + 3 (–11)

2 – 4(2)(3) = 97  

3x2 + 6x + 9 (6)

2 – 4(3)(9) = −72  

x2 − 2x + 7 (–2)

2 – 4(1)(7) = −24  

x2 + 10x + 25 (10)

2 – 4(1)(25) = 0  

x2 + 3x − 12 (3)

2 – 4(1)(–12) = 57  

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3 97 2

3x2 + 6x + 9 −72 0

x2 − 2x + 7 −24 0

x2 + 10x + 25 0 1

x2 + 3x − 12 57 2

48. CCSS PERSEVERANCE Given y = ax2 + bx + c with a ≠ 0, derive the equation for the axis of symmetry by 

completing the square and rewriting the equation in the form y = a(x − h)2 + k .

SOLUTION:  

Let    and   , then the equation becomes y = (x - h)2 + k .

For an equation in the form y = ax2 + bx + c,  the equation for the axis of symmetry is   .

So, for an equation in the form  y = (x - h)2 + k, the equation for the axis of symmetry becomes x = h.

49. REASONING Determine the number of solutions x2 + bx = c has if . Explain.

SOLUTION:  

None; Sample answer: If you add to each side of the equation and each side of the inequality, you get

. Since the left side of the last equation is a perfect square, it cannot

equal the negative number . So, there are no real solutions.

50. WHICH ONE DOESN’T BELONG? Identify the expression that does not belong with the other three. Explain your reasoning.

 

SOLUTION:  The first 3 trinomials are perfect squares.

The trinomial  is not a perfect square. So, it does not belong.

51. OPEN ENDED Write a quadratic equation for which the only solution is 4.

SOLUTION:  Find a quadratic that has two roots of 4.

Verify that the solution is 4.

So, the quadratic equation x2 − 8x + 16 = 0 has a solution of 4.

52. WRITING IN MATH Compare and contrast the following strategies for solving x2 − 5x − 7 = 0: completing the

square, graphing, and factoring.

SOLUTION:  Sample answer: Because the leading coefficient is 1, solving the equation by completing the square is simpler and yields exact answers for the solutions.   

 

To solve the equation by graphing, use a graphing calculator to graph the related function y = x2 - 5x - 7. Select the

zero option from the 2nd [CALC] menu to determine the roots. 

   

  The roots are given as decimal approximations. So, for this equation the solutions would have to be given as estimations.   There are no factors of –7 that have a sum of –5, so solving by factoring is not possible.

53. The length of a rectangle is 3 times its width. The area of the rectangle is 75 square feet. Find the length of the rectangle in feet. A 25 B 15 C 10 D 5

SOLUTION:  Let x = the width of the rectangle and let 3x = the length of the rectangle.

The length of the rectangle is 3(5) or 15 feet. Choice B is the correct answer.

54. PROBABILITY At a festival, winners of a game draw a token for a prize. There is one token for each prize. The prizes include 9 movie passes, 8 stuffed animals, 5 hats, 10 jump ropes, and 4 glow necklaces. What is the probabilitythat the first person to draw a token will win a movie pass?

F  

G

H

J  

SOLUTION:  There are 9 + 8 + 5 + 10 + 4 or 36 possible outcomes.

P(movie pass) =  or  . Choice J is the correct answer.

55. GRIDDED RESPONSE The population of a town can be modeled by P = 22,000 + 125t, where P represents the population and t represents the number of years from 2000. How many years after 2000 will the population be 26,000?

SOLUTION:  

In 32 years the population of the town will be 26,000.

56. Percy delivers pizzas for Pizza King. He is paid $6 an hour plus $2.50 for each pizza he delivers. Percy earned $280 last week. If he worked a total of 30 hours, how many pizzas did he deliver? A 250 pizzas B 184 pizzas C 40 pizzas D 34 pizzas

SOLUTION:  Let p = the number of pizzas Percy delivered.

Percy delivered 40 pizzas. Choice C is the correct answer.

Describe how the graph of each function is related to the graph of f (x) = x2.

57. g(x) = −12 + x2

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –12, the translation is

down. So, the graph is shifted down 12 units from the parent function. 

58. h(x) = (x + 2)2

SOLUTION:  

The graph of f (x) = (x – c)2 represents a translation left or right from the parent graph. Since c = –2, the translation

is to the left by 2 units. 

 

59. g(x) = 2x2 + 5

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 2 and c = 5. Since 2 > 0 and  > 1, the graph of y = 2x

2 + 5

is the graph of y = x2 vertically stretched and shifted up 5 units.

60. 

SOLUTION:  

The function can be written f (x) = a(x – b)2, where a = and b = 6. Since > 0 and  < 1, the graph of 

is the graph of y = x2 vertically compressed and shifted right 6 units.

61. g(x) = 6 + x2

SOLUTION:  

The function can be written f (x) = ax2 + c, where a =  and c = 6. Since  > 0 and   > 1, the graph of y = 6 +

x2 is the graph of y = x

2 vertically stretched and shifted up 6 units.

62. h(x) = − 1 −   x2

SOLUTION:  

The function can be written f (x) = –ax2 + c, where a =  and c = –1. Since  < 0 and   > 1, the graph of y

= –1 – x2 is the graph of y = x

2 vertically stretched, shifted down 1 unit and reflected across the x-axis.

63. RIDES A popular amusement park ride whisks riders to the top of a 250-foot tower and drops them. A function for

the height of a rider is h = −16t2 + 250, where h is the height and t is the time in seconds. The ride stops the descent

of the rider 40 feet above the ground. Write an equation that models the drop of the rider. How long does it take to fall from 250 feet to 40 feet?

SOLUTION:  

The ride stops the descent of the rider at 40 feet above ground, so h = 40. Then, the equation 40 = −16t2 + 250

models the drop of the rider.

Time cannot be negative. So, it takes about 3.6 seconds to complete the ride.

Simplify. Assume that no denominator is equal to zero.

64. 

SOLUTION:  

65. 

SOLUTION:  

66. 

SOLUTION:  

67. 

SOLUTION:  

68. 

SOLUTION:  

69. b3(m

−3)(b

−6)

SOLUTION:  

Solve each open sentence.

70. |y − 2| > 7

SOLUTION:  

The solution set is {y |y > 9 or y < −5}.

Case 1 y – 2 is positive.

and Case 2 y – 2 is negative.

 

71. |z + 5| < 3

SOLUTION:  

The solution set is {z |−8 < z < −2}.

Case 1 z +5 is positive.

and Case 2 z +5 is negative.

 

72. |2b + 7| ≤ −6

SOLUTION:  

cannot be negative. So,  cannot be less than or equal to –6. The solution set is empty.

73. |3 − 2y | ≥ 8

SOLUTION:  

The solution set is {y |y  ≥ 5.5 or y  ≤ −2.5}.

Case 1 3 – 2y is positive.

and Case 2 3 – 2y is negative.

 

74. |9 − 4m| < −1

SOLUTION:  

cannot be negative. So, cannot be less than or equal to –1. The solution set is empty.

75. |5c − 2| ≤ 13

SOLUTION:  

The solution set is {c|−2.2 ≤ c ≤ 3}.

Case 1 5c – 2 is positive.

and Case 2 5c – 2 is negative.

 

Evaluate  for each set of values. Round to the nearest tenth if necessary.

76. a = 2, b = −5, c = 2

SOLUTION:  

77. a = 1, b = 12, c = 11

SOLUTION:  

78. a = −9, b = 10, c = −1

SOLUTION:  

79. a = 1, b = 7, c = −3

SOLUTION:  

80. a = 2, b = −4, c = −6

SOLUTION:  

81. a = 3, b = 1, c = 2

SOLUTION:  

This value is not a real number.

eSolutions Manual - Powered by Cognero Page 9

9-4 Solving Quadratic Equations by Completing the Square

Find the value of c that makes each trinomial a perfect square.

1. x2 − 18x + c

SOLUTION:  In this trinomial, b = –18.

So, c must be 81 to make the trinomial a perfect square.

2. x2 + 22x + c

SOLUTION:  In this trinomial, b = 22.

So, c must be 121 to make the trinomial a perfect square.

3. x2 + 9x + c

SOLUTION:  In this trinomial, b = 9.

 

So, c must be  to make the trinomial a perfect square.

4. x2 − 7x + c

SOLUTION:  In this trinomial, b = –7.

 

So, c must be  to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

5. x2 + 4x = 6

SOLUTION:  

The solutions are about –5.2 and 1.2.

6. x2 − 8x = −9

SOLUTION:  

The solutions are about 1.4 and 6.6.

7. 4x2 + 9x − 1 = 0

SOLUTION:  

The solutions are about –2.4 and 0.1.

8. −2x2 + 10x + 22 = 4

SOLUTION:  

The solutions are about –1.4 and 6.4.

9. CCSS MODELING Collin is building a deck on the back of his family’s house. He has enough lumber for the deck to be 144 square feet. The length should be 10 feet more than its width. What should the dimensions of the deck be?

SOLUTION:  Let x = the width of the deck and x + 10 = the length of the deck.

The dimensions of the deck can not be negative. So, x = –5 + 13 or 8. The width of the deck is 8 feet and the length of the deck is 8 + 10 or 18 feet.

Find the value of c that makes each trinomial a perfect square.

10. x2 + 26x + c

SOLUTION:  In this trinomial, b = 26.

So, c must be 169 to make the trinomial a perfect square.

11. x2 − 24x + c

SOLUTION:  In this trinomial, b = –24.

So, c must be 144 to make the trinomial a perfect square.

12. x2 − 19x + c

SOLUTION:  In this trinomial, b = –19.

So, c must be  to make the trinomial a perfect square.

13. x2 + 17x + c

SOLUTION:  In this trinomial, b = 17.

So, c must be  to make the trinomial a perfect square.

14. x2 + 5x + c

SOLUTION:  In this trinomial, b = 5.

So, c must be  to make the trinomial a perfect square.

15. x2 − 13x + c

SOLUTION:  In this trinomial, b = –13.

So, c must be  to make the trinomial a perfect square.

16. x2 − 22x + c

SOLUTION:  In this trinomial, b = –22.

So, c must be 121 to make the trinomial a perfect square.

17. x2 − 15x + c

SOLUTION:  In this trinomial, b = –15.

So, c must be  to make the trinomial a perfect square.

18. x2 + 24x + c

SOLUTION:  In this trinomial, b = 24.

So, c must be 144 to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

19. x2 + 6x − 16 = 0

SOLUTION:  

  The solutions are –8 and 2.

20. x2 − 2x − 14 = 0

SOLUTION:  

  The solutions are about –2.9 and 4.9.

21. x2 − 8x − 1 = 8

SOLUTION:  

  The solutions are –1 and 9.

22. x2 + 3x + 21 = 22

SOLUTION:  

  The solutions are about –3.3 and 0.3.

23. x2 − 11x + 3 = 5

SOLUTION:  

  The solutions are about –0.2 and 11.2.

24. 5x2 − 10x = 23

SOLUTION:  

  The solutions are about –1.4 and 3.4.

25. 2x2 − 2x + 7 = 5

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

26. 3x2 + 12x + 81 = 15

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

27. 4x2 + 6x = 12

SOLUTION:  

  The solutions are about –2.6 and 1.1.

28. 4x2 + 5 = 10x

SOLUTION:  

  The solutions are about 0.7 and 1.8.

29. −2x2 + 10x = −14

SOLUTION:  

  The solutions are about –1.1 and 6.1.

30. −3x2 − 12 = 14x

SOLUTION:  

  The solutions are about –3.5 and –1.1.

31. STOCK The price p in dollars for a particular stock can be modeled by the quadratic equation p = 3.5t − 0.05t2, wh

the number of days after the stock is purchased. When is the stock worth $60?

SOLUTION:  Let p = 60.

  The stock is worth $60 on the 30th and 40th days after purchase.

GEOMETRY  Find the value of x for each figure. Round to the nearest tenth if necessary.

32. area = 45 in2

SOLUTION:  

  The height cannot be negative. So, x ≈ 6.3.

33. area = 110 ft2

SOLUTION:  

  The dimensions of the rectangle cannot be negative. So, x ≈ 5.3.

34. NUMBER THEORY The product of two consecutive even integers is 224. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers are 14 and 16 or –16 and –14.

35. CCSS PRECISION The product of two consecutive negative odd integers is 483. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers must be negative. So, they are –23 and –21.

36. GEOMETRY Find the area of the triangle.

SOLUTION:  

  The dimensions of the triangle cannot be negative. So, x = 18. The base of the triangle is 18 meters and the height is 18 + 6 or 24 meters.

The area of the triangle is 216 square meters.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

37. 0.2x2 − 0.2x − 0.4 = 0

SOLUTION:  

  The solutions are –1 and 2.

38. 0.5x2 = 2x − 0.3

SOLUTION:  

  The solutions are about 0.2 and 3.8.

39. 2x2 −

SOLUTION:  

  The solutions are about 0.2 and 0.9.

40. 

SOLUTION:  

  The solutions are –0.5 and 2.5.

41. 

SOLUTION:  

  The solutions are about –8.2 and 0.2.

42. 

SOLUTION:  

  The solutions are about –5.1 and 0.1.

43. ASTRONOMY The height of an object t seconds after it is dropped is given by the equation ,

where h0 is the initial height and g is the acceleration due to gravity. The acceleration due to gravity near the surface

of Mars is 3.73 m/s2, while on Earth it is 9.8 m/s

2. Suppose an object is dropped from an initial height of 120 meters

above the surface of each planet. a. On which planet would the object reach the ground first? b. How long would it take the object to reach the ground on each planet? Round each answer to the nearest tenth. c. Do the times that it takes the object to reach the ground seem reasonable? Explain your reasoning.

SOLUTION:  a. The object on Earth will reach the ground first because it is falling at a faster rate.   b. Mars:

So, t ≈ 8.0 seconds. Earth:

So, t ≈ 4.9 seconds.   c. Sample answer: Yes; the acceleration due to gravity is much greater on Earth than on Mars, so the time to reach the ground should be much less.

44. Find all values of c that make x2 + cx + 100 a perfect square trinomial.

SOLUTION:  

So, c can be –20 or 20 to make the trinomial a perfect square.

45. Find all values of c that make x2 + cx + 225 a perfect square trinomial.

SOLUTION:  

So, c can be –30 or 30 to make the trinomial a perfect square.

46. PAINTING Before she begins painting a picture, Donna stretches her canvas over a wood frame. The frame has a length of 60 inches and a width of 4 inches. She has enough canvas to cover 480 square inches. Donna decides to increase the dimensions of the frame. If the increase in the length is 10 times the increase in the width, what will the dimensions of the frame be?

SOLUTION:  Let x = the increase in the width and let 10x = the increase in the length. So, x + 4 = the new width and 10x + 60 = the new length.  

  The dimensions cannot be negative, so x = 2. The width is 2 + 4 or 6 inches and the length is 10(2) + 60 or 80 inches.

47. MULTIPLE REPRESENTATIONS In this problem, you will investigate a property of quadratic equations. a. TABULAR  Copy the table shown and complete the second column.

b. ALGEBRAIC Set each trinomial equal to zero, and solve the equation by completing the square. Complete the last column of the table with the number of roots of each equation.

c. VERBAL Compare the number of roots of each equation to the result in the b2 − 4ac column. Is there a

relationship between these values? If so, describe it.

d. ANALYTICAL Predict how many solutions 2x2 − 9x + 15 = 0 will have.  Verify your prediction by solving the 

equation.

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3    

3x2 + 6x + 9    

x2 − 2x + 7    

x2 + 10x + 25    

x2 + 3x − 12    

SOLUTION:  a.

b.

The trinomial x2 − 8x + 16 has 1 root.

 

The trinomial 2x

2 − 11x + 3 has 2 roots.

 

The square root of a negative number has no real roots. So, the trinomial 3x2 + 6x + 9 has 0 roots.

 

The square root of a negative number has no real roots. So, the trinomial x2 − 2x + 7.

 

The trinomial x2 + 10x + 25 has 1 root.

 

The trinomial x2 + 3x − 12 has 2 roots.

 

c. If b2 − 4ac is negative, the equation has no real solutions. If b

2 − 4ac is zero, the equation has one solution. If b2

− 4ac is positive, the equation has 2 solutions. d.

The equation 2x2 − 9x + 15 = 0 has 0 real solutions because b

2 − 4ac is negative.

 

The equation cannot be solved because the square root of a negative number has no real roots.

Trinomial b2 − 4ac Number

of Roots

x2 − 8x + 16 (–8)

2 – 4(1)(16) = 0 1

2x2 − 11x + 3 (–11)

2 – 4(2)(3) = 97  

3x2 + 6x + 9 (6)

2 – 4(3)(9) = −72  

x2 − 2x + 7 (–2)

2 – 4(1)(7) = −24  

x2 + 10x + 25 (10)

2 – 4(1)(25) = 0  

x2 + 3x − 12 (3)

2 – 4(1)(–12) = 57  

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3 97 2

3x2 + 6x + 9 −72 0

x2 − 2x + 7 −24 0

x2 + 10x + 25 0 1

x2 + 3x − 12 57 2

48. CCSS PERSEVERANCE Given y = ax2 + bx + c with a ≠ 0, derive the equation for the axis of symmetry by 

completing the square and rewriting the equation in the form y = a(x − h)2 + k .

SOLUTION:  

Let    and   , then the equation becomes y = (x - h)2 + k .

For an equation in the form y = ax2 + bx + c,  the equation for the axis of symmetry is   .

So, for an equation in the form  y = (x - h)2 + k, the equation for the axis of symmetry becomes x = h.

49. REASONING Determine the number of solutions x2 + bx = c has if . Explain.

SOLUTION:  

None; Sample answer: If you add to each side of the equation and each side of the inequality, you get

. Since the left side of the last equation is a perfect square, it cannot

equal the negative number . So, there are no real solutions.

50. WHICH ONE DOESN’T BELONG? Identify the expression that does not belong with the other three. Explain your reasoning.

 

SOLUTION:  The first 3 trinomials are perfect squares.

The trinomial  is not a perfect square. So, it does not belong.

51. OPEN ENDED Write a quadratic equation for which the only solution is 4.

SOLUTION:  Find a quadratic that has two roots of 4.

Verify that the solution is 4.

So, the quadratic equation x2 − 8x + 16 = 0 has a solution of 4.

52. WRITING IN MATH Compare and contrast the following strategies for solving x2 − 5x − 7 = 0: completing the

square, graphing, and factoring.

SOLUTION:  Sample answer: Because the leading coefficient is 1, solving the equation by completing the square is simpler and yields exact answers for the solutions.   

 

To solve the equation by graphing, use a graphing calculator to graph the related function y = x2 - 5x - 7. Select the

zero option from the 2nd [CALC] menu to determine the roots. 

   

  The roots are given as decimal approximations. So, for this equation the solutions would have to be given as estimations.   There are no factors of –7 that have a sum of –5, so solving by factoring is not possible.

53. The length of a rectangle is 3 times its width. The area of the rectangle is 75 square feet. Find the length of the rectangle in feet. A 25 B 15 C 10 D 5

SOLUTION:  Let x = the width of the rectangle and let 3x = the length of the rectangle.

The length of the rectangle is 3(5) or 15 feet. Choice B is the correct answer.

54. PROBABILITY At a festival, winners of a game draw a token for a prize. There is one token for each prize. The prizes include 9 movie passes, 8 stuffed animals, 5 hats, 10 jump ropes, and 4 glow necklaces. What is the probabilitythat the first person to draw a token will win a movie pass?

F  

G

H

J  

SOLUTION:  There are 9 + 8 + 5 + 10 + 4 or 36 possible outcomes.

P(movie pass) =  or  . Choice J is the correct answer.

55. GRIDDED RESPONSE The population of a town can be modeled by P = 22,000 + 125t, where P represents the population and t represents the number of years from 2000. How many years after 2000 will the population be 26,000?

SOLUTION:  

In 32 years the population of the town will be 26,000.

56. Percy delivers pizzas for Pizza King. He is paid $6 an hour plus $2.50 for each pizza he delivers. Percy earned $280 last week. If he worked a total of 30 hours, how many pizzas did he deliver? A 250 pizzas B 184 pizzas C 40 pizzas D 34 pizzas

SOLUTION:  Let p = the number of pizzas Percy delivered.

Percy delivered 40 pizzas. Choice C is the correct answer.

Describe how the graph of each function is related to the graph of f (x) = x2.

57. g(x) = −12 + x2

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –12, the translation is

down. So, the graph is shifted down 12 units from the parent function. 

58. h(x) = (x + 2)2

SOLUTION:  

The graph of f (x) = (x – c)2 represents a translation left or right from the parent graph. Since c = –2, the translation

is to the left by 2 units. 

 

59. g(x) = 2x2 + 5

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 2 and c = 5. Since 2 > 0 and  > 1, the graph of y = 2x

2 + 5

is the graph of y = x2 vertically stretched and shifted up 5 units.

60. 

SOLUTION:  

The function can be written f (x) = a(x – b)2, where a = and b = 6. Since > 0 and  < 1, the graph of 

is the graph of y = x2 vertically compressed and shifted right 6 units.

61. g(x) = 6 + x2

SOLUTION:  

The function can be written f (x) = ax2 + c, where a =  and c = 6. Since  > 0 and   > 1, the graph of y = 6 +

x2 is the graph of y = x

2 vertically stretched and shifted up 6 units.

62. h(x) = − 1 −   x2

SOLUTION:  

The function can be written f (x) = –ax2 + c, where a =  and c = –1. Since  < 0 and   > 1, the graph of y

= –1 – x2 is the graph of y = x

2 vertically stretched, shifted down 1 unit and reflected across the x-axis.

63. RIDES A popular amusement park ride whisks riders to the top of a 250-foot tower and drops them. A function for

the height of a rider is h = −16t2 + 250, where h is the height and t is the time in seconds. The ride stops the descent

of the rider 40 feet above the ground. Write an equation that models the drop of the rider. How long does it take to fall from 250 feet to 40 feet?

SOLUTION:  

The ride stops the descent of the rider at 40 feet above ground, so h = 40. Then, the equation 40 = −16t2 + 250

models the drop of the rider.

Time cannot be negative. So, it takes about 3.6 seconds to complete the ride.

Simplify. Assume that no denominator is equal to zero.

64. 

SOLUTION:  

65. 

SOLUTION:  

66. 

SOLUTION:  

67. 

SOLUTION:  

68. 

SOLUTION:  

69. b3(m

−3)(b

−6)

SOLUTION:  

Solve each open sentence.

70. |y − 2| > 7

SOLUTION:  

The solution set is {y |y > 9 or y < −5}.

Case 1 y – 2 is positive.

and Case 2 y – 2 is negative.

 

71. |z + 5| < 3

SOLUTION:  

The solution set is {z |−8 < z < −2}.

Case 1 z +5 is positive.

and Case 2 z +5 is negative.

 

72. |2b + 7| ≤ −6

SOLUTION:  

cannot be negative. So,  cannot be less than or equal to –6. The solution set is empty.

73. |3 − 2y | ≥ 8

SOLUTION:  

The solution set is {y |y  ≥ 5.5 or y  ≤ −2.5}.

Case 1 3 – 2y is positive.

and Case 2 3 – 2y is negative.

 

74. |9 − 4m| < −1

SOLUTION:  

cannot be negative. So, cannot be less than or equal to –1. The solution set is empty.

75. |5c − 2| ≤ 13

SOLUTION:  

The solution set is {c|−2.2 ≤ c ≤ 3}.

Case 1 5c – 2 is positive.

and Case 2 5c – 2 is negative.

 

Evaluate  for each set of values. Round to the nearest tenth if necessary.

76. a = 2, b = −5, c = 2

SOLUTION:  

77. a = 1, b = 12, c = 11

SOLUTION:  

78. a = −9, b = 10, c = −1

SOLUTION:  

79. a = 1, b = 7, c = −3

SOLUTION:  

80. a = 2, b = −4, c = −6

SOLUTION:  

81. a = 3, b = 1, c = 2

SOLUTION:  

This value is not a real number.

eSolutions Manual - Powered by Cognero Page 10

9-4 Solving Quadratic Equations by Completing the Square

Find the value of c that makes each trinomial a perfect square.

1. x2 − 18x + c

SOLUTION:  In this trinomial, b = –18.

So, c must be 81 to make the trinomial a perfect square.

2. x2 + 22x + c

SOLUTION:  In this trinomial, b = 22.

So, c must be 121 to make the trinomial a perfect square.

3. x2 + 9x + c

SOLUTION:  In this trinomial, b = 9.

 

So, c must be  to make the trinomial a perfect square.

4. x2 − 7x + c

SOLUTION:  In this trinomial, b = –7.

 

So, c must be  to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

5. x2 + 4x = 6

SOLUTION:  

The solutions are about –5.2 and 1.2.

6. x2 − 8x = −9

SOLUTION:  

The solutions are about 1.4 and 6.6.

7. 4x2 + 9x − 1 = 0

SOLUTION:  

The solutions are about –2.4 and 0.1.

8. −2x2 + 10x + 22 = 4

SOLUTION:  

The solutions are about –1.4 and 6.4.

9. CCSS MODELING Collin is building a deck on the back of his family’s house. He has enough lumber for the deck to be 144 square feet. The length should be 10 feet more than its width. What should the dimensions of the deck be?

SOLUTION:  Let x = the width of the deck and x + 10 = the length of the deck.

The dimensions of the deck can not be negative. So, x = –5 + 13 or 8. The width of the deck is 8 feet and the length of the deck is 8 + 10 or 18 feet.

Find the value of c that makes each trinomial a perfect square.

10. x2 + 26x + c

SOLUTION:  In this trinomial, b = 26.

So, c must be 169 to make the trinomial a perfect square.

11. x2 − 24x + c

SOLUTION:  In this trinomial, b = –24.

So, c must be 144 to make the trinomial a perfect square.

12. x2 − 19x + c

SOLUTION:  In this trinomial, b = –19.

So, c must be  to make the trinomial a perfect square.

13. x2 + 17x + c

SOLUTION:  In this trinomial, b = 17.

So, c must be  to make the trinomial a perfect square.

14. x2 + 5x + c

SOLUTION:  In this trinomial, b = 5.

So, c must be  to make the trinomial a perfect square.

15. x2 − 13x + c

SOLUTION:  In this trinomial, b = –13.

So, c must be  to make the trinomial a perfect square.

16. x2 − 22x + c

SOLUTION:  In this trinomial, b = –22.

So, c must be 121 to make the trinomial a perfect square.

17. x2 − 15x + c

SOLUTION:  In this trinomial, b = –15.

So, c must be  to make the trinomial a perfect square.

18. x2 + 24x + c

SOLUTION:  In this trinomial, b = 24.

So, c must be 144 to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

19. x2 + 6x − 16 = 0

SOLUTION:  

  The solutions are –8 and 2.

20. x2 − 2x − 14 = 0

SOLUTION:  

  The solutions are about –2.9 and 4.9.

21. x2 − 8x − 1 = 8

SOLUTION:  

  The solutions are –1 and 9.

22. x2 + 3x + 21 = 22

SOLUTION:  

  The solutions are about –3.3 and 0.3.

23. x2 − 11x + 3 = 5

SOLUTION:  

  The solutions are about –0.2 and 11.2.

24. 5x2 − 10x = 23

SOLUTION:  

  The solutions are about –1.4 and 3.4.

25. 2x2 − 2x + 7 = 5

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

26. 3x2 + 12x + 81 = 15

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

27. 4x2 + 6x = 12

SOLUTION:  

  The solutions are about –2.6 and 1.1.

28. 4x2 + 5 = 10x

SOLUTION:  

  The solutions are about 0.7 and 1.8.

29. −2x2 + 10x = −14

SOLUTION:  

  The solutions are about –1.1 and 6.1.

30. −3x2 − 12 = 14x

SOLUTION:  

  The solutions are about –3.5 and –1.1.

31. STOCK The price p in dollars for a particular stock can be modeled by the quadratic equation p = 3.5t − 0.05t2, wh

the number of days after the stock is purchased. When is the stock worth $60?

SOLUTION:  Let p = 60.

  The stock is worth $60 on the 30th and 40th days after purchase.

GEOMETRY  Find the value of x for each figure. Round to the nearest tenth if necessary.

32. area = 45 in2

SOLUTION:  

  The height cannot be negative. So, x ≈ 6.3.

33. area = 110 ft2

SOLUTION:  

  The dimensions of the rectangle cannot be negative. So, x ≈ 5.3.

34. NUMBER THEORY The product of two consecutive even integers is 224. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers are 14 and 16 or –16 and –14.

35. CCSS PRECISION The product of two consecutive negative odd integers is 483. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers must be negative. So, they are –23 and –21.

36. GEOMETRY Find the area of the triangle.

SOLUTION:  

  The dimensions of the triangle cannot be negative. So, x = 18. The base of the triangle is 18 meters and the height is 18 + 6 or 24 meters.

The area of the triangle is 216 square meters.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

37. 0.2x2 − 0.2x − 0.4 = 0

SOLUTION:  

  The solutions are –1 and 2.

38. 0.5x2 = 2x − 0.3

SOLUTION:  

  The solutions are about 0.2 and 3.8.

39. 2x2 −

SOLUTION:  

  The solutions are about 0.2 and 0.9.

40. 

SOLUTION:  

  The solutions are –0.5 and 2.5.

41. 

SOLUTION:  

  The solutions are about –8.2 and 0.2.

42. 

SOLUTION:  

  The solutions are about –5.1 and 0.1.

43. ASTRONOMY The height of an object t seconds after it is dropped is given by the equation ,

where h0 is the initial height and g is the acceleration due to gravity. The acceleration due to gravity near the surface

of Mars is 3.73 m/s2, while on Earth it is 9.8 m/s

2. Suppose an object is dropped from an initial height of 120 meters

above the surface of each planet. a. On which planet would the object reach the ground first? b. How long would it take the object to reach the ground on each planet? Round each answer to the nearest tenth. c. Do the times that it takes the object to reach the ground seem reasonable? Explain your reasoning.

SOLUTION:  a. The object on Earth will reach the ground first because it is falling at a faster rate.   b. Mars:

So, t ≈ 8.0 seconds. Earth:

So, t ≈ 4.9 seconds.   c. Sample answer: Yes; the acceleration due to gravity is much greater on Earth than on Mars, so the time to reach the ground should be much less.

44. Find all values of c that make x2 + cx + 100 a perfect square trinomial.

SOLUTION:  

So, c can be –20 or 20 to make the trinomial a perfect square.

45. Find all values of c that make x2 + cx + 225 a perfect square trinomial.

SOLUTION:  

So, c can be –30 or 30 to make the trinomial a perfect square.

46. PAINTING Before she begins painting a picture, Donna stretches her canvas over a wood frame. The frame has a length of 60 inches and a width of 4 inches. She has enough canvas to cover 480 square inches. Donna decides to increase the dimensions of the frame. If the increase in the length is 10 times the increase in the width, what will the dimensions of the frame be?

SOLUTION:  Let x = the increase in the width and let 10x = the increase in the length. So, x + 4 = the new width and 10x + 60 = the new length.  

  The dimensions cannot be negative, so x = 2. The width is 2 + 4 or 6 inches and the length is 10(2) + 60 or 80 inches.

47. MULTIPLE REPRESENTATIONS In this problem, you will investigate a property of quadratic equations. a. TABULAR  Copy the table shown and complete the second column.

b. ALGEBRAIC Set each trinomial equal to zero, and solve the equation by completing the square. Complete the last column of the table with the number of roots of each equation.

c. VERBAL Compare the number of roots of each equation to the result in the b2 − 4ac column. Is there a

relationship between these values? If so, describe it.

d. ANALYTICAL Predict how many solutions 2x2 − 9x + 15 = 0 will have.  Verify your prediction by solving the 

equation.

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3    

3x2 + 6x + 9    

x2 − 2x + 7    

x2 + 10x + 25    

x2 + 3x − 12    

SOLUTION:  a.

b.

The trinomial x2 − 8x + 16 has 1 root.

 

The trinomial 2x

2 − 11x + 3 has 2 roots.

 

The square root of a negative number has no real roots. So, the trinomial 3x2 + 6x + 9 has 0 roots.

 

The square root of a negative number has no real roots. So, the trinomial x2 − 2x + 7.

 

The trinomial x2 + 10x + 25 has 1 root.

 

The trinomial x2 + 3x − 12 has 2 roots.

 

c. If b2 − 4ac is negative, the equation has no real solutions. If b

2 − 4ac is zero, the equation has one solution. If b2

− 4ac is positive, the equation has 2 solutions. d.

The equation 2x2 − 9x + 15 = 0 has 0 real solutions because b

2 − 4ac is negative.

 

The equation cannot be solved because the square root of a negative number has no real roots.

Trinomial b2 − 4ac Number

of Roots

x2 − 8x + 16 (–8)

2 – 4(1)(16) = 0 1

2x2 − 11x + 3 (–11)

2 – 4(2)(3) = 97  

3x2 + 6x + 9 (6)

2 – 4(3)(9) = −72  

x2 − 2x + 7 (–2)

2 – 4(1)(7) = −24  

x2 + 10x + 25 (10)

2 – 4(1)(25) = 0  

x2 + 3x − 12 (3)

2 – 4(1)(–12) = 57  

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3 97 2

3x2 + 6x + 9 −72 0

x2 − 2x + 7 −24 0

x2 + 10x + 25 0 1

x2 + 3x − 12 57 2

48. CCSS PERSEVERANCE Given y = ax2 + bx + c with a ≠ 0, derive the equation for the axis of symmetry by 

completing the square and rewriting the equation in the form y = a(x − h)2 + k .

SOLUTION:  

Let    and   , then the equation becomes y = (x - h)2 + k .

For an equation in the form y = ax2 + bx + c,  the equation for the axis of symmetry is   .

So, for an equation in the form  y = (x - h)2 + k, the equation for the axis of symmetry becomes x = h.

49. REASONING Determine the number of solutions x2 + bx = c has if . Explain.

SOLUTION:  

None; Sample answer: If you add to each side of the equation and each side of the inequality, you get

. Since the left side of the last equation is a perfect square, it cannot

equal the negative number . So, there are no real solutions.

50. WHICH ONE DOESN’T BELONG? Identify the expression that does not belong with the other three. Explain your reasoning.

 

SOLUTION:  The first 3 trinomials are perfect squares.

The trinomial  is not a perfect square. So, it does not belong.

51. OPEN ENDED Write a quadratic equation for which the only solution is 4.

SOLUTION:  Find a quadratic that has two roots of 4.

Verify that the solution is 4.

So, the quadratic equation x2 − 8x + 16 = 0 has a solution of 4.

52. WRITING IN MATH Compare and contrast the following strategies for solving x2 − 5x − 7 = 0: completing the

square, graphing, and factoring.

SOLUTION:  Sample answer: Because the leading coefficient is 1, solving the equation by completing the square is simpler and yields exact answers for the solutions.   

 

To solve the equation by graphing, use a graphing calculator to graph the related function y = x2 - 5x - 7. Select the

zero option from the 2nd [CALC] menu to determine the roots. 

   

  The roots are given as decimal approximations. So, for this equation the solutions would have to be given as estimations.   There are no factors of –7 that have a sum of –5, so solving by factoring is not possible.

53. The length of a rectangle is 3 times its width. The area of the rectangle is 75 square feet. Find the length of the rectangle in feet. A 25 B 15 C 10 D 5

SOLUTION:  Let x = the width of the rectangle and let 3x = the length of the rectangle.

The length of the rectangle is 3(5) or 15 feet. Choice B is the correct answer.

54. PROBABILITY At a festival, winners of a game draw a token for a prize. There is one token for each prize. The prizes include 9 movie passes, 8 stuffed animals, 5 hats, 10 jump ropes, and 4 glow necklaces. What is the probabilitythat the first person to draw a token will win a movie pass?

F  

G

H

J  

SOLUTION:  There are 9 + 8 + 5 + 10 + 4 or 36 possible outcomes.

P(movie pass) =  or  . Choice J is the correct answer.

55. GRIDDED RESPONSE The population of a town can be modeled by P = 22,000 + 125t, where P represents the population and t represents the number of years from 2000. How many years after 2000 will the population be 26,000?

SOLUTION:  

In 32 years the population of the town will be 26,000.

56. Percy delivers pizzas for Pizza King. He is paid $6 an hour plus $2.50 for each pizza he delivers. Percy earned $280 last week. If he worked a total of 30 hours, how many pizzas did he deliver? A 250 pizzas B 184 pizzas C 40 pizzas D 34 pizzas

SOLUTION:  Let p = the number of pizzas Percy delivered.

Percy delivered 40 pizzas. Choice C is the correct answer.

Describe how the graph of each function is related to the graph of f (x) = x2.

57. g(x) = −12 + x2

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –12, the translation is

down. So, the graph is shifted down 12 units from the parent function. 

58. h(x) = (x + 2)2

SOLUTION:  

The graph of f (x) = (x – c)2 represents a translation left or right from the parent graph. Since c = –2, the translation

is to the left by 2 units. 

 

59. g(x) = 2x2 + 5

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 2 and c = 5. Since 2 > 0 and  > 1, the graph of y = 2x

2 + 5

is the graph of y = x2 vertically stretched and shifted up 5 units.

60. 

SOLUTION:  

The function can be written f (x) = a(x – b)2, where a = and b = 6. Since > 0 and  < 1, the graph of 

is the graph of y = x2 vertically compressed and shifted right 6 units.

61. g(x) = 6 + x2

SOLUTION:  

The function can be written f (x) = ax2 + c, where a =  and c = 6. Since  > 0 and   > 1, the graph of y = 6 +

x2 is the graph of y = x

2 vertically stretched and shifted up 6 units.

62. h(x) = − 1 −   x2

SOLUTION:  

The function can be written f (x) = –ax2 + c, where a =  and c = –1. Since  < 0 and   > 1, the graph of y

= –1 – x2 is the graph of y = x

2 vertically stretched, shifted down 1 unit and reflected across the x-axis.

63. RIDES A popular amusement park ride whisks riders to the top of a 250-foot tower and drops them. A function for

the height of a rider is h = −16t2 + 250, where h is the height and t is the time in seconds. The ride stops the descent

of the rider 40 feet above the ground. Write an equation that models the drop of the rider. How long does it take to fall from 250 feet to 40 feet?

SOLUTION:  

The ride stops the descent of the rider at 40 feet above ground, so h = 40. Then, the equation 40 = −16t2 + 250

models the drop of the rider.

Time cannot be negative. So, it takes about 3.6 seconds to complete the ride.

Simplify. Assume that no denominator is equal to zero.

64. 

SOLUTION:  

65. 

SOLUTION:  

66. 

SOLUTION:  

67. 

SOLUTION:  

68. 

SOLUTION:  

69. b3(m

−3)(b

−6)

SOLUTION:  

Solve each open sentence.

70. |y − 2| > 7

SOLUTION:  

The solution set is {y |y > 9 or y < −5}.

Case 1 y – 2 is positive.

and Case 2 y – 2 is negative.

 

71. |z + 5| < 3

SOLUTION:  

The solution set is {z |−8 < z < −2}.

Case 1 z +5 is positive.

and Case 2 z +5 is negative.

 

72. |2b + 7| ≤ −6

SOLUTION:  

cannot be negative. So,  cannot be less than or equal to –6. The solution set is empty.

73. |3 − 2y | ≥ 8

SOLUTION:  

The solution set is {y |y  ≥ 5.5 or y  ≤ −2.5}.

Case 1 3 – 2y is positive.

and Case 2 3 – 2y is negative.

 

74. |9 − 4m| < −1

SOLUTION:  

cannot be negative. So, cannot be less than or equal to –1. The solution set is empty.

75. |5c − 2| ≤ 13

SOLUTION:  

The solution set is {c|−2.2 ≤ c ≤ 3}.

Case 1 5c – 2 is positive.

and Case 2 5c – 2 is negative.

 

Evaluate  for each set of values. Round to the nearest tenth if necessary.

76. a = 2, b = −5, c = 2

SOLUTION:  

77. a = 1, b = 12, c = 11

SOLUTION:  

78. a = −9, b = 10, c = −1

SOLUTION:  

79. a = 1, b = 7, c = −3

SOLUTION:  

80. a = 2, b = −4, c = −6

SOLUTION:  

81. a = 3, b = 1, c = 2

SOLUTION:  

This value is not a real number.

eSolutions Manual - Powered by Cognero Page 11

9-4 Solving Quadratic Equations by Completing the Square

Find the value of c that makes each trinomial a perfect square.

1. x2 − 18x + c

SOLUTION:  In this trinomial, b = –18.

So, c must be 81 to make the trinomial a perfect square.

2. x2 + 22x + c

SOLUTION:  In this trinomial, b = 22.

So, c must be 121 to make the trinomial a perfect square.

3. x2 + 9x + c

SOLUTION:  In this trinomial, b = 9.

 

So, c must be  to make the trinomial a perfect square.

4. x2 − 7x + c

SOLUTION:  In this trinomial, b = –7.

 

So, c must be  to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

5. x2 + 4x = 6

SOLUTION:  

The solutions are about –5.2 and 1.2.

6. x2 − 8x = −9

SOLUTION:  

The solutions are about 1.4 and 6.6.

7. 4x2 + 9x − 1 = 0

SOLUTION:  

The solutions are about –2.4 and 0.1.

8. −2x2 + 10x + 22 = 4

SOLUTION:  

The solutions are about –1.4 and 6.4.

9. CCSS MODELING Collin is building a deck on the back of his family’s house. He has enough lumber for the deck to be 144 square feet. The length should be 10 feet more than its width. What should the dimensions of the deck be?

SOLUTION:  Let x = the width of the deck and x + 10 = the length of the deck.

The dimensions of the deck can not be negative. So, x = –5 + 13 or 8. The width of the deck is 8 feet and the length of the deck is 8 + 10 or 18 feet.

Find the value of c that makes each trinomial a perfect square.

10. x2 + 26x + c

SOLUTION:  In this trinomial, b = 26.

So, c must be 169 to make the trinomial a perfect square.

11. x2 − 24x + c

SOLUTION:  In this trinomial, b = –24.

So, c must be 144 to make the trinomial a perfect square.

12. x2 − 19x + c

SOLUTION:  In this trinomial, b = –19.

So, c must be  to make the trinomial a perfect square.

13. x2 + 17x + c

SOLUTION:  In this trinomial, b = 17.

So, c must be  to make the trinomial a perfect square.

14. x2 + 5x + c

SOLUTION:  In this trinomial, b = 5.

So, c must be  to make the trinomial a perfect square.

15. x2 − 13x + c

SOLUTION:  In this trinomial, b = –13.

So, c must be  to make the trinomial a perfect square.

16. x2 − 22x + c

SOLUTION:  In this trinomial, b = –22.

So, c must be 121 to make the trinomial a perfect square.

17. x2 − 15x + c

SOLUTION:  In this trinomial, b = –15.

So, c must be  to make the trinomial a perfect square.

18. x2 + 24x + c

SOLUTION:  In this trinomial, b = 24.

So, c must be 144 to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

19. x2 + 6x − 16 = 0

SOLUTION:  

  The solutions are –8 and 2.

20. x2 − 2x − 14 = 0

SOLUTION:  

  The solutions are about –2.9 and 4.9.

21. x2 − 8x − 1 = 8

SOLUTION:  

  The solutions are –1 and 9.

22. x2 + 3x + 21 = 22

SOLUTION:  

  The solutions are about –3.3 and 0.3.

23. x2 − 11x + 3 = 5

SOLUTION:  

  The solutions are about –0.2 and 11.2.

24. 5x2 − 10x = 23

SOLUTION:  

  The solutions are about –1.4 and 3.4.

25. 2x2 − 2x + 7 = 5

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

26. 3x2 + 12x + 81 = 15

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

27. 4x2 + 6x = 12

SOLUTION:  

  The solutions are about –2.6 and 1.1.

28. 4x2 + 5 = 10x

SOLUTION:  

  The solutions are about 0.7 and 1.8.

29. −2x2 + 10x = −14

SOLUTION:  

  The solutions are about –1.1 and 6.1.

30. −3x2 − 12 = 14x

SOLUTION:  

  The solutions are about –3.5 and –1.1.

31. STOCK The price p in dollars for a particular stock can be modeled by the quadratic equation p = 3.5t − 0.05t2, wh

the number of days after the stock is purchased. When is the stock worth $60?

SOLUTION:  Let p = 60.

  The stock is worth $60 on the 30th and 40th days after purchase.

GEOMETRY  Find the value of x for each figure. Round to the nearest tenth if necessary.

32. area = 45 in2

SOLUTION:  

  The height cannot be negative. So, x ≈ 6.3.

33. area = 110 ft2

SOLUTION:  

  The dimensions of the rectangle cannot be negative. So, x ≈ 5.3.

34. NUMBER THEORY The product of two consecutive even integers is 224. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers are 14 and 16 or –16 and –14.

35. CCSS PRECISION The product of two consecutive negative odd integers is 483. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers must be negative. So, they are –23 and –21.

36. GEOMETRY Find the area of the triangle.

SOLUTION:  

  The dimensions of the triangle cannot be negative. So, x = 18. The base of the triangle is 18 meters and the height is 18 + 6 or 24 meters.

The area of the triangle is 216 square meters.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

37. 0.2x2 − 0.2x − 0.4 = 0

SOLUTION:  

  The solutions are –1 and 2.

38. 0.5x2 = 2x − 0.3

SOLUTION:  

  The solutions are about 0.2 and 3.8.

39. 2x2 −

SOLUTION:  

  The solutions are about 0.2 and 0.9.

40. 

SOLUTION:  

  The solutions are –0.5 and 2.5.

41. 

SOLUTION:  

  The solutions are about –8.2 and 0.2.

42. 

SOLUTION:  

  The solutions are about –5.1 and 0.1.

43. ASTRONOMY The height of an object t seconds after it is dropped is given by the equation ,

where h0 is the initial height and g is the acceleration due to gravity. The acceleration due to gravity near the surface

of Mars is 3.73 m/s2, while on Earth it is 9.8 m/s

2. Suppose an object is dropped from an initial height of 120 meters

above the surface of each planet. a. On which planet would the object reach the ground first? b. How long would it take the object to reach the ground on each planet? Round each answer to the nearest tenth. c. Do the times that it takes the object to reach the ground seem reasonable? Explain your reasoning.

SOLUTION:  a. The object on Earth will reach the ground first because it is falling at a faster rate.   b. Mars:

So, t ≈ 8.0 seconds. Earth:

So, t ≈ 4.9 seconds.   c. Sample answer: Yes; the acceleration due to gravity is much greater on Earth than on Mars, so the time to reach the ground should be much less.

44. Find all values of c that make x2 + cx + 100 a perfect square trinomial.

SOLUTION:  

So, c can be –20 or 20 to make the trinomial a perfect square.

45. Find all values of c that make x2 + cx + 225 a perfect square trinomial.

SOLUTION:  

So, c can be –30 or 30 to make the trinomial a perfect square.

46. PAINTING Before she begins painting a picture, Donna stretches her canvas over a wood frame. The frame has a length of 60 inches and a width of 4 inches. She has enough canvas to cover 480 square inches. Donna decides to increase the dimensions of the frame. If the increase in the length is 10 times the increase in the width, what will the dimensions of the frame be?

SOLUTION:  Let x = the increase in the width and let 10x = the increase in the length. So, x + 4 = the new width and 10x + 60 = the new length.  

  The dimensions cannot be negative, so x = 2. The width is 2 + 4 or 6 inches and the length is 10(2) + 60 or 80 inches.

47. MULTIPLE REPRESENTATIONS In this problem, you will investigate a property of quadratic equations. a. TABULAR  Copy the table shown and complete the second column.

b. ALGEBRAIC Set each trinomial equal to zero, and solve the equation by completing the square. Complete the last column of the table with the number of roots of each equation.

c. VERBAL Compare the number of roots of each equation to the result in the b2 − 4ac column. Is there a

relationship between these values? If so, describe it.

d. ANALYTICAL Predict how many solutions 2x2 − 9x + 15 = 0 will have.  Verify your prediction by solving the 

equation.

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3    

3x2 + 6x + 9    

x2 − 2x + 7    

x2 + 10x + 25    

x2 + 3x − 12    

SOLUTION:  a.

b.

The trinomial x2 − 8x + 16 has 1 root.

 

The trinomial 2x

2 − 11x + 3 has 2 roots.

 

The square root of a negative number has no real roots. So, the trinomial 3x2 + 6x + 9 has 0 roots.

 

The square root of a negative number has no real roots. So, the trinomial x2 − 2x + 7.

 

The trinomial x2 + 10x + 25 has 1 root.

 

The trinomial x2 + 3x − 12 has 2 roots.

 

c. If b2 − 4ac is negative, the equation has no real solutions. If b

2 − 4ac is zero, the equation has one solution. If b2

− 4ac is positive, the equation has 2 solutions. d.

The equation 2x2 − 9x + 15 = 0 has 0 real solutions because b

2 − 4ac is negative.

 

The equation cannot be solved because the square root of a negative number has no real roots.

Trinomial b2 − 4ac Number

of Roots

x2 − 8x + 16 (–8)

2 – 4(1)(16) = 0 1

2x2 − 11x + 3 (–11)

2 – 4(2)(3) = 97  

3x2 + 6x + 9 (6)

2 – 4(3)(9) = −72  

x2 − 2x + 7 (–2)

2 – 4(1)(7) = −24  

x2 + 10x + 25 (10)

2 – 4(1)(25) = 0  

x2 + 3x − 12 (3)

2 – 4(1)(–12) = 57  

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3 97 2

3x2 + 6x + 9 −72 0

x2 − 2x + 7 −24 0

x2 + 10x + 25 0 1

x2 + 3x − 12 57 2

48. CCSS PERSEVERANCE Given y = ax2 + bx + c with a ≠ 0, derive the equation for the axis of symmetry by 

completing the square and rewriting the equation in the form y = a(x − h)2 + k .

SOLUTION:  

Let    and   , then the equation becomes y = (x - h)2 + k .

For an equation in the form y = ax2 + bx + c,  the equation for the axis of symmetry is   .

So, for an equation in the form  y = (x - h)2 + k, the equation for the axis of symmetry becomes x = h.

49. REASONING Determine the number of solutions x2 + bx = c has if . Explain.

SOLUTION:  

None; Sample answer: If you add to each side of the equation and each side of the inequality, you get

. Since the left side of the last equation is a perfect square, it cannot

equal the negative number . So, there are no real solutions.

50. WHICH ONE DOESN’T BELONG? Identify the expression that does not belong with the other three. Explain your reasoning.

 

SOLUTION:  The first 3 trinomials are perfect squares.

The trinomial  is not a perfect square. So, it does not belong.

51. OPEN ENDED Write a quadratic equation for which the only solution is 4.

SOLUTION:  Find a quadratic that has two roots of 4.

Verify that the solution is 4.

So, the quadratic equation x2 − 8x + 16 = 0 has a solution of 4.

52. WRITING IN MATH Compare and contrast the following strategies for solving x2 − 5x − 7 = 0: completing the

square, graphing, and factoring.

SOLUTION:  Sample answer: Because the leading coefficient is 1, solving the equation by completing the square is simpler and yields exact answers for the solutions.   

 

To solve the equation by graphing, use a graphing calculator to graph the related function y = x2 - 5x - 7. Select the

zero option from the 2nd [CALC] menu to determine the roots. 

   

  The roots are given as decimal approximations. So, for this equation the solutions would have to be given as estimations.   There are no factors of –7 that have a sum of –5, so solving by factoring is not possible.

53. The length of a rectangle is 3 times its width. The area of the rectangle is 75 square feet. Find the length of the rectangle in feet. A 25 B 15 C 10 D 5

SOLUTION:  Let x = the width of the rectangle and let 3x = the length of the rectangle.

The length of the rectangle is 3(5) or 15 feet. Choice B is the correct answer.

54. PROBABILITY At a festival, winners of a game draw a token for a prize. There is one token for each prize. The prizes include 9 movie passes, 8 stuffed animals, 5 hats, 10 jump ropes, and 4 glow necklaces. What is the probabilitythat the first person to draw a token will win a movie pass?

F  

G

H

J  

SOLUTION:  There are 9 + 8 + 5 + 10 + 4 or 36 possible outcomes.

P(movie pass) =  or  . Choice J is the correct answer.

55. GRIDDED RESPONSE The population of a town can be modeled by P = 22,000 + 125t, where P represents the population and t represents the number of years from 2000. How many years after 2000 will the population be 26,000?

SOLUTION:  

In 32 years the population of the town will be 26,000.

56. Percy delivers pizzas for Pizza King. He is paid $6 an hour plus $2.50 for each pizza he delivers. Percy earned $280 last week. If he worked a total of 30 hours, how many pizzas did he deliver? A 250 pizzas B 184 pizzas C 40 pizzas D 34 pizzas

SOLUTION:  Let p = the number of pizzas Percy delivered.

Percy delivered 40 pizzas. Choice C is the correct answer.

Describe how the graph of each function is related to the graph of f (x) = x2.

57. g(x) = −12 + x2

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –12, the translation is

down. So, the graph is shifted down 12 units from the parent function. 

58. h(x) = (x + 2)2

SOLUTION:  

The graph of f (x) = (x – c)2 represents a translation left or right from the parent graph. Since c = –2, the translation

is to the left by 2 units. 

 

59. g(x) = 2x2 + 5

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 2 and c = 5. Since 2 > 0 and  > 1, the graph of y = 2x

2 + 5

is the graph of y = x2 vertically stretched and shifted up 5 units.

60. 

SOLUTION:  

The function can be written f (x) = a(x – b)2, where a = and b = 6. Since > 0 and  < 1, the graph of 

is the graph of y = x2 vertically compressed and shifted right 6 units.

61. g(x) = 6 + x2

SOLUTION:  

The function can be written f (x) = ax2 + c, where a =  and c = 6. Since  > 0 and   > 1, the graph of y = 6 +

x2 is the graph of y = x

2 vertically stretched and shifted up 6 units.

62. h(x) = − 1 −   x2

SOLUTION:  

The function can be written f (x) = –ax2 + c, where a =  and c = –1. Since  < 0 and   > 1, the graph of y

= –1 – x2 is the graph of y = x

2 vertically stretched, shifted down 1 unit and reflected across the x-axis.

63. RIDES A popular amusement park ride whisks riders to the top of a 250-foot tower and drops them. A function for

the height of a rider is h = −16t2 + 250, where h is the height and t is the time in seconds. The ride stops the descent

of the rider 40 feet above the ground. Write an equation that models the drop of the rider. How long does it take to fall from 250 feet to 40 feet?

SOLUTION:  

The ride stops the descent of the rider at 40 feet above ground, so h = 40. Then, the equation 40 = −16t2 + 250

models the drop of the rider.

Time cannot be negative. So, it takes about 3.6 seconds to complete the ride.

Simplify. Assume that no denominator is equal to zero.

64. 

SOLUTION:  

65. 

SOLUTION:  

66. 

SOLUTION:  

67. 

SOLUTION:  

68. 

SOLUTION:  

69. b3(m

−3)(b

−6)

SOLUTION:  

Solve each open sentence.

70. |y − 2| > 7

SOLUTION:  

The solution set is {y |y > 9 or y < −5}.

Case 1 y – 2 is positive.

and Case 2 y – 2 is negative.

 

71. |z + 5| < 3

SOLUTION:  

The solution set is {z |−8 < z < −2}.

Case 1 z +5 is positive.

and Case 2 z +5 is negative.

 

72. |2b + 7| ≤ −6

SOLUTION:  

cannot be negative. So,  cannot be less than or equal to –6. The solution set is empty.

73. |3 − 2y | ≥ 8

SOLUTION:  

The solution set is {y |y  ≥ 5.5 or y  ≤ −2.5}.

Case 1 3 – 2y is positive.

and Case 2 3 – 2y is negative.

 

74. |9 − 4m| < −1

SOLUTION:  

cannot be negative. So, cannot be less than or equal to –1. The solution set is empty.

75. |5c − 2| ≤ 13

SOLUTION:  

The solution set is {c|−2.2 ≤ c ≤ 3}.

Case 1 5c – 2 is positive.

and Case 2 5c – 2 is negative.

 

Evaluate  for each set of values. Round to the nearest tenth if necessary.

76. a = 2, b = −5, c = 2

SOLUTION:  

77. a = 1, b = 12, c = 11

SOLUTION:  

78. a = −9, b = 10, c = −1

SOLUTION:  

79. a = 1, b = 7, c = −3

SOLUTION:  

80. a = 2, b = −4, c = −6

SOLUTION:  

81. a = 3, b = 1, c = 2

SOLUTION:  

This value is not a real number.

eSolutions Manual - Powered by Cognero Page 12

9-4 Solving Quadratic Equations by Completing the Square

Find the value of c that makes each trinomial a perfect square.

1. x2 − 18x + c

SOLUTION:  In this trinomial, b = –18.

So, c must be 81 to make the trinomial a perfect square.

2. x2 + 22x + c

SOLUTION:  In this trinomial, b = 22.

So, c must be 121 to make the trinomial a perfect square.

3. x2 + 9x + c

SOLUTION:  In this trinomial, b = 9.

 

So, c must be  to make the trinomial a perfect square.

4. x2 − 7x + c

SOLUTION:  In this trinomial, b = –7.

 

So, c must be  to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

5. x2 + 4x = 6

SOLUTION:  

The solutions are about –5.2 and 1.2.

6. x2 − 8x = −9

SOLUTION:  

The solutions are about 1.4 and 6.6.

7. 4x2 + 9x − 1 = 0

SOLUTION:  

The solutions are about –2.4 and 0.1.

8. −2x2 + 10x + 22 = 4

SOLUTION:  

The solutions are about –1.4 and 6.4.

9. CCSS MODELING Collin is building a deck on the back of his family’s house. He has enough lumber for the deck to be 144 square feet. The length should be 10 feet more than its width. What should the dimensions of the deck be?

SOLUTION:  Let x = the width of the deck and x + 10 = the length of the deck.

The dimensions of the deck can not be negative. So, x = –5 + 13 or 8. The width of the deck is 8 feet and the length of the deck is 8 + 10 or 18 feet.

Find the value of c that makes each trinomial a perfect square.

10. x2 + 26x + c

SOLUTION:  In this trinomial, b = 26.

So, c must be 169 to make the trinomial a perfect square.

11. x2 − 24x + c

SOLUTION:  In this trinomial, b = –24.

So, c must be 144 to make the trinomial a perfect square.

12. x2 − 19x + c

SOLUTION:  In this trinomial, b = –19.

So, c must be  to make the trinomial a perfect square.

13. x2 + 17x + c

SOLUTION:  In this trinomial, b = 17.

So, c must be  to make the trinomial a perfect square.

14. x2 + 5x + c

SOLUTION:  In this trinomial, b = 5.

So, c must be  to make the trinomial a perfect square.

15. x2 − 13x + c

SOLUTION:  In this trinomial, b = –13.

So, c must be  to make the trinomial a perfect square.

16. x2 − 22x + c

SOLUTION:  In this trinomial, b = –22.

So, c must be 121 to make the trinomial a perfect square.

17. x2 − 15x + c

SOLUTION:  In this trinomial, b = –15.

So, c must be  to make the trinomial a perfect square.

18. x2 + 24x + c

SOLUTION:  In this trinomial, b = 24.

So, c must be 144 to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

19. x2 + 6x − 16 = 0

SOLUTION:  

  The solutions are –8 and 2.

20. x2 − 2x − 14 = 0

SOLUTION:  

  The solutions are about –2.9 and 4.9.

21. x2 − 8x − 1 = 8

SOLUTION:  

  The solutions are –1 and 9.

22. x2 + 3x + 21 = 22

SOLUTION:  

  The solutions are about –3.3 and 0.3.

23. x2 − 11x + 3 = 5

SOLUTION:  

  The solutions are about –0.2 and 11.2.

24. 5x2 − 10x = 23

SOLUTION:  

  The solutions are about –1.4 and 3.4.

25. 2x2 − 2x + 7 = 5

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

26. 3x2 + 12x + 81 = 15

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

27. 4x2 + 6x = 12

SOLUTION:  

  The solutions are about –2.6 and 1.1.

28. 4x2 + 5 = 10x

SOLUTION:  

  The solutions are about 0.7 and 1.8.

29. −2x2 + 10x = −14

SOLUTION:  

  The solutions are about –1.1 and 6.1.

30. −3x2 − 12 = 14x

SOLUTION:  

  The solutions are about –3.5 and –1.1.

31. STOCK The price p in dollars for a particular stock can be modeled by the quadratic equation p = 3.5t − 0.05t2, wh

the number of days after the stock is purchased. When is the stock worth $60?

SOLUTION:  Let p = 60.

  The stock is worth $60 on the 30th and 40th days after purchase.

GEOMETRY  Find the value of x for each figure. Round to the nearest tenth if necessary.

32. area = 45 in2

SOLUTION:  

  The height cannot be negative. So, x ≈ 6.3.

33. area = 110 ft2

SOLUTION:  

  The dimensions of the rectangle cannot be negative. So, x ≈ 5.3.

34. NUMBER THEORY The product of two consecutive even integers is 224. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers are 14 and 16 or –16 and –14.

35. CCSS PRECISION The product of two consecutive negative odd integers is 483. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers must be negative. So, they are –23 and –21.

36. GEOMETRY Find the area of the triangle.

SOLUTION:  

  The dimensions of the triangle cannot be negative. So, x = 18. The base of the triangle is 18 meters and the height is 18 + 6 or 24 meters.

The area of the triangle is 216 square meters.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

37. 0.2x2 − 0.2x − 0.4 = 0

SOLUTION:  

  The solutions are –1 and 2.

38. 0.5x2 = 2x − 0.3

SOLUTION:  

  The solutions are about 0.2 and 3.8.

39. 2x2 −

SOLUTION:  

  The solutions are about 0.2 and 0.9.

40. 

SOLUTION:  

  The solutions are –0.5 and 2.5.

41. 

SOLUTION:  

  The solutions are about –8.2 and 0.2.

42. 

SOLUTION:  

  The solutions are about –5.1 and 0.1.

43. ASTRONOMY The height of an object t seconds after it is dropped is given by the equation ,

where h0 is the initial height and g is the acceleration due to gravity. The acceleration due to gravity near the surface

of Mars is 3.73 m/s2, while on Earth it is 9.8 m/s

2. Suppose an object is dropped from an initial height of 120 meters

above the surface of each planet. a. On which planet would the object reach the ground first? b. How long would it take the object to reach the ground on each planet? Round each answer to the nearest tenth. c. Do the times that it takes the object to reach the ground seem reasonable? Explain your reasoning.

SOLUTION:  a. The object on Earth will reach the ground first because it is falling at a faster rate.   b. Mars:

So, t ≈ 8.0 seconds. Earth:

So, t ≈ 4.9 seconds.   c. Sample answer: Yes; the acceleration due to gravity is much greater on Earth than on Mars, so the time to reach the ground should be much less.

44. Find all values of c that make x2 + cx + 100 a perfect square trinomial.

SOLUTION:  

So, c can be –20 or 20 to make the trinomial a perfect square.

45. Find all values of c that make x2 + cx + 225 a perfect square trinomial.

SOLUTION:  

So, c can be –30 or 30 to make the trinomial a perfect square.

46. PAINTING Before she begins painting a picture, Donna stretches her canvas over a wood frame. The frame has a length of 60 inches and a width of 4 inches. She has enough canvas to cover 480 square inches. Donna decides to increase the dimensions of the frame. If the increase in the length is 10 times the increase in the width, what will the dimensions of the frame be?

SOLUTION:  Let x = the increase in the width and let 10x = the increase in the length. So, x + 4 = the new width and 10x + 60 = the new length.  

  The dimensions cannot be negative, so x = 2. The width is 2 + 4 or 6 inches and the length is 10(2) + 60 or 80 inches.

47. MULTIPLE REPRESENTATIONS In this problem, you will investigate a property of quadratic equations. a. TABULAR  Copy the table shown and complete the second column.

b. ALGEBRAIC Set each trinomial equal to zero, and solve the equation by completing the square. Complete the last column of the table with the number of roots of each equation.

c. VERBAL Compare the number of roots of each equation to the result in the b2 − 4ac column. Is there a

relationship between these values? If so, describe it.

d. ANALYTICAL Predict how many solutions 2x2 − 9x + 15 = 0 will have.  Verify your prediction by solving the 

equation.

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3    

3x2 + 6x + 9    

x2 − 2x + 7    

x2 + 10x + 25    

x2 + 3x − 12    

SOLUTION:  a.

b.

The trinomial x2 − 8x + 16 has 1 root.

 

The trinomial 2x

2 − 11x + 3 has 2 roots.

 

The square root of a negative number has no real roots. So, the trinomial 3x2 + 6x + 9 has 0 roots.

 

The square root of a negative number has no real roots. So, the trinomial x2 − 2x + 7.

 

The trinomial x2 + 10x + 25 has 1 root.

 

The trinomial x2 + 3x − 12 has 2 roots.

 

c. If b2 − 4ac is negative, the equation has no real solutions. If b

2 − 4ac is zero, the equation has one solution. If b2

− 4ac is positive, the equation has 2 solutions. d.

The equation 2x2 − 9x + 15 = 0 has 0 real solutions because b

2 − 4ac is negative.

 

The equation cannot be solved because the square root of a negative number has no real roots.

Trinomial b2 − 4ac Number

of Roots

x2 − 8x + 16 (–8)

2 – 4(1)(16) = 0 1

2x2 − 11x + 3 (–11)

2 – 4(2)(3) = 97  

3x2 + 6x + 9 (6)

2 – 4(3)(9) = −72  

x2 − 2x + 7 (–2)

2 – 4(1)(7) = −24  

x2 + 10x + 25 (10)

2 – 4(1)(25) = 0  

x2 + 3x − 12 (3)

2 – 4(1)(–12) = 57  

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3 97 2

3x2 + 6x + 9 −72 0

x2 − 2x + 7 −24 0

x2 + 10x + 25 0 1

x2 + 3x − 12 57 2

48. CCSS PERSEVERANCE Given y = ax2 + bx + c with a ≠ 0, derive the equation for the axis of symmetry by 

completing the square and rewriting the equation in the form y = a(x − h)2 + k .

SOLUTION:  

Let    and   , then the equation becomes y = (x - h)2 + k .

For an equation in the form y = ax2 + bx + c,  the equation for the axis of symmetry is   .

So, for an equation in the form  y = (x - h)2 + k, the equation for the axis of symmetry becomes x = h.

49. REASONING Determine the number of solutions x2 + bx = c has if . Explain.

SOLUTION:  

None; Sample answer: If you add to each side of the equation and each side of the inequality, you get

. Since the left side of the last equation is a perfect square, it cannot

equal the negative number . So, there are no real solutions.

50. WHICH ONE DOESN’T BELONG? Identify the expression that does not belong with the other three. Explain your reasoning.

 

SOLUTION:  The first 3 trinomials are perfect squares.

The trinomial  is not a perfect square. So, it does not belong.

51. OPEN ENDED Write a quadratic equation for which the only solution is 4.

SOLUTION:  Find a quadratic that has two roots of 4.

Verify that the solution is 4.

So, the quadratic equation x2 − 8x + 16 = 0 has a solution of 4.

52. WRITING IN MATH Compare and contrast the following strategies for solving x2 − 5x − 7 = 0: completing the

square, graphing, and factoring.

SOLUTION:  Sample answer: Because the leading coefficient is 1, solving the equation by completing the square is simpler and yields exact answers for the solutions.   

 

To solve the equation by graphing, use a graphing calculator to graph the related function y = x2 - 5x - 7. Select the

zero option from the 2nd [CALC] menu to determine the roots. 

   

  The roots are given as decimal approximations. So, for this equation the solutions would have to be given as estimations.   There are no factors of –7 that have a sum of –5, so solving by factoring is not possible.

53. The length of a rectangle is 3 times its width. The area of the rectangle is 75 square feet. Find the length of the rectangle in feet. A 25 B 15 C 10 D 5

SOLUTION:  Let x = the width of the rectangle and let 3x = the length of the rectangle.

The length of the rectangle is 3(5) or 15 feet. Choice B is the correct answer.

54. PROBABILITY At a festival, winners of a game draw a token for a prize. There is one token for each prize. The prizes include 9 movie passes, 8 stuffed animals, 5 hats, 10 jump ropes, and 4 glow necklaces. What is the probabilitythat the first person to draw a token will win a movie pass?

F  

G

H

J  

SOLUTION:  There are 9 + 8 + 5 + 10 + 4 or 36 possible outcomes.

P(movie pass) =  or  . Choice J is the correct answer.

55. GRIDDED RESPONSE The population of a town can be modeled by P = 22,000 + 125t, where P represents the population and t represents the number of years from 2000. How many years after 2000 will the population be 26,000?

SOLUTION:  

In 32 years the population of the town will be 26,000.

56. Percy delivers pizzas for Pizza King. He is paid $6 an hour plus $2.50 for each pizza he delivers. Percy earned $280 last week. If he worked a total of 30 hours, how many pizzas did he deliver? A 250 pizzas B 184 pizzas C 40 pizzas D 34 pizzas

SOLUTION:  Let p = the number of pizzas Percy delivered.

Percy delivered 40 pizzas. Choice C is the correct answer.

Describe how the graph of each function is related to the graph of f (x) = x2.

57. g(x) = −12 + x2

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –12, the translation is

down. So, the graph is shifted down 12 units from the parent function. 

58. h(x) = (x + 2)2

SOLUTION:  

The graph of f (x) = (x – c)2 represents a translation left or right from the parent graph. Since c = –2, the translation

is to the left by 2 units. 

 

59. g(x) = 2x2 + 5

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 2 and c = 5. Since 2 > 0 and  > 1, the graph of y = 2x

2 + 5

is the graph of y = x2 vertically stretched and shifted up 5 units.

60. 

SOLUTION:  

The function can be written f (x) = a(x – b)2, where a = and b = 6. Since > 0 and  < 1, the graph of 

is the graph of y = x2 vertically compressed and shifted right 6 units.

61. g(x) = 6 + x2

SOLUTION:  

The function can be written f (x) = ax2 + c, where a =  and c = 6. Since  > 0 and   > 1, the graph of y = 6 +

x2 is the graph of y = x

2 vertically stretched and shifted up 6 units.

62. h(x) = − 1 −   x2

SOLUTION:  

The function can be written f (x) = –ax2 + c, where a =  and c = –1. Since  < 0 and   > 1, the graph of y

= –1 – x2 is the graph of y = x

2 vertically stretched, shifted down 1 unit and reflected across the x-axis.

63. RIDES A popular amusement park ride whisks riders to the top of a 250-foot tower and drops them. A function for

the height of a rider is h = −16t2 + 250, where h is the height and t is the time in seconds. The ride stops the descent

of the rider 40 feet above the ground. Write an equation that models the drop of the rider. How long does it take to fall from 250 feet to 40 feet?

SOLUTION:  

The ride stops the descent of the rider at 40 feet above ground, so h = 40. Then, the equation 40 = −16t2 + 250

models the drop of the rider.

Time cannot be negative. So, it takes about 3.6 seconds to complete the ride.

Simplify. Assume that no denominator is equal to zero.

64. 

SOLUTION:  

65. 

SOLUTION:  

66. 

SOLUTION:  

67. 

SOLUTION:  

68. 

SOLUTION:  

69. b3(m

−3)(b

−6)

SOLUTION:  

Solve each open sentence.

70. |y − 2| > 7

SOLUTION:  

The solution set is {y |y > 9 or y < −5}.

Case 1 y – 2 is positive.

and Case 2 y – 2 is negative.

 

71. |z + 5| < 3

SOLUTION:  

The solution set is {z |−8 < z < −2}.

Case 1 z +5 is positive.

and Case 2 z +5 is negative.

 

72. |2b + 7| ≤ −6

SOLUTION:  

cannot be negative. So,  cannot be less than or equal to –6. The solution set is empty.

73. |3 − 2y | ≥ 8

SOLUTION:  

The solution set is {y |y  ≥ 5.5 or y  ≤ −2.5}.

Case 1 3 – 2y is positive.

and Case 2 3 – 2y is negative.

 

74. |9 − 4m| < −1

SOLUTION:  

cannot be negative. So, cannot be less than or equal to –1. The solution set is empty.

75. |5c − 2| ≤ 13

SOLUTION:  

The solution set is {c|−2.2 ≤ c ≤ 3}.

Case 1 5c – 2 is positive.

and Case 2 5c – 2 is negative.

 

Evaluate  for each set of values. Round to the nearest tenth if necessary.

76. a = 2, b = −5, c = 2

SOLUTION:  

77. a = 1, b = 12, c = 11

SOLUTION:  

78. a = −9, b = 10, c = −1

SOLUTION:  

79. a = 1, b = 7, c = −3

SOLUTION:  

80. a = 2, b = −4, c = −6

SOLUTION:  

81. a = 3, b = 1, c = 2

SOLUTION:  

This value is not a real number.

eSolutions Manual - Powered by Cognero Page 13

9-4 Solving Quadratic Equations by Completing the Square

Find the value of c that makes each trinomial a perfect square.

1. x2 − 18x + c

SOLUTION:  In this trinomial, b = –18.

So, c must be 81 to make the trinomial a perfect square.

2. x2 + 22x + c

SOLUTION:  In this trinomial, b = 22.

So, c must be 121 to make the trinomial a perfect square.

3. x2 + 9x + c

SOLUTION:  In this trinomial, b = 9.

 

So, c must be  to make the trinomial a perfect square.

4. x2 − 7x + c

SOLUTION:  In this trinomial, b = –7.

 

So, c must be  to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

5. x2 + 4x = 6

SOLUTION:  

The solutions are about –5.2 and 1.2.

6. x2 − 8x = −9

SOLUTION:  

The solutions are about 1.4 and 6.6.

7. 4x2 + 9x − 1 = 0

SOLUTION:  

The solutions are about –2.4 and 0.1.

8. −2x2 + 10x + 22 = 4

SOLUTION:  

The solutions are about –1.4 and 6.4.

9. CCSS MODELING Collin is building a deck on the back of his family’s house. He has enough lumber for the deck to be 144 square feet. The length should be 10 feet more than its width. What should the dimensions of the deck be?

SOLUTION:  Let x = the width of the deck and x + 10 = the length of the deck.

The dimensions of the deck can not be negative. So, x = –5 + 13 or 8. The width of the deck is 8 feet and the length of the deck is 8 + 10 or 18 feet.

Find the value of c that makes each trinomial a perfect square.

10. x2 + 26x + c

SOLUTION:  In this trinomial, b = 26.

So, c must be 169 to make the trinomial a perfect square.

11. x2 − 24x + c

SOLUTION:  In this trinomial, b = –24.

So, c must be 144 to make the trinomial a perfect square.

12. x2 − 19x + c

SOLUTION:  In this trinomial, b = –19.

So, c must be  to make the trinomial a perfect square.

13. x2 + 17x + c

SOLUTION:  In this trinomial, b = 17.

So, c must be  to make the trinomial a perfect square.

14. x2 + 5x + c

SOLUTION:  In this trinomial, b = 5.

So, c must be  to make the trinomial a perfect square.

15. x2 − 13x + c

SOLUTION:  In this trinomial, b = –13.

So, c must be  to make the trinomial a perfect square.

16. x2 − 22x + c

SOLUTION:  In this trinomial, b = –22.

So, c must be 121 to make the trinomial a perfect square.

17. x2 − 15x + c

SOLUTION:  In this trinomial, b = –15.

So, c must be  to make the trinomial a perfect square.

18. x2 + 24x + c

SOLUTION:  In this trinomial, b = 24.

So, c must be 144 to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

19. x2 + 6x − 16 = 0

SOLUTION:  

  The solutions are –8 and 2.

20. x2 − 2x − 14 = 0

SOLUTION:  

  The solutions are about –2.9 and 4.9.

21. x2 − 8x − 1 = 8

SOLUTION:  

  The solutions are –1 and 9.

22. x2 + 3x + 21 = 22

SOLUTION:  

  The solutions are about –3.3 and 0.3.

23. x2 − 11x + 3 = 5

SOLUTION:  

  The solutions are about –0.2 and 11.2.

24. 5x2 − 10x = 23

SOLUTION:  

  The solutions are about –1.4 and 3.4.

25. 2x2 − 2x + 7 = 5

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

26. 3x2 + 12x + 81 = 15

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

27. 4x2 + 6x = 12

SOLUTION:  

  The solutions are about –2.6 and 1.1.

28. 4x2 + 5 = 10x

SOLUTION:  

  The solutions are about 0.7 and 1.8.

29. −2x2 + 10x = −14

SOLUTION:  

  The solutions are about –1.1 and 6.1.

30. −3x2 − 12 = 14x

SOLUTION:  

  The solutions are about –3.5 and –1.1.

31. STOCK The price p in dollars for a particular stock can be modeled by the quadratic equation p = 3.5t − 0.05t2, wh

the number of days after the stock is purchased. When is the stock worth $60?

SOLUTION:  Let p = 60.

  The stock is worth $60 on the 30th and 40th days after purchase.

GEOMETRY  Find the value of x for each figure. Round to the nearest tenth if necessary.

32. area = 45 in2

SOLUTION:  

  The height cannot be negative. So, x ≈ 6.3.

33. area = 110 ft2

SOLUTION:  

  The dimensions of the rectangle cannot be negative. So, x ≈ 5.3.

34. NUMBER THEORY The product of two consecutive even integers is 224. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers are 14 and 16 or –16 and –14.

35. CCSS PRECISION The product of two consecutive negative odd integers is 483. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers must be negative. So, they are –23 and –21.

36. GEOMETRY Find the area of the triangle.

SOLUTION:  

  The dimensions of the triangle cannot be negative. So, x = 18. The base of the triangle is 18 meters and the height is 18 + 6 or 24 meters.

The area of the triangle is 216 square meters.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

37. 0.2x2 − 0.2x − 0.4 = 0

SOLUTION:  

  The solutions are –1 and 2.

38. 0.5x2 = 2x − 0.3

SOLUTION:  

  The solutions are about 0.2 and 3.8.

39. 2x2 −

SOLUTION:  

  The solutions are about 0.2 and 0.9.

40. 

SOLUTION:  

  The solutions are –0.5 and 2.5.

41. 

SOLUTION:  

  The solutions are about –8.2 and 0.2.

42. 

SOLUTION:  

  The solutions are about –5.1 and 0.1.

43. ASTRONOMY The height of an object t seconds after it is dropped is given by the equation ,

where h0 is the initial height and g is the acceleration due to gravity. The acceleration due to gravity near the surface

of Mars is 3.73 m/s2, while on Earth it is 9.8 m/s

2. Suppose an object is dropped from an initial height of 120 meters

above the surface of each planet. a. On which planet would the object reach the ground first? b. How long would it take the object to reach the ground on each planet? Round each answer to the nearest tenth. c. Do the times that it takes the object to reach the ground seem reasonable? Explain your reasoning.

SOLUTION:  a. The object on Earth will reach the ground first because it is falling at a faster rate.   b. Mars:

So, t ≈ 8.0 seconds. Earth:

So, t ≈ 4.9 seconds.   c. Sample answer: Yes; the acceleration due to gravity is much greater on Earth than on Mars, so the time to reach the ground should be much less.

44. Find all values of c that make x2 + cx + 100 a perfect square trinomial.

SOLUTION:  

So, c can be –20 or 20 to make the trinomial a perfect square.

45. Find all values of c that make x2 + cx + 225 a perfect square trinomial.

SOLUTION:  

So, c can be –30 or 30 to make the trinomial a perfect square.

46. PAINTING Before she begins painting a picture, Donna stretches her canvas over a wood frame. The frame has a length of 60 inches and a width of 4 inches. She has enough canvas to cover 480 square inches. Donna decides to increase the dimensions of the frame. If the increase in the length is 10 times the increase in the width, what will the dimensions of the frame be?

SOLUTION:  Let x = the increase in the width and let 10x = the increase in the length. So, x + 4 = the new width and 10x + 60 = the new length.  

  The dimensions cannot be negative, so x = 2. The width is 2 + 4 or 6 inches and the length is 10(2) + 60 or 80 inches.

47. MULTIPLE REPRESENTATIONS In this problem, you will investigate a property of quadratic equations. a. TABULAR  Copy the table shown and complete the second column.

b. ALGEBRAIC Set each trinomial equal to zero, and solve the equation by completing the square. Complete the last column of the table with the number of roots of each equation.

c. VERBAL Compare the number of roots of each equation to the result in the b2 − 4ac column. Is there a

relationship between these values? If so, describe it.

d. ANALYTICAL Predict how many solutions 2x2 − 9x + 15 = 0 will have.  Verify your prediction by solving the 

equation.

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3    

3x2 + 6x + 9    

x2 − 2x + 7    

x2 + 10x + 25    

x2 + 3x − 12    

SOLUTION:  a.

b.

The trinomial x2 − 8x + 16 has 1 root.

 

The trinomial 2x

2 − 11x + 3 has 2 roots.

 

The square root of a negative number has no real roots. So, the trinomial 3x2 + 6x + 9 has 0 roots.

 

The square root of a negative number has no real roots. So, the trinomial x2 − 2x + 7.

 

The trinomial x2 + 10x + 25 has 1 root.

 

The trinomial x2 + 3x − 12 has 2 roots.

 

c. If b2 − 4ac is negative, the equation has no real solutions. If b

2 − 4ac is zero, the equation has one solution. If b2

− 4ac is positive, the equation has 2 solutions. d.

The equation 2x2 − 9x + 15 = 0 has 0 real solutions because b

2 − 4ac is negative.

 

The equation cannot be solved because the square root of a negative number has no real roots.

Trinomial b2 − 4ac Number

of Roots

x2 − 8x + 16 (–8)

2 – 4(1)(16) = 0 1

2x2 − 11x + 3 (–11)

2 – 4(2)(3) = 97  

3x2 + 6x + 9 (6)

2 – 4(3)(9) = −72  

x2 − 2x + 7 (–2)

2 – 4(1)(7) = −24  

x2 + 10x + 25 (10)

2 – 4(1)(25) = 0  

x2 + 3x − 12 (3)

2 – 4(1)(–12) = 57  

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3 97 2

3x2 + 6x + 9 −72 0

x2 − 2x + 7 −24 0

x2 + 10x + 25 0 1

x2 + 3x − 12 57 2

48. CCSS PERSEVERANCE Given y = ax2 + bx + c with a ≠ 0, derive the equation for the axis of symmetry by 

completing the square and rewriting the equation in the form y = a(x − h)2 + k .

SOLUTION:  

Let    and   , then the equation becomes y = (x - h)2 + k .

For an equation in the form y = ax2 + bx + c,  the equation for the axis of symmetry is   .

So, for an equation in the form  y = (x - h)2 + k, the equation for the axis of symmetry becomes x = h.

49. REASONING Determine the number of solutions x2 + bx = c has if . Explain.

SOLUTION:  

None; Sample answer: If you add to each side of the equation and each side of the inequality, you get

. Since the left side of the last equation is a perfect square, it cannot

equal the negative number . So, there are no real solutions.

50. WHICH ONE DOESN’T BELONG? Identify the expression that does not belong with the other three. Explain your reasoning.

 

SOLUTION:  The first 3 trinomials are perfect squares.

The trinomial  is not a perfect square. So, it does not belong.

51. OPEN ENDED Write a quadratic equation for which the only solution is 4.

SOLUTION:  Find a quadratic that has two roots of 4.

Verify that the solution is 4.

So, the quadratic equation x2 − 8x + 16 = 0 has a solution of 4.

52. WRITING IN MATH Compare and contrast the following strategies for solving x2 − 5x − 7 = 0: completing the

square, graphing, and factoring.

SOLUTION:  Sample answer: Because the leading coefficient is 1, solving the equation by completing the square is simpler and yields exact answers for the solutions.   

 

To solve the equation by graphing, use a graphing calculator to graph the related function y = x2 - 5x - 7. Select the

zero option from the 2nd [CALC] menu to determine the roots. 

   

  The roots are given as decimal approximations. So, for this equation the solutions would have to be given as estimations.   There are no factors of –7 that have a sum of –5, so solving by factoring is not possible.

53. The length of a rectangle is 3 times its width. The area of the rectangle is 75 square feet. Find the length of the rectangle in feet. A 25 B 15 C 10 D 5

SOLUTION:  Let x = the width of the rectangle and let 3x = the length of the rectangle.

The length of the rectangle is 3(5) or 15 feet. Choice B is the correct answer.

54. PROBABILITY At a festival, winners of a game draw a token for a prize. There is one token for each prize. The prizes include 9 movie passes, 8 stuffed animals, 5 hats, 10 jump ropes, and 4 glow necklaces. What is the probabilitythat the first person to draw a token will win a movie pass?

F  

G

H

J  

SOLUTION:  There are 9 + 8 + 5 + 10 + 4 or 36 possible outcomes.

P(movie pass) =  or  . Choice J is the correct answer.

55. GRIDDED RESPONSE The population of a town can be modeled by P = 22,000 + 125t, where P represents the population and t represents the number of years from 2000. How many years after 2000 will the population be 26,000?

SOLUTION:  

In 32 years the population of the town will be 26,000.

56. Percy delivers pizzas for Pizza King. He is paid $6 an hour plus $2.50 for each pizza he delivers. Percy earned $280 last week. If he worked a total of 30 hours, how many pizzas did he deliver? A 250 pizzas B 184 pizzas C 40 pizzas D 34 pizzas

SOLUTION:  Let p = the number of pizzas Percy delivered.

Percy delivered 40 pizzas. Choice C is the correct answer.

Describe how the graph of each function is related to the graph of f (x) = x2.

57. g(x) = −12 + x2

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –12, the translation is

down. So, the graph is shifted down 12 units from the parent function. 

58. h(x) = (x + 2)2

SOLUTION:  

The graph of f (x) = (x – c)2 represents a translation left or right from the parent graph. Since c = –2, the translation

is to the left by 2 units. 

 

59. g(x) = 2x2 + 5

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 2 and c = 5. Since 2 > 0 and  > 1, the graph of y = 2x

2 + 5

is the graph of y = x2 vertically stretched and shifted up 5 units.

60. 

SOLUTION:  

The function can be written f (x) = a(x – b)2, where a = and b = 6. Since > 0 and  < 1, the graph of 

is the graph of y = x2 vertically compressed and shifted right 6 units.

61. g(x) = 6 + x2

SOLUTION:  

The function can be written f (x) = ax2 + c, where a =  and c = 6. Since  > 0 and   > 1, the graph of y = 6 +

x2 is the graph of y = x

2 vertically stretched and shifted up 6 units.

62. h(x) = − 1 −   x2

SOLUTION:  

The function can be written f (x) = –ax2 + c, where a =  and c = –1. Since  < 0 and   > 1, the graph of y

= –1 – x2 is the graph of y = x

2 vertically stretched, shifted down 1 unit and reflected across the x-axis.

63. RIDES A popular amusement park ride whisks riders to the top of a 250-foot tower and drops them. A function for

the height of a rider is h = −16t2 + 250, where h is the height and t is the time in seconds. The ride stops the descent

of the rider 40 feet above the ground. Write an equation that models the drop of the rider. How long does it take to fall from 250 feet to 40 feet?

SOLUTION:  

The ride stops the descent of the rider at 40 feet above ground, so h = 40. Then, the equation 40 = −16t2 + 250

models the drop of the rider.

Time cannot be negative. So, it takes about 3.6 seconds to complete the ride.

Simplify. Assume that no denominator is equal to zero.

64. 

SOLUTION:  

65. 

SOLUTION:  

66. 

SOLUTION:  

67. 

SOLUTION:  

68. 

SOLUTION:  

69. b3(m

−3)(b

−6)

SOLUTION:  

Solve each open sentence.

70. |y − 2| > 7

SOLUTION:  

The solution set is {y |y > 9 or y < −5}.

Case 1 y – 2 is positive.

and Case 2 y – 2 is negative.

 

71. |z + 5| < 3

SOLUTION:  

The solution set is {z |−8 < z < −2}.

Case 1 z +5 is positive.

and Case 2 z +5 is negative.

 

72. |2b + 7| ≤ −6

SOLUTION:  

cannot be negative. So,  cannot be less than or equal to –6. The solution set is empty.

73. |3 − 2y | ≥ 8

SOLUTION:  

The solution set is {y |y  ≥ 5.5 or y  ≤ −2.5}.

Case 1 3 – 2y is positive.

and Case 2 3 – 2y is negative.

 

74. |9 − 4m| < −1

SOLUTION:  

cannot be negative. So, cannot be less than or equal to –1. The solution set is empty.

75. |5c − 2| ≤ 13

SOLUTION:  

The solution set is {c|−2.2 ≤ c ≤ 3}.

Case 1 5c – 2 is positive.

and Case 2 5c – 2 is negative.

 

Evaluate  for each set of values. Round to the nearest tenth if necessary.

76. a = 2, b = −5, c = 2

SOLUTION:  

77. a = 1, b = 12, c = 11

SOLUTION:  

78. a = −9, b = 10, c = −1

SOLUTION:  

79. a = 1, b = 7, c = −3

SOLUTION:  

80. a = 2, b = −4, c = −6

SOLUTION:  

81. a = 3, b = 1, c = 2

SOLUTION:  

This value is not a real number.

eSolutions Manual - Powered by Cognero Page 14

9-4 Solving Quadratic Equations by Completing the Square

Find the value of c that makes each trinomial a perfect square.

1. x2 − 18x + c

SOLUTION:  In this trinomial, b = –18.

So, c must be 81 to make the trinomial a perfect square.

2. x2 + 22x + c

SOLUTION:  In this trinomial, b = 22.

So, c must be 121 to make the trinomial a perfect square.

3. x2 + 9x + c

SOLUTION:  In this trinomial, b = 9.

 

So, c must be  to make the trinomial a perfect square.

4. x2 − 7x + c

SOLUTION:  In this trinomial, b = –7.

 

So, c must be  to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

5. x2 + 4x = 6

SOLUTION:  

The solutions are about –5.2 and 1.2.

6. x2 − 8x = −9

SOLUTION:  

The solutions are about 1.4 and 6.6.

7. 4x2 + 9x − 1 = 0

SOLUTION:  

The solutions are about –2.4 and 0.1.

8. −2x2 + 10x + 22 = 4

SOLUTION:  

The solutions are about –1.4 and 6.4.

9. CCSS MODELING Collin is building a deck on the back of his family’s house. He has enough lumber for the deck to be 144 square feet. The length should be 10 feet more than its width. What should the dimensions of the deck be?

SOLUTION:  Let x = the width of the deck and x + 10 = the length of the deck.

The dimensions of the deck can not be negative. So, x = –5 + 13 or 8. The width of the deck is 8 feet and the length of the deck is 8 + 10 or 18 feet.

Find the value of c that makes each trinomial a perfect square.

10. x2 + 26x + c

SOLUTION:  In this trinomial, b = 26.

So, c must be 169 to make the trinomial a perfect square.

11. x2 − 24x + c

SOLUTION:  In this trinomial, b = –24.

So, c must be 144 to make the trinomial a perfect square.

12. x2 − 19x + c

SOLUTION:  In this trinomial, b = –19.

So, c must be  to make the trinomial a perfect square.

13. x2 + 17x + c

SOLUTION:  In this trinomial, b = 17.

So, c must be  to make the trinomial a perfect square.

14. x2 + 5x + c

SOLUTION:  In this trinomial, b = 5.

So, c must be  to make the trinomial a perfect square.

15. x2 − 13x + c

SOLUTION:  In this trinomial, b = –13.

So, c must be  to make the trinomial a perfect square.

16. x2 − 22x + c

SOLUTION:  In this trinomial, b = –22.

So, c must be 121 to make the trinomial a perfect square.

17. x2 − 15x + c

SOLUTION:  In this trinomial, b = –15.

So, c must be  to make the trinomial a perfect square.

18. x2 + 24x + c

SOLUTION:  In this trinomial, b = 24.

So, c must be 144 to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

19. x2 + 6x − 16 = 0

SOLUTION:  

  The solutions are –8 and 2.

20. x2 − 2x − 14 = 0

SOLUTION:  

  The solutions are about –2.9 and 4.9.

21. x2 − 8x − 1 = 8

SOLUTION:  

  The solutions are –1 and 9.

22. x2 + 3x + 21 = 22

SOLUTION:  

  The solutions are about –3.3 and 0.3.

23. x2 − 11x + 3 = 5

SOLUTION:  

  The solutions are about –0.2 and 11.2.

24. 5x2 − 10x = 23

SOLUTION:  

  The solutions are about –1.4 and 3.4.

25. 2x2 − 2x + 7 = 5

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

26. 3x2 + 12x + 81 = 15

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

27. 4x2 + 6x = 12

SOLUTION:  

  The solutions are about –2.6 and 1.1.

28. 4x2 + 5 = 10x

SOLUTION:  

  The solutions are about 0.7 and 1.8.

29. −2x2 + 10x = −14

SOLUTION:  

  The solutions are about –1.1 and 6.1.

30. −3x2 − 12 = 14x

SOLUTION:  

  The solutions are about –3.5 and –1.1.

31. STOCK The price p in dollars for a particular stock can be modeled by the quadratic equation p = 3.5t − 0.05t2, wh

the number of days after the stock is purchased. When is the stock worth $60?

SOLUTION:  Let p = 60.

  The stock is worth $60 on the 30th and 40th days after purchase.

GEOMETRY  Find the value of x for each figure. Round to the nearest tenth if necessary.

32. area = 45 in2

SOLUTION:  

  The height cannot be negative. So, x ≈ 6.3.

33. area = 110 ft2

SOLUTION:  

  The dimensions of the rectangle cannot be negative. So, x ≈ 5.3.

34. NUMBER THEORY The product of two consecutive even integers is 224. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers are 14 and 16 or –16 and –14.

35. CCSS PRECISION The product of two consecutive negative odd integers is 483. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers must be negative. So, they are –23 and –21.

36. GEOMETRY Find the area of the triangle.

SOLUTION:  

  The dimensions of the triangle cannot be negative. So, x = 18. The base of the triangle is 18 meters and the height is 18 + 6 or 24 meters.

The area of the triangle is 216 square meters.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

37. 0.2x2 − 0.2x − 0.4 = 0

SOLUTION:  

  The solutions are –1 and 2.

38. 0.5x2 = 2x − 0.3

SOLUTION:  

  The solutions are about 0.2 and 3.8.

39. 2x2 −

SOLUTION:  

  The solutions are about 0.2 and 0.9.

40. 

SOLUTION:  

  The solutions are –0.5 and 2.5.

41. 

SOLUTION:  

  The solutions are about –8.2 and 0.2.

42. 

SOLUTION:  

  The solutions are about –5.1 and 0.1.

43. ASTRONOMY The height of an object t seconds after it is dropped is given by the equation ,

where h0 is the initial height and g is the acceleration due to gravity. The acceleration due to gravity near the surface

of Mars is 3.73 m/s2, while on Earth it is 9.8 m/s

2. Suppose an object is dropped from an initial height of 120 meters

above the surface of each planet. a. On which planet would the object reach the ground first? b. How long would it take the object to reach the ground on each planet? Round each answer to the nearest tenth. c. Do the times that it takes the object to reach the ground seem reasonable? Explain your reasoning.

SOLUTION:  a. The object on Earth will reach the ground first because it is falling at a faster rate.   b. Mars:

So, t ≈ 8.0 seconds. Earth:

So, t ≈ 4.9 seconds.   c. Sample answer: Yes; the acceleration due to gravity is much greater on Earth than on Mars, so the time to reach the ground should be much less.

44. Find all values of c that make x2 + cx + 100 a perfect square trinomial.

SOLUTION:  

So, c can be –20 or 20 to make the trinomial a perfect square.

45. Find all values of c that make x2 + cx + 225 a perfect square trinomial.

SOLUTION:  

So, c can be –30 or 30 to make the trinomial a perfect square.

46. PAINTING Before she begins painting a picture, Donna stretches her canvas over a wood frame. The frame has a length of 60 inches and a width of 4 inches. She has enough canvas to cover 480 square inches. Donna decides to increase the dimensions of the frame. If the increase in the length is 10 times the increase in the width, what will the dimensions of the frame be?

SOLUTION:  Let x = the increase in the width and let 10x = the increase in the length. So, x + 4 = the new width and 10x + 60 = the new length.  

  The dimensions cannot be negative, so x = 2. The width is 2 + 4 or 6 inches and the length is 10(2) + 60 or 80 inches.

47. MULTIPLE REPRESENTATIONS In this problem, you will investigate a property of quadratic equations. a. TABULAR  Copy the table shown and complete the second column.

b. ALGEBRAIC Set each trinomial equal to zero, and solve the equation by completing the square. Complete the last column of the table with the number of roots of each equation.

c. VERBAL Compare the number of roots of each equation to the result in the b2 − 4ac column. Is there a

relationship between these values? If so, describe it.

d. ANALYTICAL Predict how many solutions 2x2 − 9x + 15 = 0 will have.  Verify your prediction by solving the 

equation.

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3    

3x2 + 6x + 9    

x2 − 2x + 7    

x2 + 10x + 25    

x2 + 3x − 12    

SOLUTION:  a.

b.

The trinomial x2 − 8x + 16 has 1 root.

 

The trinomial 2x

2 − 11x + 3 has 2 roots.

 

The square root of a negative number has no real roots. So, the trinomial 3x2 + 6x + 9 has 0 roots.

 

The square root of a negative number has no real roots. So, the trinomial x2 − 2x + 7.

 

The trinomial x2 + 10x + 25 has 1 root.

 

The trinomial x2 + 3x − 12 has 2 roots.

 

c. If b2 − 4ac is negative, the equation has no real solutions. If b

2 − 4ac is zero, the equation has one solution. If b2

− 4ac is positive, the equation has 2 solutions. d.

The equation 2x2 − 9x + 15 = 0 has 0 real solutions because b

2 − 4ac is negative.

 

The equation cannot be solved because the square root of a negative number has no real roots.

Trinomial b2 − 4ac Number

of Roots

x2 − 8x + 16 (–8)

2 – 4(1)(16) = 0 1

2x2 − 11x + 3 (–11)

2 – 4(2)(3) = 97  

3x2 + 6x + 9 (6)

2 – 4(3)(9) = −72  

x2 − 2x + 7 (–2)

2 – 4(1)(7) = −24  

x2 + 10x + 25 (10)

2 – 4(1)(25) = 0  

x2 + 3x − 12 (3)

2 – 4(1)(–12) = 57  

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3 97 2

3x2 + 6x + 9 −72 0

x2 − 2x + 7 −24 0

x2 + 10x + 25 0 1

x2 + 3x − 12 57 2

48. CCSS PERSEVERANCE Given y = ax2 + bx + c with a ≠ 0, derive the equation for the axis of symmetry by 

completing the square and rewriting the equation in the form y = a(x − h)2 + k .

SOLUTION:  

Let    and   , then the equation becomes y = (x - h)2 + k .

For an equation in the form y = ax2 + bx + c,  the equation for the axis of symmetry is   .

So, for an equation in the form  y = (x - h)2 + k, the equation for the axis of symmetry becomes x = h.

49. REASONING Determine the number of solutions x2 + bx = c has if . Explain.

SOLUTION:  

None; Sample answer: If you add to each side of the equation and each side of the inequality, you get

. Since the left side of the last equation is a perfect square, it cannot

equal the negative number . So, there are no real solutions.

50. WHICH ONE DOESN’T BELONG? Identify the expression that does not belong with the other three. Explain your reasoning.

 

SOLUTION:  The first 3 trinomials are perfect squares.

The trinomial  is not a perfect square. So, it does not belong.

51. OPEN ENDED Write a quadratic equation for which the only solution is 4.

SOLUTION:  Find a quadratic that has two roots of 4.

Verify that the solution is 4.

So, the quadratic equation x2 − 8x + 16 = 0 has a solution of 4.

52. WRITING IN MATH Compare and contrast the following strategies for solving x2 − 5x − 7 = 0: completing the

square, graphing, and factoring.

SOLUTION:  Sample answer: Because the leading coefficient is 1, solving the equation by completing the square is simpler and yields exact answers for the solutions.   

 

To solve the equation by graphing, use a graphing calculator to graph the related function y = x2 - 5x - 7. Select the

zero option from the 2nd [CALC] menu to determine the roots. 

   

  The roots are given as decimal approximations. So, for this equation the solutions would have to be given as estimations.   There are no factors of –7 that have a sum of –5, so solving by factoring is not possible.

53. The length of a rectangle is 3 times its width. The area of the rectangle is 75 square feet. Find the length of the rectangle in feet. A 25 B 15 C 10 D 5

SOLUTION:  Let x = the width of the rectangle and let 3x = the length of the rectangle.

The length of the rectangle is 3(5) or 15 feet. Choice B is the correct answer.

54. PROBABILITY At a festival, winners of a game draw a token for a prize. There is one token for each prize. The prizes include 9 movie passes, 8 stuffed animals, 5 hats, 10 jump ropes, and 4 glow necklaces. What is the probabilitythat the first person to draw a token will win a movie pass?

F  

G

H

J  

SOLUTION:  There are 9 + 8 + 5 + 10 + 4 or 36 possible outcomes.

P(movie pass) =  or  . Choice J is the correct answer.

55. GRIDDED RESPONSE The population of a town can be modeled by P = 22,000 + 125t, where P represents the population and t represents the number of years from 2000. How many years after 2000 will the population be 26,000?

SOLUTION:  

In 32 years the population of the town will be 26,000.

56. Percy delivers pizzas for Pizza King. He is paid $6 an hour plus $2.50 for each pizza he delivers. Percy earned $280 last week. If he worked a total of 30 hours, how many pizzas did he deliver? A 250 pizzas B 184 pizzas C 40 pizzas D 34 pizzas

SOLUTION:  Let p = the number of pizzas Percy delivered.

Percy delivered 40 pizzas. Choice C is the correct answer.

Describe how the graph of each function is related to the graph of f (x) = x2.

57. g(x) = −12 + x2

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –12, the translation is

down. So, the graph is shifted down 12 units from the parent function. 

58. h(x) = (x + 2)2

SOLUTION:  

The graph of f (x) = (x – c)2 represents a translation left or right from the parent graph. Since c = –2, the translation

is to the left by 2 units. 

 

59. g(x) = 2x2 + 5

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 2 and c = 5. Since 2 > 0 and  > 1, the graph of y = 2x

2 + 5

is the graph of y = x2 vertically stretched and shifted up 5 units.

60. 

SOLUTION:  

The function can be written f (x) = a(x – b)2, where a = and b = 6. Since > 0 and  < 1, the graph of 

is the graph of y = x2 vertically compressed and shifted right 6 units.

61. g(x) = 6 + x2

SOLUTION:  

The function can be written f (x) = ax2 + c, where a =  and c = 6. Since  > 0 and   > 1, the graph of y = 6 +

x2 is the graph of y = x

2 vertically stretched and shifted up 6 units.

62. h(x) = − 1 −   x2

SOLUTION:  

The function can be written f (x) = –ax2 + c, where a =  and c = –1. Since  < 0 and   > 1, the graph of y

= –1 – x2 is the graph of y = x

2 vertically stretched, shifted down 1 unit and reflected across the x-axis.

63. RIDES A popular amusement park ride whisks riders to the top of a 250-foot tower and drops them. A function for

the height of a rider is h = −16t2 + 250, where h is the height and t is the time in seconds. The ride stops the descent

of the rider 40 feet above the ground. Write an equation that models the drop of the rider. How long does it take to fall from 250 feet to 40 feet?

SOLUTION:  

The ride stops the descent of the rider at 40 feet above ground, so h = 40. Then, the equation 40 = −16t2 + 250

models the drop of the rider.

Time cannot be negative. So, it takes about 3.6 seconds to complete the ride.

Simplify. Assume that no denominator is equal to zero.

64. 

SOLUTION:  

65. 

SOLUTION:  

66. 

SOLUTION:  

67. 

SOLUTION:  

68. 

SOLUTION:  

69. b3(m

−3)(b

−6)

SOLUTION:  

Solve each open sentence.

70. |y − 2| > 7

SOLUTION:  

The solution set is {y |y > 9 or y < −5}.

Case 1 y – 2 is positive.

and Case 2 y – 2 is negative.

 

71. |z + 5| < 3

SOLUTION:  

The solution set is {z |−8 < z < −2}.

Case 1 z +5 is positive.

and Case 2 z +5 is negative.

 

72. |2b + 7| ≤ −6

SOLUTION:  

cannot be negative. So,  cannot be less than or equal to –6. The solution set is empty.

73. |3 − 2y | ≥ 8

SOLUTION:  

The solution set is {y |y  ≥ 5.5 or y  ≤ −2.5}.

Case 1 3 – 2y is positive.

and Case 2 3 – 2y is negative.

 

74. |9 − 4m| < −1

SOLUTION:  

cannot be negative. So, cannot be less than or equal to –1. The solution set is empty.

75. |5c − 2| ≤ 13

SOLUTION:  

The solution set is {c|−2.2 ≤ c ≤ 3}.

Case 1 5c – 2 is positive.

and Case 2 5c – 2 is negative.

 

Evaluate  for each set of values. Round to the nearest tenth if necessary.

76. a = 2, b = −5, c = 2

SOLUTION:  

77. a = 1, b = 12, c = 11

SOLUTION:  

78. a = −9, b = 10, c = −1

SOLUTION:  

79. a = 1, b = 7, c = −3

SOLUTION:  

80. a = 2, b = −4, c = −6

SOLUTION:  

81. a = 3, b = 1, c = 2

SOLUTION:  

This value is not a real number.

eSolutions Manual - Powered by Cognero Page 15

9-4 Solving Quadratic Equations by Completing the Square

Find the value of c that makes each trinomial a perfect square.

1. x2 − 18x + c

SOLUTION:  In this trinomial, b = –18.

So, c must be 81 to make the trinomial a perfect square.

2. x2 + 22x + c

SOLUTION:  In this trinomial, b = 22.

So, c must be 121 to make the trinomial a perfect square.

3. x2 + 9x + c

SOLUTION:  In this trinomial, b = 9.

 

So, c must be  to make the trinomial a perfect square.

4. x2 − 7x + c

SOLUTION:  In this trinomial, b = –7.

 

So, c must be  to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

5. x2 + 4x = 6

SOLUTION:  

The solutions are about –5.2 and 1.2.

6. x2 − 8x = −9

SOLUTION:  

The solutions are about 1.4 and 6.6.

7. 4x2 + 9x − 1 = 0

SOLUTION:  

The solutions are about –2.4 and 0.1.

8. −2x2 + 10x + 22 = 4

SOLUTION:  

The solutions are about –1.4 and 6.4.

9. CCSS MODELING Collin is building a deck on the back of his family’s house. He has enough lumber for the deck to be 144 square feet. The length should be 10 feet more than its width. What should the dimensions of the deck be?

SOLUTION:  Let x = the width of the deck and x + 10 = the length of the deck.

The dimensions of the deck can not be negative. So, x = –5 + 13 or 8. The width of the deck is 8 feet and the length of the deck is 8 + 10 or 18 feet.

Find the value of c that makes each trinomial a perfect square.

10. x2 + 26x + c

SOLUTION:  In this trinomial, b = 26.

So, c must be 169 to make the trinomial a perfect square.

11. x2 − 24x + c

SOLUTION:  In this trinomial, b = –24.

So, c must be 144 to make the trinomial a perfect square.

12. x2 − 19x + c

SOLUTION:  In this trinomial, b = –19.

So, c must be  to make the trinomial a perfect square.

13. x2 + 17x + c

SOLUTION:  In this trinomial, b = 17.

So, c must be  to make the trinomial a perfect square.

14. x2 + 5x + c

SOLUTION:  In this trinomial, b = 5.

So, c must be  to make the trinomial a perfect square.

15. x2 − 13x + c

SOLUTION:  In this trinomial, b = –13.

So, c must be  to make the trinomial a perfect square.

16. x2 − 22x + c

SOLUTION:  In this trinomial, b = –22.

So, c must be 121 to make the trinomial a perfect square.

17. x2 − 15x + c

SOLUTION:  In this trinomial, b = –15.

So, c must be  to make the trinomial a perfect square.

18. x2 + 24x + c

SOLUTION:  In this trinomial, b = 24.

So, c must be 144 to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

19. x2 + 6x − 16 = 0

SOLUTION:  

  The solutions are –8 and 2.

20. x2 − 2x − 14 = 0

SOLUTION:  

  The solutions are about –2.9 and 4.9.

21. x2 − 8x − 1 = 8

SOLUTION:  

  The solutions are –1 and 9.

22. x2 + 3x + 21 = 22

SOLUTION:  

  The solutions are about –3.3 and 0.3.

23. x2 − 11x + 3 = 5

SOLUTION:  

  The solutions are about –0.2 and 11.2.

24. 5x2 − 10x = 23

SOLUTION:  

  The solutions are about –1.4 and 3.4.

25. 2x2 − 2x + 7 = 5

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

26. 3x2 + 12x + 81 = 15

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

27. 4x2 + 6x = 12

SOLUTION:  

  The solutions are about –2.6 and 1.1.

28. 4x2 + 5 = 10x

SOLUTION:  

  The solutions are about 0.7 and 1.8.

29. −2x2 + 10x = −14

SOLUTION:  

  The solutions are about –1.1 and 6.1.

30. −3x2 − 12 = 14x

SOLUTION:  

  The solutions are about –3.5 and –1.1.

31. STOCK The price p in dollars for a particular stock can be modeled by the quadratic equation p = 3.5t − 0.05t2, wh

the number of days after the stock is purchased. When is the stock worth $60?

SOLUTION:  Let p = 60.

  The stock is worth $60 on the 30th and 40th days after purchase.

GEOMETRY  Find the value of x for each figure. Round to the nearest tenth if necessary.

32. area = 45 in2

SOLUTION:  

  The height cannot be negative. So, x ≈ 6.3.

33. area = 110 ft2

SOLUTION:  

  The dimensions of the rectangle cannot be negative. So, x ≈ 5.3.

34. NUMBER THEORY The product of two consecutive even integers is 224. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers are 14 and 16 or –16 and –14.

35. CCSS PRECISION The product of two consecutive negative odd integers is 483. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers must be negative. So, they are –23 and –21.

36. GEOMETRY Find the area of the triangle.

SOLUTION:  

  The dimensions of the triangle cannot be negative. So, x = 18. The base of the triangle is 18 meters and the height is 18 + 6 or 24 meters.

The area of the triangle is 216 square meters.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

37. 0.2x2 − 0.2x − 0.4 = 0

SOLUTION:  

  The solutions are –1 and 2.

38. 0.5x2 = 2x − 0.3

SOLUTION:  

  The solutions are about 0.2 and 3.8.

39. 2x2 −

SOLUTION:  

  The solutions are about 0.2 and 0.9.

40. 

SOLUTION:  

  The solutions are –0.5 and 2.5.

41. 

SOLUTION:  

  The solutions are about –8.2 and 0.2.

42. 

SOLUTION:  

  The solutions are about –5.1 and 0.1.

43. ASTRONOMY The height of an object t seconds after it is dropped is given by the equation ,

where h0 is the initial height and g is the acceleration due to gravity. The acceleration due to gravity near the surface

of Mars is 3.73 m/s2, while on Earth it is 9.8 m/s

2. Suppose an object is dropped from an initial height of 120 meters

above the surface of each planet. a. On which planet would the object reach the ground first? b. How long would it take the object to reach the ground on each planet? Round each answer to the nearest tenth. c. Do the times that it takes the object to reach the ground seem reasonable? Explain your reasoning.

SOLUTION:  a. The object on Earth will reach the ground first because it is falling at a faster rate.   b. Mars:

So, t ≈ 8.0 seconds. Earth:

So, t ≈ 4.9 seconds.   c. Sample answer: Yes; the acceleration due to gravity is much greater on Earth than on Mars, so the time to reach the ground should be much less.

44. Find all values of c that make x2 + cx + 100 a perfect square trinomial.

SOLUTION:  

So, c can be –20 or 20 to make the trinomial a perfect square.

45. Find all values of c that make x2 + cx + 225 a perfect square trinomial.

SOLUTION:  

So, c can be –30 or 30 to make the trinomial a perfect square.

46. PAINTING Before she begins painting a picture, Donna stretches her canvas over a wood frame. The frame has a length of 60 inches and a width of 4 inches. She has enough canvas to cover 480 square inches. Donna decides to increase the dimensions of the frame. If the increase in the length is 10 times the increase in the width, what will the dimensions of the frame be?

SOLUTION:  Let x = the increase in the width and let 10x = the increase in the length. So, x + 4 = the new width and 10x + 60 = the new length.  

  The dimensions cannot be negative, so x = 2. The width is 2 + 4 or 6 inches and the length is 10(2) + 60 or 80 inches.

47. MULTIPLE REPRESENTATIONS In this problem, you will investigate a property of quadratic equations. a. TABULAR  Copy the table shown and complete the second column.

b. ALGEBRAIC Set each trinomial equal to zero, and solve the equation by completing the square. Complete the last column of the table with the number of roots of each equation.

c. VERBAL Compare the number of roots of each equation to the result in the b2 − 4ac column. Is there a

relationship between these values? If so, describe it.

d. ANALYTICAL Predict how many solutions 2x2 − 9x + 15 = 0 will have.  Verify your prediction by solving the 

equation.

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3    

3x2 + 6x + 9    

x2 − 2x + 7    

x2 + 10x + 25    

x2 + 3x − 12    

SOLUTION:  a.

b.

The trinomial x2 − 8x + 16 has 1 root.

 

The trinomial 2x

2 − 11x + 3 has 2 roots.

 

The square root of a negative number has no real roots. So, the trinomial 3x2 + 6x + 9 has 0 roots.

 

The square root of a negative number has no real roots. So, the trinomial x2 − 2x + 7.

 

The trinomial x2 + 10x + 25 has 1 root.

 

The trinomial x2 + 3x − 12 has 2 roots.

 

c. If b2 − 4ac is negative, the equation has no real solutions. If b

2 − 4ac is zero, the equation has one solution. If b2

− 4ac is positive, the equation has 2 solutions. d.

The equation 2x2 − 9x + 15 = 0 has 0 real solutions because b

2 − 4ac is negative.

 

The equation cannot be solved because the square root of a negative number has no real roots.

Trinomial b2 − 4ac Number

of Roots

x2 − 8x + 16 (–8)

2 – 4(1)(16) = 0 1

2x2 − 11x + 3 (–11)

2 – 4(2)(3) = 97  

3x2 + 6x + 9 (6)

2 – 4(3)(9) = −72  

x2 − 2x + 7 (–2)

2 – 4(1)(7) = −24  

x2 + 10x + 25 (10)

2 – 4(1)(25) = 0  

x2 + 3x − 12 (3)

2 – 4(1)(–12) = 57  

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3 97 2

3x2 + 6x + 9 −72 0

x2 − 2x + 7 −24 0

x2 + 10x + 25 0 1

x2 + 3x − 12 57 2

48. CCSS PERSEVERANCE Given y = ax2 + bx + c with a ≠ 0, derive the equation for the axis of symmetry by 

completing the square and rewriting the equation in the form y = a(x − h)2 + k .

SOLUTION:  

Let    and   , then the equation becomes y = (x - h)2 + k .

For an equation in the form y = ax2 + bx + c,  the equation for the axis of symmetry is   .

So, for an equation in the form  y = (x - h)2 + k, the equation for the axis of symmetry becomes x = h.

49. REASONING Determine the number of solutions x2 + bx = c has if . Explain.

SOLUTION:  

None; Sample answer: If you add to each side of the equation and each side of the inequality, you get

. Since the left side of the last equation is a perfect square, it cannot

equal the negative number . So, there are no real solutions.

50. WHICH ONE DOESN’T BELONG? Identify the expression that does not belong with the other three. Explain your reasoning.

 

SOLUTION:  The first 3 trinomials are perfect squares.

The trinomial  is not a perfect square. So, it does not belong.

51. OPEN ENDED Write a quadratic equation for which the only solution is 4.

SOLUTION:  Find a quadratic that has two roots of 4.

Verify that the solution is 4.

So, the quadratic equation x2 − 8x + 16 = 0 has a solution of 4.

52. WRITING IN MATH Compare and contrast the following strategies for solving x2 − 5x − 7 = 0: completing the

square, graphing, and factoring.

SOLUTION:  Sample answer: Because the leading coefficient is 1, solving the equation by completing the square is simpler and yields exact answers for the solutions.   

 

To solve the equation by graphing, use a graphing calculator to graph the related function y = x2 - 5x - 7. Select the

zero option from the 2nd [CALC] menu to determine the roots. 

   

  The roots are given as decimal approximations. So, for this equation the solutions would have to be given as estimations.   There are no factors of –7 that have a sum of –5, so solving by factoring is not possible.

53. The length of a rectangle is 3 times its width. The area of the rectangle is 75 square feet. Find the length of the rectangle in feet. A 25 B 15 C 10 D 5

SOLUTION:  Let x = the width of the rectangle and let 3x = the length of the rectangle.

The length of the rectangle is 3(5) or 15 feet. Choice B is the correct answer.

54. PROBABILITY At a festival, winners of a game draw a token for a prize. There is one token for each prize. The prizes include 9 movie passes, 8 stuffed animals, 5 hats, 10 jump ropes, and 4 glow necklaces. What is the probabilitythat the first person to draw a token will win a movie pass?

F  

G

H

J  

SOLUTION:  There are 9 + 8 + 5 + 10 + 4 or 36 possible outcomes.

P(movie pass) =  or  . Choice J is the correct answer.

55. GRIDDED RESPONSE The population of a town can be modeled by P = 22,000 + 125t, where P represents the population and t represents the number of years from 2000. How many years after 2000 will the population be 26,000?

SOLUTION:  

In 32 years the population of the town will be 26,000.

56. Percy delivers pizzas for Pizza King. He is paid $6 an hour plus $2.50 for each pizza he delivers. Percy earned $280 last week. If he worked a total of 30 hours, how many pizzas did he deliver? A 250 pizzas B 184 pizzas C 40 pizzas D 34 pizzas

SOLUTION:  Let p = the number of pizzas Percy delivered.

Percy delivered 40 pizzas. Choice C is the correct answer.

Describe how the graph of each function is related to the graph of f (x) = x2.

57. g(x) = −12 + x2

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –12, the translation is

down. So, the graph is shifted down 12 units from the parent function. 

58. h(x) = (x + 2)2

SOLUTION:  

The graph of f (x) = (x – c)2 represents a translation left or right from the parent graph. Since c = –2, the translation

is to the left by 2 units. 

 

59. g(x) = 2x2 + 5

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 2 and c = 5. Since 2 > 0 and  > 1, the graph of y = 2x

2 + 5

is the graph of y = x2 vertically stretched and shifted up 5 units.

60. 

SOLUTION:  

The function can be written f (x) = a(x – b)2, where a = and b = 6. Since > 0 and  < 1, the graph of 

is the graph of y = x2 vertically compressed and shifted right 6 units.

61. g(x) = 6 + x2

SOLUTION:  

The function can be written f (x) = ax2 + c, where a =  and c = 6. Since  > 0 and   > 1, the graph of y = 6 +

x2 is the graph of y = x

2 vertically stretched and shifted up 6 units.

62. h(x) = − 1 −   x2

SOLUTION:  

The function can be written f (x) = –ax2 + c, where a =  and c = –1. Since  < 0 and   > 1, the graph of y

= –1 – x2 is the graph of y = x

2 vertically stretched, shifted down 1 unit and reflected across the x-axis.

63. RIDES A popular amusement park ride whisks riders to the top of a 250-foot tower and drops them. A function for

the height of a rider is h = −16t2 + 250, where h is the height and t is the time in seconds. The ride stops the descent

of the rider 40 feet above the ground. Write an equation that models the drop of the rider. How long does it take to fall from 250 feet to 40 feet?

SOLUTION:  

The ride stops the descent of the rider at 40 feet above ground, so h = 40. Then, the equation 40 = −16t2 + 250

models the drop of the rider.

Time cannot be negative. So, it takes about 3.6 seconds to complete the ride.

Simplify. Assume that no denominator is equal to zero.

64. 

SOLUTION:  

65. 

SOLUTION:  

66. 

SOLUTION:  

67. 

SOLUTION:  

68. 

SOLUTION:  

69. b3(m

−3)(b

−6)

SOLUTION:  

Solve each open sentence.

70. |y − 2| > 7

SOLUTION:  

The solution set is {y |y > 9 or y < −5}.

Case 1 y – 2 is positive.

and Case 2 y – 2 is negative.

 

71. |z + 5| < 3

SOLUTION:  

The solution set is {z |−8 < z < −2}.

Case 1 z +5 is positive.

and Case 2 z +5 is negative.

 

72. |2b + 7| ≤ −6

SOLUTION:  

cannot be negative. So,  cannot be less than or equal to –6. The solution set is empty.

73. |3 − 2y | ≥ 8

SOLUTION:  

The solution set is {y |y  ≥ 5.5 or y  ≤ −2.5}.

Case 1 3 – 2y is positive.

and Case 2 3 – 2y is negative.

 

74. |9 − 4m| < −1

SOLUTION:  

cannot be negative. So, cannot be less than or equal to –1. The solution set is empty.

75. |5c − 2| ≤ 13

SOLUTION:  

The solution set is {c|−2.2 ≤ c ≤ 3}.

Case 1 5c – 2 is positive.

and Case 2 5c – 2 is negative.

 

Evaluate  for each set of values. Round to the nearest tenth if necessary.

76. a = 2, b = −5, c = 2

SOLUTION:  

77. a = 1, b = 12, c = 11

SOLUTION:  

78. a = −9, b = 10, c = −1

SOLUTION:  

79. a = 1, b = 7, c = −3

SOLUTION:  

80. a = 2, b = −4, c = −6

SOLUTION:  

81. a = 3, b = 1, c = 2

SOLUTION:  

This value is not a real number.

eSolutions Manual - Powered by Cognero Page 16

9-4 Solving Quadratic Equations by Completing the Square

Find the value of c that makes each trinomial a perfect square.

1. x2 − 18x + c

SOLUTION:  In this trinomial, b = –18.

So, c must be 81 to make the trinomial a perfect square.

2. x2 + 22x + c

SOLUTION:  In this trinomial, b = 22.

So, c must be 121 to make the trinomial a perfect square.

3. x2 + 9x + c

SOLUTION:  In this trinomial, b = 9.

 

So, c must be  to make the trinomial a perfect square.

4. x2 − 7x + c

SOLUTION:  In this trinomial, b = –7.

 

So, c must be  to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

5. x2 + 4x = 6

SOLUTION:  

The solutions are about –5.2 and 1.2.

6. x2 − 8x = −9

SOLUTION:  

The solutions are about 1.4 and 6.6.

7. 4x2 + 9x − 1 = 0

SOLUTION:  

The solutions are about –2.4 and 0.1.

8. −2x2 + 10x + 22 = 4

SOLUTION:  

The solutions are about –1.4 and 6.4.

9. CCSS MODELING Collin is building a deck on the back of his family’s house. He has enough lumber for the deck to be 144 square feet. The length should be 10 feet more than its width. What should the dimensions of the deck be?

SOLUTION:  Let x = the width of the deck and x + 10 = the length of the deck.

The dimensions of the deck can not be negative. So, x = –5 + 13 or 8. The width of the deck is 8 feet and the length of the deck is 8 + 10 or 18 feet.

Find the value of c that makes each trinomial a perfect square.

10. x2 + 26x + c

SOLUTION:  In this trinomial, b = 26.

So, c must be 169 to make the trinomial a perfect square.

11. x2 − 24x + c

SOLUTION:  In this trinomial, b = –24.

So, c must be 144 to make the trinomial a perfect square.

12. x2 − 19x + c

SOLUTION:  In this trinomial, b = –19.

So, c must be  to make the trinomial a perfect square.

13. x2 + 17x + c

SOLUTION:  In this trinomial, b = 17.

So, c must be  to make the trinomial a perfect square.

14. x2 + 5x + c

SOLUTION:  In this trinomial, b = 5.

So, c must be  to make the trinomial a perfect square.

15. x2 − 13x + c

SOLUTION:  In this trinomial, b = –13.

So, c must be  to make the trinomial a perfect square.

16. x2 − 22x + c

SOLUTION:  In this trinomial, b = –22.

So, c must be 121 to make the trinomial a perfect square.

17. x2 − 15x + c

SOLUTION:  In this trinomial, b = –15.

So, c must be  to make the trinomial a perfect square.

18. x2 + 24x + c

SOLUTION:  In this trinomial, b = 24.

So, c must be 144 to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

19. x2 + 6x − 16 = 0

SOLUTION:  

  The solutions are –8 and 2.

20. x2 − 2x − 14 = 0

SOLUTION:  

  The solutions are about –2.9 and 4.9.

21. x2 − 8x − 1 = 8

SOLUTION:  

  The solutions are –1 and 9.

22. x2 + 3x + 21 = 22

SOLUTION:  

  The solutions are about –3.3 and 0.3.

23. x2 − 11x + 3 = 5

SOLUTION:  

  The solutions are about –0.2 and 11.2.

24. 5x2 − 10x = 23

SOLUTION:  

  The solutions are about –1.4 and 3.4.

25. 2x2 − 2x + 7 = 5

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

26. 3x2 + 12x + 81 = 15

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

27. 4x2 + 6x = 12

SOLUTION:  

  The solutions are about –2.6 and 1.1.

28. 4x2 + 5 = 10x

SOLUTION:  

  The solutions are about 0.7 and 1.8.

29. −2x2 + 10x = −14

SOLUTION:  

  The solutions are about –1.1 and 6.1.

30. −3x2 − 12 = 14x

SOLUTION:  

  The solutions are about –3.5 and –1.1.

31. STOCK The price p in dollars for a particular stock can be modeled by the quadratic equation p = 3.5t − 0.05t2, wh

the number of days after the stock is purchased. When is the stock worth $60?

SOLUTION:  Let p = 60.

  The stock is worth $60 on the 30th and 40th days after purchase.

GEOMETRY  Find the value of x for each figure. Round to the nearest tenth if necessary.

32. area = 45 in2

SOLUTION:  

  The height cannot be negative. So, x ≈ 6.3.

33. area = 110 ft2

SOLUTION:  

  The dimensions of the rectangle cannot be negative. So, x ≈ 5.3.

34. NUMBER THEORY The product of two consecutive even integers is 224. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers are 14 and 16 or –16 and –14.

35. CCSS PRECISION The product of two consecutive negative odd integers is 483. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers must be negative. So, they are –23 and –21.

36. GEOMETRY Find the area of the triangle.

SOLUTION:  

  The dimensions of the triangle cannot be negative. So, x = 18. The base of the triangle is 18 meters and the height is 18 + 6 or 24 meters.

The area of the triangle is 216 square meters.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

37. 0.2x2 − 0.2x − 0.4 = 0

SOLUTION:  

  The solutions are –1 and 2.

38. 0.5x2 = 2x − 0.3

SOLUTION:  

  The solutions are about 0.2 and 3.8.

39. 2x2 −

SOLUTION:  

  The solutions are about 0.2 and 0.9.

40. 

SOLUTION:  

  The solutions are –0.5 and 2.5.

41. 

SOLUTION:  

  The solutions are about –8.2 and 0.2.

42. 

SOLUTION:  

  The solutions are about –5.1 and 0.1.

43. ASTRONOMY The height of an object t seconds after it is dropped is given by the equation ,

where h0 is the initial height and g is the acceleration due to gravity. The acceleration due to gravity near the surface

of Mars is 3.73 m/s2, while on Earth it is 9.8 m/s

2. Suppose an object is dropped from an initial height of 120 meters

above the surface of each planet. a. On which planet would the object reach the ground first? b. How long would it take the object to reach the ground on each planet? Round each answer to the nearest tenth. c. Do the times that it takes the object to reach the ground seem reasonable? Explain your reasoning.

SOLUTION:  a. The object on Earth will reach the ground first because it is falling at a faster rate.   b. Mars:

So, t ≈ 8.0 seconds. Earth:

So, t ≈ 4.9 seconds.   c. Sample answer: Yes; the acceleration due to gravity is much greater on Earth than on Mars, so the time to reach the ground should be much less.

44. Find all values of c that make x2 + cx + 100 a perfect square trinomial.

SOLUTION:  

So, c can be –20 or 20 to make the trinomial a perfect square.

45. Find all values of c that make x2 + cx + 225 a perfect square trinomial.

SOLUTION:  

So, c can be –30 or 30 to make the trinomial a perfect square.

46. PAINTING Before she begins painting a picture, Donna stretches her canvas over a wood frame. The frame has a length of 60 inches and a width of 4 inches. She has enough canvas to cover 480 square inches. Donna decides to increase the dimensions of the frame. If the increase in the length is 10 times the increase in the width, what will the dimensions of the frame be?

SOLUTION:  Let x = the increase in the width and let 10x = the increase in the length. So, x + 4 = the new width and 10x + 60 = the new length.  

  The dimensions cannot be negative, so x = 2. The width is 2 + 4 or 6 inches and the length is 10(2) + 60 or 80 inches.

47. MULTIPLE REPRESENTATIONS In this problem, you will investigate a property of quadratic equations. a. TABULAR  Copy the table shown and complete the second column.

b. ALGEBRAIC Set each trinomial equal to zero, and solve the equation by completing the square. Complete the last column of the table with the number of roots of each equation.

c. VERBAL Compare the number of roots of each equation to the result in the b2 − 4ac column. Is there a

relationship between these values? If so, describe it.

d. ANALYTICAL Predict how many solutions 2x2 − 9x + 15 = 0 will have.  Verify your prediction by solving the 

equation.

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3    

3x2 + 6x + 9    

x2 − 2x + 7    

x2 + 10x + 25    

x2 + 3x − 12    

SOLUTION:  a.

b.

The trinomial x2 − 8x + 16 has 1 root.

 

The trinomial 2x

2 − 11x + 3 has 2 roots.

 

The square root of a negative number has no real roots. So, the trinomial 3x2 + 6x + 9 has 0 roots.

 

The square root of a negative number has no real roots. So, the trinomial x2 − 2x + 7.

 

The trinomial x2 + 10x + 25 has 1 root.

 

The trinomial x2 + 3x − 12 has 2 roots.

 

c. If b2 − 4ac is negative, the equation has no real solutions. If b

2 − 4ac is zero, the equation has one solution. If b2

− 4ac is positive, the equation has 2 solutions. d.

The equation 2x2 − 9x + 15 = 0 has 0 real solutions because b

2 − 4ac is negative.

 

The equation cannot be solved because the square root of a negative number has no real roots.

Trinomial b2 − 4ac Number

of Roots

x2 − 8x + 16 (–8)

2 – 4(1)(16) = 0 1

2x2 − 11x + 3 (–11)

2 – 4(2)(3) = 97  

3x2 + 6x + 9 (6)

2 – 4(3)(9) = −72  

x2 − 2x + 7 (–2)

2 – 4(1)(7) = −24  

x2 + 10x + 25 (10)

2 – 4(1)(25) = 0  

x2 + 3x − 12 (3)

2 – 4(1)(–12) = 57  

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3 97 2

3x2 + 6x + 9 −72 0

x2 − 2x + 7 −24 0

x2 + 10x + 25 0 1

x2 + 3x − 12 57 2

48. CCSS PERSEVERANCE Given y = ax2 + bx + c with a ≠ 0, derive the equation for the axis of symmetry by 

completing the square and rewriting the equation in the form y = a(x − h)2 + k .

SOLUTION:  

Let    and   , then the equation becomes y = (x - h)2 + k .

For an equation in the form y = ax2 + bx + c,  the equation for the axis of symmetry is   .

So, for an equation in the form  y = (x - h)2 + k, the equation for the axis of symmetry becomes x = h.

49. REASONING Determine the number of solutions x2 + bx = c has if . Explain.

SOLUTION:  

None; Sample answer: If you add to each side of the equation and each side of the inequality, you get

. Since the left side of the last equation is a perfect square, it cannot

equal the negative number . So, there are no real solutions.

50. WHICH ONE DOESN’T BELONG? Identify the expression that does not belong with the other three. Explain your reasoning.

 

SOLUTION:  The first 3 trinomials are perfect squares.

The trinomial  is not a perfect square. So, it does not belong.

51. OPEN ENDED Write a quadratic equation for which the only solution is 4.

SOLUTION:  Find a quadratic that has two roots of 4.

Verify that the solution is 4.

So, the quadratic equation x2 − 8x + 16 = 0 has a solution of 4.

52. WRITING IN MATH Compare and contrast the following strategies for solving x2 − 5x − 7 = 0: completing the

square, graphing, and factoring.

SOLUTION:  Sample answer: Because the leading coefficient is 1, solving the equation by completing the square is simpler and yields exact answers for the solutions.   

 

To solve the equation by graphing, use a graphing calculator to graph the related function y = x2 - 5x - 7. Select the

zero option from the 2nd [CALC] menu to determine the roots. 

   

  The roots are given as decimal approximations. So, for this equation the solutions would have to be given as estimations.   There are no factors of –7 that have a sum of –5, so solving by factoring is not possible.

53. The length of a rectangle is 3 times its width. The area of the rectangle is 75 square feet. Find the length of the rectangle in feet. A 25 B 15 C 10 D 5

SOLUTION:  Let x = the width of the rectangle and let 3x = the length of the rectangle.

The length of the rectangle is 3(5) or 15 feet. Choice B is the correct answer.

54. PROBABILITY At a festival, winners of a game draw a token for a prize. There is one token for each prize. The prizes include 9 movie passes, 8 stuffed animals, 5 hats, 10 jump ropes, and 4 glow necklaces. What is the probabilitythat the first person to draw a token will win a movie pass?

F  

G

H

J  

SOLUTION:  There are 9 + 8 + 5 + 10 + 4 or 36 possible outcomes.

P(movie pass) =  or  . Choice J is the correct answer.

55. GRIDDED RESPONSE The population of a town can be modeled by P = 22,000 + 125t, where P represents the population and t represents the number of years from 2000. How many years after 2000 will the population be 26,000?

SOLUTION:  

In 32 years the population of the town will be 26,000.

56. Percy delivers pizzas for Pizza King. He is paid $6 an hour plus $2.50 for each pizza he delivers. Percy earned $280 last week. If he worked a total of 30 hours, how many pizzas did he deliver? A 250 pizzas B 184 pizzas C 40 pizzas D 34 pizzas

SOLUTION:  Let p = the number of pizzas Percy delivered.

Percy delivered 40 pizzas. Choice C is the correct answer.

Describe how the graph of each function is related to the graph of f (x) = x2.

57. g(x) = −12 + x2

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –12, the translation is

down. So, the graph is shifted down 12 units from the parent function. 

58. h(x) = (x + 2)2

SOLUTION:  

The graph of f (x) = (x – c)2 represents a translation left or right from the parent graph. Since c = –2, the translation

is to the left by 2 units. 

 

59. g(x) = 2x2 + 5

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 2 and c = 5. Since 2 > 0 and  > 1, the graph of y = 2x

2 + 5

is the graph of y = x2 vertically stretched and shifted up 5 units.

60. 

SOLUTION:  

The function can be written f (x) = a(x – b)2, where a = and b = 6. Since > 0 and  < 1, the graph of 

is the graph of y = x2 vertically compressed and shifted right 6 units.

61. g(x) = 6 + x2

SOLUTION:  

The function can be written f (x) = ax2 + c, where a =  and c = 6. Since  > 0 and   > 1, the graph of y = 6 +

x2 is the graph of y = x

2 vertically stretched and shifted up 6 units.

62. h(x) = − 1 −   x2

SOLUTION:  

The function can be written f (x) = –ax2 + c, where a =  and c = –1. Since  < 0 and   > 1, the graph of y

= –1 – x2 is the graph of y = x

2 vertically stretched, shifted down 1 unit and reflected across the x-axis.

63. RIDES A popular amusement park ride whisks riders to the top of a 250-foot tower and drops them. A function for

the height of a rider is h = −16t2 + 250, where h is the height and t is the time in seconds. The ride stops the descent

of the rider 40 feet above the ground. Write an equation that models the drop of the rider. How long does it take to fall from 250 feet to 40 feet?

SOLUTION:  

The ride stops the descent of the rider at 40 feet above ground, so h = 40. Then, the equation 40 = −16t2 + 250

models the drop of the rider.

Time cannot be negative. So, it takes about 3.6 seconds to complete the ride.

Simplify. Assume that no denominator is equal to zero.

64. 

SOLUTION:  

65. 

SOLUTION:  

66. 

SOLUTION:  

67. 

SOLUTION:  

68. 

SOLUTION:  

69. b3(m

−3)(b

−6)

SOLUTION:  

Solve each open sentence.

70. |y − 2| > 7

SOLUTION:  

The solution set is {y |y > 9 or y < −5}.

Case 1 y – 2 is positive.

and Case 2 y – 2 is negative.

 

71. |z + 5| < 3

SOLUTION:  

The solution set is {z |−8 < z < −2}.

Case 1 z +5 is positive.

and Case 2 z +5 is negative.

 

72. |2b + 7| ≤ −6

SOLUTION:  

cannot be negative. So,  cannot be less than or equal to –6. The solution set is empty.

73. |3 − 2y | ≥ 8

SOLUTION:  

The solution set is {y |y  ≥ 5.5 or y  ≤ −2.5}.

Case 1 3 – 2y is positive.

and Case 2 3 – 2y is negative.

 

74. |9 − 4m| < −1

SOLUTION:  

cannot be negative. So, cannot be less than or equal to –1. The solution set is empty.

75. |5c − 2| ≤ 13

SOLUTION:  

The solution set is {c|−2.2 ≤ c ≤ 3}.

Case 1 5c – 2 is positive.

and Case 2 5c – 2 is negative.

 

Evaluate  for each set of values. Round to the nearest tenth if necessary.

76. a = 2, b = −5, c = 2

SOLUTION:  

77. a = 1, b = 12, c = 11

SOLUTION:  

78. a = −9, b = 10, c = −1

SOLUTION:  

79. a = 1, b = 7, c = −3

SOLUTION:  

80. a = 2, b = −4, c = −6

SOLUTION:  

81. a = 3, b = 1, c = 2

SOLUTION:  

This value is not a real number.

eSolutions Manual - Powered by Cognero Page 17

9-4 Solving Quadratic Equations by Completing the Square

Find the value of c that makes each trinomial a perfect square.

1. x2 − 18x + c

SOLUTION:  In this trinomial, b = –18.

So, c must be 81 to make the trinomial a perfect square.

2. x2 + 22x + c

SOLUTION:  In this trinomial, b = 22.

So, c must be 121 to make the trinomial a perfect square.

3. x2 + 9x + c

SOLUTION:  In this trinomial, b = 9.

 

So, c must be  to make the trinomial a perfect square.

4. x2 − 7x + c

SOLUTION:  In this trinomial, b = –7.

 

So, c must be  to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

5. x2 + 4x = 6

SOLUTION:  

The solutions are about –5.2 and 1.2.

6. x2 − 8x = −9

SOLUTION:  

The solutions are about 1.4 and 6.6.

7. 4x2 + 9x − 1 = 0

SOLUTION:  

The solutions are about –2.4 and 0.1.

8. −2x2 + 10x + 22 = 4

SOLUTION:  

The solutions are about –1.4 and 6.4.

9. CCSS MODELING Collin is building a deck on the back of his family’s house. He has enough lumber for the deck to be 144 square feet. The length should be 10 feet more than its width. What should the dimensions of the deck be?

SOLUTION:  Let x = the width of the deck and x + 10 = the length of the deck.

The dimensions of the deck can not be negative. So, x = –5 + 13 or 8. The width of the deck is 8 feet and the length of the deck is 8 + 10 or 18 feet.

Find the value of c that makes each trinomial a perfect square.

10. x2 + 26x + c

SOLUTION:  In this trinomial, b = 26.

So, c must be 169 to make the trinomial a perfect square.

11. x2 − 24x + c

SOLUTION:  In this trinomial, b = –24.

So, c must be 144 to make the trinomial a perfect square.

12. x2 − 19x + c

SOLUTION:  In this trinomial, b = –19.

So, c must be  to make the trinomial a perfect square.

13. x2 + 17x + c

SOLUTION:  In this trinomial, b = 17.

So, c must be  to make the trinomial a perfect square.

14. x2 + 5x + c

SOLUTION:  In this trinomial, b = 5.

So, c must be  to make the trinomial a perfect square.

15. x2 − 13x + c

SOLUTION:  In this trinomial, b = –13.

So, c must be  to make the trinomial a perfect square.

16. x2 − 22x + c

SOLUTION:  In this trinomial, b = –22.

So, c must be 121 to make the trinomial a perfect square.

17. x2 − 15x + c

SOLUTION:  In this trinomial, b = –15.

So, c must be  to make the trinomial a perfect square.

18. x2 + 24x + c

SOLUTION:  In this trinomial, b = 24.

So, c must be 144 to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

19. x2 + 6x − 16 = 0

SOLUTION:  

  The solutions are –8 and 2.

20. x2 − 2x − 14 = 0

SOLUTION:  

  The solutions are about –2.9 and 4.9.

21. x2 − 8x − 1 = 8

SOLUTION:  

  The solutions are –1 and 9.

22. x2 + 3x + 21 = 22

SOLUTION:  

  The solutions are about –3.3 and 0.3.

23. x2 − 11x + 3 = 5

SOLUTION:  

  The solutions are about –0.2 and 11.2.

24. 5x2 − 10x = 23

SOLUTION:  

  The solutions are about –1.4 and 3.4.

25. 2x2 − 2x + 7 = 5

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

26. 3x2 + 12x + 81 = 15

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

27. 4x2 + 6x = 12

SOLUTION:  

  The solutions are about –2.6 and 1.1.

28. 4x2 + 5 = 10x

SOLUTION:  

  The solutions are about 0.7 and 1.8.

29. −2x2 + 10x = −14

SOLUTION:  

  The solutions are about –1.1 and 6.1.

30. −3x2 − 12 = 14x

SOLUTION:  

  The solutions are about –3.5 and –1.1.

31. STOCK The price p in dollars for a particular stock can be modeled by the quadratic equation p = 3.5t − 0.05t2, wh

the number of days after the stock is purchased. When is the stock worth $60?

SOLUTION:  Let p = 60.

  The stock is worth $60 on the 30th and 40th days after purchase.

GEOMETRY  Find the value of x for each figure. Round to the nearest tenth if necessary.

32. area = 45 in2

SOLUTION:  

  The height cannot be negative. So, x ≈ 6.3.

33. area = 110 ft2

SOLUTION:  

  The dimensions of the rectangle cannot be negative. So, x ≈ 5.3.

34. NUMBER THEORY The product of two consecutive even integers is 224. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers are 14 and 16 or –16 and –14.

35. CCSS PRECISION The product of two consecutive negative odd integers is 483. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers must be negative. So, they are –23 and –21.

36. GEOMETRY Find the area of the triangle.

SOLUTION:  

  The dimensions of the triangle cannot be negative. So, x = 18. The base of the triangle is 18 meters and the height is 18 + 6 or 24 meters.

The area of the triangle is 216 square meters.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

37. 0.2x2 − 0.2x − 0.4 = 0

SOLUTION:  

  The solutions are –1 and 2.

38. 0.5x2 = 2x − 0.3

SOLUTION:  

  The solutions are about 0.2 and 3.8.

39. 2x2 −

SOLUTION:  

  The solutions are about 0.2 and 0.9.

40. 

SOLUTION:  

  The solutions are –0.5 and 2.5.

41. 

SOLUTION:  

  The solutions are about –8.2 and 0.2.

42. 

SOLUTION:  

  The solutions are about –5.1 and 0.1.

43. ASTRONOMY The height of an object t seconds after it is dropped is given by the equation ,

where h0 is the initial height and g is the acceleration due to gravity. The acceleration due to gravity near the surface

of Mars is 3.73 m/s2, while on Earth it is 9.8 m/s

2. Suppose an object is dropped from an initial height of 120 meters

above the surface of each planet. a. On which planet would the object reach the ground first? b. How long would it take the object to reach the ground on each planet? Round each answer to the nearest tenth. c. Do the times that it takes the object to reach the ground seem reasonable? Explain your reasoning.

SOLUTION:  a. The object on Earth will reach the ground first because it is falling at a faster rate.   b. Mars:

So, t ≈ 8.0 seconds. Earth:

So, t ≈ 4.9 seconds.   c. Sample answer: Yes; the acceleration due to gravity is much greater on Earth than on Mars, so the time to reach the ground should be much less.

44. Find all values of c that make x2 + cx + 100 a perfect square trinomial.

SOLUTION:  

So, c can be –20 or 20 to make the trinomial a perfect square.

45. Find all values of c that make x2 + cx + 225 a perfect square trinomial.

SOLUTION:  

So, c can be –30 or 30 to make the trinomial a perfect square.

46. PAINTING Before she begins painting a picture, Donna stretches her canvas over a wood frame. The frame has a length of 60 inches and a width of 4 inches. She has enough canvas to cover 480 square inches. Donna decides to increase the dimensions of the frame. If the increase in the length is 10 times the increase in the width, what will the dimensions of the frame be?

SOLUTION:  Let x = the increase in the width and let 10x = the increase in the length. So, x + 4 = the new width and 10x + 60 = the new length.  

  The dimensions cannot be negative, so x = 2. The width is 2 + 4 or 6 inches and the length is 10(2) + 60 or 80 inches.

47. MULTIPLE REPRESENTATIONS In this problem, you will investigate a property of quadratic equations. a. TABULAR  Copy the table shown and complete the second column.

b. ALGEBRAIC Set each trinomial equal to zero, and solve the equation by completing the square. Complete the last column of the table with the number of roots of each equation.

c. VERBAL Compare the number of roots of each equation to the result in the b2 − 4ac column. Is there a

relationship between these values? If so, describe it.

d. ANALYTICAL Predict how many solutions 2x2 − 9x + 15 = 0 will have.  Verify your prediction by solving the 

equation.

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3    

3x2 + 6x + 9    

x2 − 2x + 7    

x2 + 10x + 25    

x2 + 3x − 12    

SOLUTION:  a.

b.

The trinomial x2 − 8x + 16 has 1 root.

 

The trinomial 2x

2 − 11x + 3 has 2 roots.

 

The square root of a negative number has no real roots. So, the trinomial 3x2 + 6x + 9 has 0 roots.

 

The square root of a negative number has no real roots. So, the trinomial x2 − 2x + 7.

 

The trinomial x2 + 10x + 25 has 1 root.

 

The trinomial x2 + 3x − 12 has 2 roots.

 

c. If b2 − 4ac is negative, the equation has no real solutions. If b

2 − 4ac is zero, the equation has one solution. If b2

− 4ac is positive, the equation has 2 solutions. d.

The equation 2x2 − 9x + 15 = 0 has 0 real solutions because b

2 − 4ac is negative.

 

The equation cannot be solved because the square root of a negative number has no real roots.

Trinomial b2 − 4ac Number

of Roots

x2 − 8x + 16 (–8)

2 – 4(1)(16) = 0 1

2x2 − 11x + 3 (–11)

2 – 4(2)(3) = 97  

3x2 + 6x + 9 (6)

2 – 4(3)(9) = −72  

x2 − 2x + 7 (–2)

2 – 4(1)(7) = −24  

x2 + 10x + 25 (10)

2 – 4(1)(25) = 0  

x2 + 3x − 12 (3)

2 – 4(1)(–12) = 57  

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3 97 2

3x2 + 6x + 9 −72 0

x2 − 2x + 7 −24 0

x2 + 10x + 25 0 1

x2 + 3x − 12 57 2

48. CCSS PERSEVERANCE Given y = ax2 + bx + c with a ≠ 0, derive the equation for the axis of symmetry by 

completing the square and rewriting the equation in the form y = a(x − h)2 + k .

SOLUTION:  

Let    and   , then the equation becomes y = (x - h)2 + k .

For an equation in the form y = ax2 + bx + c,  the equation for the axis of symmetry is   .

So, for an equation in the form  y = (x - h)2 + k, the equation for the axis of symmetry becomes x = h.

49. REASONING Determine the number of solutions x2 + bx = c has if . Explain.

SOLUTION:  

None; Sample answer: If you add to each side of the equation and each side of the inequality, you get

. Since the left side of the last equation is a perfect square, it cannot

equal the negative number . So, there are no real solutions.

50. WHICH ONE DOESN’T BELONG? Identify the expression that does not belong with the other three. Explain your reasoning.

 

SOLUTION:  The first 3 trinomials are perfect squares.

The trinomial  is not a perfect square. So, it does not belong.

51. OPEN ENDED Write a quadratic equation for which the only solution is 4.

SOLUTION:  Find a quadratic that has two roots of 4.

Verify that the solution is 4.

So, the quadratic equation x2 − 8x + 16 = 0 has a solution of 4.

52. WRITING IN MATH Compare and contrast the following strategies for solving x2 − 5x − 7 = 0: completing the

square, graphing, and factoring.

SOLUTION:  Sample answer: Because the leading coefficient is 1, solving the equation by completing the square is simpler and yields exact answers for the solutions.   

 

To solve the equation by graphing, use a graphing calculator to graph the related function y = x2 - 5x - 7. Select the

zero option from the 2nd [CALC] menu to determine the roots. 

   

  The roots are given as decimal approximations. So, for this equation the solutions would have to be given as estimations.   There are no factors of –7 that have a sum of –5, so solving by factoring is not possible.

53. The length of a rectangle is 3 times its width. The area of the rectangle is 75 square feet. Find the length of the rectangle in feet. A 25 B 15 C 10 D 5

SOLUTION:  Let x = the width of the rectangle and let 3x = the length of the rectangle.

The length of the rectangle is 3(5) or 15 feet. Choice B is the correct answer.

54. PROBABILITY At a festival, winners of a game draw a token for a prize. There is one token for each prize. The prizes include 9 movie passes, 8 stuffed animals, 5 hats, 10 jump ropes, and 4 glow necklaces. What is the probabilitythat the first person to draw a token will win a movie pass?

F  

G

H

J  

SOLUTION:  There are 9 + 8 + 5 + 10 + 4 or 36 possible outcomes.

P(movie pass) =  or  . Choice J is the correct answer.

55. GRIDDED RESPONSE The population of a town can be modeled by P = 22,000 + 125t, where P represents the population and t represents the number of years from 2000. How many years after 2000 will the population be 26,000?

SOLUTION:  

In 32 years the population of the town will be 26,000.

56. Percy delivers pizzas for Pizza King. He is paid $6 an hour plus $2.50 for each pizza he delivers. Percy earned $280 last week. If he worked a total of 30 hours, how many pizzas did he deliver? A 250 pizzas B 184 pizzas C 40 pizzas D 34 pizzas

SOLUTION:  Let p = the number of pizzas Percy delivered.

Percy delivered 40 pizzas. Choice C is the correct answer.

Describe how the graph of each function is related to the graph of f (x) = x2.

57. g(x) = −12 + x2

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –12, the translation is

down. So, the graph is shifted down 12 units from the parent function. 

58. h(x) = (x + 2)2

SOLUTION:  

The graph of f (x) = (x – c)2 represents a translation left or right from the parent graph. Since c = –2, the translation

is to the left by 2 units. 

 

59. g(x) = 2x2 + 5

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 2 and c = 5. Since 2 > 0 and  > 1, the graph of y = 2x

2 + 5

is the graph of y = x2 vertically stretched and shifted up 5 units.

60. 

SOLUTION:  

The function can be written f (x) = a(x – b)2, where a = and b = 6. Since > 0 and  < 1, the graph of 

is the graph of y = x2 vertically compressed and shifted right 6 units.

61. g(x) = 6 + x2

SOLUTION:  

The function can be written f (x) = ax2 + c, where a =  and c = 6. Since  > 0 and   > 1, the graph of y = 6 +

x2 is the graph of y = x

2 vertically stretched and shifted up 6 units.

62. h(x) = − 1 −   x2

SOLUTION:  

The function can be written f (x) = –ax2 + c, where a =  and c = –1. Since  < 0 and   > 1, the graph of y

= –1 – x2 is the graph of y = x

2 vertically stretched, shifted down 1 unit and reflected across the x-axis.

63. RIDES A popular amusement park ride whisks riders to the top of a 250-foot tower and drops them. A function for

the height of a rider is h = −16t2 + 250, where h is the height and t is the time in seconds. The ride stops the descent

of the rider 40 feet above the ground. Write an equation that models the drop of the rider. How long does it take to fall from 250 feet to 40 feet?

SOLUTION:  

The ride stops the descent of the rider at 40 feet above ground, so h = 40. Then, the equation 40 = −16t2 + 250

models the drop of the rider.

Time cannot be negative. So, it takes about 3.6 seconds to complete the ride.

Simplify. Assume that no denominator is equal to zero.

64. 

SOLUTION:  

65. 

SOLUTION:  

66. 

SOLUTION:  

67. 

SOLUTION:  

68. 

SOLUTION:  

69. b3(m

−3)(b

−6)

SOLUTION:  

Solve each open sentence.

70. |y − 2| > 7

SOLUTION:  

The solution set is {y |y > 9 or y < −5}.

Case 1 y – 2 is positive.

and Case 2 y – 2 is negative.

 

71. |z + 5| < 3

SOLUTION:  

The solution set is {z |−8 < z < −2}.

Case 1 z +5 is positive.

and Case 2 z +5 is negative.

 

72. |2b + 7| ≤ −6

SOLUTION:  

cannot be negative. So,  cannot be less than or equal to –6. The solution set is empty.

73. |3 − 2y | ≥ 8

SOLUTION:  

The solution set is {y |y  ≥ 5.5 or y  ≤ −2.5}.

Case 1 3 – 2y is positive.

and Case 2 3 – 2y is negative.

 

74. |9 − 4m| < −1

SOLUTION:  

cannot be negative. So, cannot be less than or equal to –1. The solution set is empty.

75. |5c − 2| ≤ 13

SOLUTION:  

The solution set is {c|−2.2 ≤ c ≤ 3}.

Case 1 5c – 2 is positive.

and Case 2 5c – 2 is negative.

 

Evaluate  for each set of values. Round to the nearest tenth if necessary.

76. a = 2, b = −5, c = 2

SOLUTION:  

77. a = 1, b = 12, c = 11

SOLUTION:  

78. a = −9, b = 10, c = −1

SOLUTION:  

79. a = 1, b = 7, c = −3

SOLUTION:  

80. a = 2, b = −4, c = −6

SOLUTION:  

81. a = 3, b = 1, c = 2

SOLUTION:  

This value is not a real number.

eSolutions Manual - Powered by Cognero Page 18

9-4 Solving Quadratic Equations by Completing the Square

Find the value of c that makes each trinomial a perfect square.

1. x2 − 18x + c

SOLUTION:  In this trinomial, b = –18.

So, c must be 81 to make the trinomial a perfect square.

2. x2 + 22x + c

SOLUTION:  In this trinomial, b = 22.

So, c must be 121 to make the trinomial a perfect square.

3. x2 + 9x + c

SOLUTION:  In this trinomial, b = 9.

 

So, c must be  to make the trinomial a perfect square.

4. x2 − 7x + c

SOLUTION:  In this trinomial, b = –7.

 

So, c must be  to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

5. x2 + 4x = 6

SOLUTION:  

The solutions are about –5.2 and 1.2.

6. x2 − 8x = −9

SOLUTION:  

The solutions are about 1.4 and 6.6.

7. 4x2 + 9x − 1 = 0

SOLUTION:  

The solutions are about –2.4 and 0.1.

8. −2x2 + 10x + 22 = 4

SOLUTION:  

The solutions are about –1.4 and 6.4.

9. CCSS MODELING Collin is building a deck on the back of his family’s house. He has enough lumber for the deck to be 144 square feet. The length should be 10 feet more than its width. What should the dimensions of the deck be?

SOLUTION:  Let x = the width of the deck and x + 10 = the length of the deck.

The dimensions of the deck can not be negative. So, x = –5 + 13 or 8. The width of the deck is 8 feet and the length of the deck is 8 + 10 or 18 feet.

Find the value of c that makes each trinomial a perfect square.

10. x2 + 26x + c

SOLUTION:  In this trinomial, b = 26.

So, c must be 169 to make the trinomial a perfect square.

11. x2 − 24x + c

SOLUTION:  In this trinomial, b = –24.

So, c must be 144 to make the trinomial a perfect square.

12. x2 − 19x + c

SOLUTION:  In this trinomial, b = –19.

So, c must be  to make the trinomial a perfect square.

13. x2 + 17x + c

SOLUTION:  In this trinomial, b = 17.

So, c must be  to make the trinomial a perfect square.

14. x2 + 5x + c

SOLUTION:  In this trinomial, b = 5.

So, c must be  to make the trinomial a perfect square.

15. x2 − 13x + c

SOLUTION:  In this trinomial, b = –13.

So, c must be  to make the trinomial a perfect square.

16. x2 − 22x + c

SOLUTION:  In this trinomial, b = –22.

So, c must be 121 to make the trinomial a perfect square.

17. x2 − 15x + c

SOLUTION:  In this trinomial, b = –15.

So, c must be  to make the trinomial a perfect square.

18. x2 + 24x + c

SOLUTION:  In this trinomial, b = 24.

So, c must be 144 to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

19. x2 + 6x − 16 = 0

SOLUTION:  

  The solutions are –8 and 2.

20. x2 − 2x − 14 = 0

SOLUTION:  

  The solutions are about –2.9 and 4.9.

21. x2 − 8x − 1 = 8

SOLUTION:  

  The solutions are –1 and 9.

22. x2 + 3x + 21 = 22

SOLUTION:  

  The solutions are about –3.3 and 0.3.

23. x2 − 11x + 3 = 5

SOLUTION:  

  The solutions are about –0.2 and 11.2.

24. 5x2 − 10x = 23

SOLUTION:  

  The solutions are about –1.4 and 3.4.

25. 2x2 − 2x + 7 = 5

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

26. 3x2 + 12x + 81 = 15

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

27. 4x2 + 6x = 12

SOLUTION:  

  The solutions are about –2.6 and 1.1.

28. 4x2 + 5 = 10x

SOLUTION:  

  The solutions are about 0.7 and 1.8.

29. −2x2 + 10x = −14

SOLUTION:  

  The solutions are about –1.1 and 6.1.

30. −3x2 − 12 = 14x

SOLUTION:  

  The solutions are about –3.5 and –1.1.

31. STOCK The price p in dollars for a particular stock can be modeled by the quadratic equation p = 3.5t − 0.05t2, wh

the number of days after the stock is purchased. When is the stock worth $60?

SOLUTION:  Let p = 60.

  The stock is worth $60 on the 30th and 40th days after purchase.

GEOMETRY  Find the value of x for each figure. Round to the nearest tenth if necessary.

32. area = 45 in2

SOLUTION:  

  The height cannot be negative. So, x ≈ 6.3.

33. area = 110 ft2

SOLUTION:  

  The dimensions of the rectangle cannot be negative. So, x ≈ 5.3.

34. NUMBER THEORY The product of two consecutive even integers is 224. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers are 14 and 16 or –16 and –14.

35. CCSS PRECISION The product of two consecutive negative odd integers is 483. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers must be negative. So, they are –23 and –21.

36. GEOMETRY Find the area of the triangle.

SOLUTION:  

  The dimensions of the triangle cannot be negative. So, x = 18. The base of the triangle is 18 meters and the height is 18 + 6 or 24 meters.

The area of the triangle is 216 square meters.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

37. 0.2x2 − 0.2x − 0.4 = 0

SOLUTION:  

  The solutions are –1 and 2.

38. 0.5x2 = 2x − 0.3

SOLUTION:  

  The solutions are about 0.2 and 3.8.

39. 2x2 −

SOLUTION:  

  The solutions are about 0.2 and 0.9.

40. 

SOLUTION:  

  The solutions are –0.5 and 2.5.

41. 

SOLUTION:  

  The solutions are about –8.2 and 0.2.

42. 

SOLUTION:  

  The solutions are about –5.1 and 0.1.

43. ASTRONOMY The height of an object t seconds after it is dropped is given by the equation ,

where h0 is the initial height and g is the acceleration due to gravity. The acceleration due to gravity near the surface

of Mars is 3.73 m/s2, while on Earth it is 9.8 m/s

2. Suppose an object is dropped from an initial height of 120 meters

above the surface of each planet. a. On which planet would the object reach the ground first? b. How long would it take the object to reach the ground on each planet? Round each answer to the nearest tenth. c. Do the times that it takes the object to reach the ground seem reasonable? Explain your reasoning.

SOLUTION:  a. The object on Earth will reach the ground first because it is falling at a faster rate.   b. Mars:

So, t ≈ 8.0 seconds. Earth:

So, t ≈ 4.9 seconds.   c. Sample answer: Yes; the acceleration due to gravity is much greater on Earth than on Mars, so the time to reach the ground should be much less.

44. Find all values of c that make x2 + cx + 100 a perfect square trinomial.

SOLUTION:  

So, c can be –20 or 20 to make the trinomial a perfect square.

45. Find all values of c that make x2 + cx + 225 a perfect square trinomial.

SOLUTION:  

So, c can be –30 or 30 to make the trinomial a perfect square.

46. PAINTING Before she begins painting a picture, Donna stretches her canvas over a wood frame. The frame has a length of 60 inches and a width of 4 inches. She has enough canvas to cover 480 square inches. Donna decides to increase the dimensions of the frame. If the increase in the length is 10 times the increase in the width, what will the dimensions of the frame be?

SOLUTION:  Let x = the increase in the width and let 10x = the increase in the length. So, x + 4 = the new width and 10x + 60 = the new length.  

  The dimensions cannot be negative, so x = 2. The width is 2 + 4 or 6 inches and the length is 10(2) + 60 or 80 inches.

47. MULTIPLE REPRESENTATIONS In this problem, you will investigate a property of quadratic equations. a. TABULAR  Copy the table shown and complete the second column.

b. ALGEBRAIC Set each trinomial equal to zero, and solve the equation by completing the square. Complete the last column of the table with the number of roots of each equation.

c. VERBAL Compare the number of roots of each equation to the result in the b2 − 4ac column. Is there a

relationship between these values? If so, describe it.

d. ANALYTICAL Predict how many solutions 2x2 − 9x + 15 = 0 will have.  Verify your prediction by solving the 

equation.

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3    

3x2 + 6x + 9    

x2 − 2x + 7    

x2 + 10x + 25    

x2 + 3x − 12    

SOLUTION:  a.

b.

The trinomial x2 − 8x + 16 has 1 root.

 

The trinomial 2x

2 − 11x + 3 has 2 roots.

 

The square root of a negative number has no real roots. So, the trinomial 3x2 + 6x + 9 has 0 roots.

 

The square root of a negative number has no real roots. So, the trinomial x2 − 2x + 7.

 

The trinomial x2 + 10x + 25 has 1 root.

 

The trinomial x2 + 3x − 12 has 2 roots.

 

c. If b2 − 4ac is negative, the equation has no real solutions. If b

2 − 4ac is zero, the equation has one solution. If b2

− 4ac is positive, the equation has 2 solutions. d.

The equation 2x2 − 9x + 15 = 0 has 0 real solutions because b

2 − 4ac is negative.

 

The equation cannot be solved because the square root of a negative number has no real roots.

Trinomial b2 − 4ac Number

of Roots

x2 − 8x + 16 (–8)

2 – 4(1)(16) = 0 1

2x2 − 11x + 3 (–11)

2 – 4(2)(3) = 97  

3x2 + 6x + 9 (6)

2 – 4(3)(9) = −72  

x2 − 2x + 7 (–2)

2 – 4(1)(7) = −24  

x2 + 10x + 25 (10)

2 – 4(1)(25) = 0  

x2 + 3x − 12 (3)

2 – 4(1)(–12) = 57  

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3 97 2

3x2 + 6x + 9 −72 0

x2 − 2x + 7 −24 0

x2 + 10x + 25 0 1

x2 + 3x − 12 57 2

48. CCSS PERSEVERANCE Given y = ax2 + bx + c with a ≠ 0, derive the equation for the axis of symmetry by 

completing the square and rewriting the equation in the form y = a(x − h)2 + k .

SOLUTION:  

Let    and   , then the equation becomes y = (x - h)2 + k .

For an equation in the form y = ax2 + bx + c,  the equation for the axis of symmetry is   .

So, for an equation in the form  y = (x - h)2 + k, the equation for the axis of symmetry becomes x = h.

49. REASONING Determine the number of solutions x2 + bx = c has if . Explain.

SOLUTION:  

None; Sample answer: If you add to each side of the equation and each side of the inequality, you get

. Since the left side of the last equation is a perfect square, it cannot

equal the negative number . So, there are no real solutions.

50. WHICH ONE DOESN’T BELONG? Identify the expression that does not belong with the other three. Explain your reasoning.

 

SOLUTION:  The first 3 trinomials are perfect squares.

The trinomial  is not a perfect square. So, it does not belong.

51. OPEN ENDED Write a quadratic equation for which the only solution is 4.

SOLUTION:  Find a quadratic that has two roots of 4.

Verify that the solution is 4.

So, the quadratic equation x2 − 8x + 16 = 0 has a solution of 4.

52. WRITING IN MATH Compare and contrast the following strategies for solving x2 − 5x − 7 = 0: completing the

square, graphing, and factoring.

SOLUTION:  Sample answer: Because the leading coefficient is 1, solving the equation by completing the square is simpler and yields exact answers for the solutions.   

 

To solve the equation by graphing, use a graphing calculator to graph the related function y = x2 - 5x - 7. Select the

zero option from the 2nd [CALC] menu to determine the roots. 

   

  The roots are given as decimal approximations. So, for this equation the solutions would have to be given as estimations.   There are no factors of –7 that have a sum of –5, so solving by factoring is not possible.

53. The length of a rectangle is 3 times its width. The area of the rectangle is 75 square feet. Find the length of the rectangle in feet. A 25 B 15 C 10 D 5

SOLUTION:  Let x = the width of the rectangle and let 3x = the length of the rectangle.

The length of the rectangle is 3(5) or 15 feet. Choice B is the correct answer.

54. PROBABILITY At a festival, winners of a game draw a token for a prize. There is one token for each prize. The prizes include 9 movie passes, 8 stuffed animals, 5 hats, 10 jump ropes, and 4 glow necklaces. What is the probabilitythat the first person to draw a token will win a movie pass?

F  

G

H

J  

SOLUTION:  There are 9 + 8 + 5 + 10 + 4 or 36 possible outcomes.

P(movie pass) =  or  . Choice J is the correct answer.

55. GRIDDED RESPONSE The population of a town can be modeled by P = 22,000 + 125t, where P represents the population and t represents the number of years from 2000. How many years after 2000 will the population be 26,000?

SOLUTION:  

In 32 years the population of the town will be 26,000.

56. Percy delivers pizzas for Pizza King. He is paid $6 an hour plus $2.50 for each pizza he delivers. Percy earned $280 last week. If he worked a total of 30 hours, how many pizzas did he deliver? A 250 pizzas B 184 pizzas C 40 pizzas D 34 pizzas

SOLUTION:  Let p = the number of pizzas Percy delivered.

Percy delivered 40 pizzas. Choice C is the correct answer.

Describe how the graph of each function is related to the graph of f (x) = x2.

57. g(x) = −12 + x2

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –12, the translation is

down. So, the graph is shifted down 12 units from the parent function. 

58. h(x) = (x + 2)2

SOLUTION:  

The graph of f (x) = (x – c)2 represents a translation left or right from the parent graph. Since c = –2, the translation

is to the left by 2 units. 

 

59. g(x) = 2x2 + 5

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 2 and c = 5. Since 2 > 0 and  > 1, the graph of y = 2x

2 + 5

is the graph of y = x2 vertically stretched and shifted up 5 units.

60. 

SOLUTION:  

The function can be written f (x) = a(x – b)2, where a = and b = 6. Since > 0 and  < 1, the graph of 

is the graph of y = x2 vertically compressed and shifted right 6 units.

61. g(x) = 6 + x2

SOLUTION:  

The function can be written f (x) = ax2 + c, where a =  and c = 6. Since  > 0 and   > 1, the graph of y = 6 +

x2 is the graph of y = x

2 vertically stretched and shifted up 6 units.

62. h(x) = − 1 −   x2

SOLUTION:  

The function can be written f (x) = –ax2 + c, where a =  and c = –1. Since  < 0 and   > 1, the graph of y

= –1 – x2 is the graph of y = x

2 vertically stretched, shifted down 1 unit and reflected across the x-axis.

63. RIDES A popular amusement park ride whisks riders to the top of a 250-foot tower and drops them. A function for

the height of a rider is h = −16t2 + 250, where h is the height and t is the time in seconds. The ride stops the descent

of the rider 40 feet above the ground. Write an equation that models the drop of the rider. How long does it take to fall from 250 feet to 40 feet?

SOLUTION:  

The ride stops the descent of the rider at 40 feet above ground, so h = 40. Then, the equation 40 = −16t2 + 250

models the drop of the rider.

Time cannot be negative. So, it takes about 3.6 seconds to complete the ride.

Simplify. Assume that no denominator is equal to zero.

64. 

SOLUTION:  

65. 

SOLUTION:  

66. 

SOLUTION:  

67. 

SOLUTION:  

68. 

SOLUTION:  

69. b3(m

−3)(b

−6)

SOLUTION:  

Solve each open sentence.

70. |y − 2| > 7

SOLUTION:  

The solution set is {y |y > 9 or y < −5}.

Case 1 y – 2 is positive.

and Case 2 y – 2 is negative.

 

71. |z + 5| < 3

SOLUTION:  

The solution set is {z |−8 < z < −2}.

Case 1 z +5 is positive.

and Case 2 z +5 is negative.

 

72. |2b + 7| ≤ −6

SOLUTION:  

cannot be negative. So,  cannot be less than or equal to –6. The solution set is empty.

73. |3 − 2y | ≥ 8

SOLUTION:  

The solution set is {y |y  ≥ 5.5 or y  ≤ −2.5}.

Case 1 3 – 2y is positive.

and Case 2 3 – 2y is negative.

 

74. |9 − 4m| < −1

SOLUTION:  

cannot be negative. So, cannot be less than or equal to –1. The solution set is empty.

75. |5c − 2| ≤ 13

SOLUTION:  

The solution set is {c|−2.2 ≤ c ≤ 3}.

Case 1 5c – 2 is positive.

and Case 2 5c – 2 is negative.

 

Evaluate  for each set of values. Round to the nearest tenth if necessary.

76. a = 2, b = −5, c = 2

SOLUTION:  

77. a = 1, b = 12, c = 11

SOLUTION:  

78. a = −9, b = 10, c = −1

SOLUTION:  

79. a = 1, b = 7, c = −3

SOLUTION:  

80. a = 2, b = −4, c = −6

SOLUTION:  

81. a = 3, b = 1, c = 2

SOLUTION:  

This value is not a real number.

eSolutions Manual - Powered by Cognero Page 19

9-4 Solving Quadratic Equations by Completing the Square

Find the value of c that makes each trinomial a perfect square.

1. x2 − 18x + c

SOLUTION:  In this trinomial, b = –18.

So, c must be 81 to make the trinomial a perfect square.

2. x2 + 22x + c

SOLUTION:  In this trinomial, b = 22.

So, c must be 121 to make the trinomial a perfect square.

3. x2 + 9x + c

SOLUTION:  In this trinomial, b = 9.

 

So, c must be  to make the trinomial a perfect square.

4. x2 − 7x + c

SOLUTION:  In this trinomial, b = –7.

 

So, c must be  to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

5. x2 + 4x = 6

SOLUTION:  

The solutions are about –5.2 and 1.2.

6. x2 − 8x = −9

SOLUTION:  

The solutions are about 1.4 and 6.6.

7. 4x2 + 9x − 1 = 0

SOLUTION:  

The solutions are about –2.4 and 0.1.

8. −2x2 + 10x + 22 = 4

SOLUTION:  

The solutions are about –1.4 and 6.4.

9. CCSS MODELING Collin is building a deck on the back of his family’s house. He has enough lumber for the deck to be 144 square feet. The length should be 10 feet more than its width. What should the dimensions of the deck be?

SOLUTION:  Let x = the width of the deck and x + 10 = the length of the deck.

The dimensions of the deck can not be negative. So, x = –5 + 13 or 8. The width of the deck is 8 feet and the length of the deck is 8 + 10 or 18 feet.

Find the value of c that makes each trinomial a perfect square.

10. x2 + 26x + c

SOLUTION:  In this trinomial, b = 26.

So, c must be 169 to make the trinomial a perfect square.

11. x2 − 24x + c

SOLUTION:  In this trinomial, b = –24.

So, c must be 144 to make the trinomial a perfect square.

12. x2 − 19x + c

SOLUTION:  In this trinomial, b = –19.

So, c must be  to make the trinomial a perfect square.

13. x2 + 17x + c

SOLUTION:  In this trinomial, b = 17.

So, c must be  to make the trinomial a perfect square.

14. x2 + 5x + c

SOLUTION:  In this trinomial, b = 5.

So, c must be  to make the trinomial a perfect square.

15. x2 − 13x + c

SOLUTION:  In this trinomial, b = –13.

So, c must be  to make the trinomial a perfect square.

16. x2 − 22x + c

SOLUTION:  In this trinomial, b = –22.

So, c must be 121 to make the trinomial a perfect square.

17. x2 − 15x + c

SOLUTION:  In this trinomial, b = –15.

So, c must be  to make the trinomial a perfect square.

18. x2 + 24x + c

SOLUTION:  In this trinomial, b = 24.

So, c must be 144 to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

19. x2 + 6x − 16 = 0

SOLUTION:  

  The solutions are –8 and 2.

20. x2 − 2x − 14 = 0

SOLUTION:  

  The solutions are about –2.9 and 4.9.

21. x2 − 8x − 1 = 8

SOLUTION:  

  The solutions are –1 and 9.

22. x2 + 3x + 21 = 22

SOLUTION:  

  The solutions are about –3.3 and 0.3.

23. x2 − 11x + 3 = 5

SOLUTION:  

  The solutions are about –0.2 and 11.2.

24. 5x2 − 10x = 23

SOLUTION:  

  The solutions are about –1.4 and 3.4.

25. 2x2 − 2x + 7 = 5

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

26. 3x2 + 12x + 81 = 15

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

27. 4x2 + 6x = 12

SOLUTION:  

  The solutions are about –2.6 and 1.1.

28. 4x2 + 5 = 10x

SOLUTION:  

  The solutions are about 0.7 and 1.8.

29. −2x2 + 10x = −14

SOLUTION:  

  The solutions are about –1.1 and 6.1.

30. −3x2 − 12 = 14x

SOLUTION:  

  The solutions are about –3.5 and –1.1.

31. STOCK The price p in dollars for a particular stock can be modeled by the quadratic equation p = 3.5t − 0.05t2, wh

the number of days after the stock is purchased. When is the stock worth $60?

SOLUTION:  Let p = 60.

  The stock is worth $60 on the 30th and 40th days after purchase.

GEOMETRY  Find the value of x for each figure. Round to the nearest tenth if necessary.

32. area = 45 in2

SOLUTION:  

  The height cannot be negative. So, x ≈ 6.3.

33. area = 110 ft2

SOLUTION:  

  The dimensions of the rectangle cannot be negative. So, x ≈ 5.3.

34. NUMBER THEORY The product of two consecutive even integers is 224. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers are 14 and 16 or –16 and –14.

35. CCSS PRECISION The product of two consecutive negative odd integers is 483. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers must be negative. So, they are –23 and –21.

36. GEOMETRY Find the area of the triangle.

SOLUTION:  

  The dimensions of the triangle cannot be negative. So, x = 18. The base of the triangle is 18 meters and the height is 18 + 6 or 24 meters.

The area of the triangle is 216 square meters.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

37. 0.2x2 − 0.2x − 0.4 = 0

SOLUTION:  

  The solutions are –1 and 2.

38. 0.5x2 = 2x − 0.3

SOLUTION:  

  The solutions are about 0.2 and 3.8.

39. 2x2 −

SOLUTION:  

  The solutions are about 0.2 and 0.9.

40. 

SOLUTION:  

  The solutions are –0.5 and 2.5.

41. 

SOLUTION:  

  The solutions are about –8.2 and 0.2.

42. 

SOLUTION:  

  The solutions are about –5.1 and 0.1.

43. ASTRONOMY The height of an object t seconds after it is dropped is given by the equation ,

where h0 is the initial height and g is the acceleration due to gravity. The acceleration due to gravity near the surface

of Mars is 3.73 m/s2, while on Earth it is 9.8 m/s

2. Suppose an object is dropped from an initial height of 120 meters

above the surface of each planet. a. On which planet would the object reach the ground first? b. How long would it take the object to reach the ground on each planet? Round each answer to the nearest tenth. c. Do the times that it takes the object to reach the ground seem reasonable? Explain your reasoning.

SOLUTION:  a. The object on Earth will reach the ground first because it is falling at a faster rate.   b. Mars:

So, t ≈ 8.0 seconds. Earth:

So, t ≈ 4.9 seconds.   c. Sample answer: Yes; the acceleration due to gravity is much greater on Earth than on Mars, so the time to reach the ground should be much less.

44. Find all values of c that make x2 + cx + 100 a perfect square trinomial.

SOLUTION:  

So, c can be –20 or 20 to make the trinomial a perfect square.

45. Find all values of c that make x2 + cx + 225 a perfect square trinomial.

SOLUTION:  

So, c can be –30 or 30 to make the trinomial a perfect square.

46. PAINTING Before she begins painting a picture, Donna stretches her canvas over a wood frame. The frame has a length of 60 inches and a width of 4 inches. She has enough canvas to cover 480 square inches. Donna decides to increase the dimensions of the frame. If the increase in the length is 10 times the increase in the width, what will the dimensions of the frame be?

SOLUTION:  Let x = the increase in the width and let 10x = the increase in the length. So, x + 4 = the new width and 10x + 60 = the new length.  

  The dimensions cannot be negative, so x = 2. The width is 2 + 4 or 6 inches and the length is 10(2) + 60 or 80 inches.

47. MULTIPLE REPRESENTATIONS In this problem, you will investigate a property of quadratic equations. a. TABULAR  Copy the table shown and complete the second column.

b. ALGEBRAIC Set each trinomial equal to zero, and solve the equation by completing the square. Complete the last column of the table with the number of roots of each equation.

c. VERBAL Compare the number of roots of each equation to the result in the b2 − 4ac column. Is there a

relationship between these values? If so, describe it.

d. ANALYTICAL Predict how many solutions 2x2 − 9x + 15 = 0 will have.  Verify your prediction by solving the 

equation.

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3    

3x2 + 6x + 9    

x2 − 2x + 7    

x2 + 10x + 25    

x2 + 3x − 12    

SOLUTION:  a.

b.

The trinomial x2 − 8x + 16 has 1 root.

 

The trinomial 2x

2 − 11x + 3 has 2 roots.

 

The square root of a negative number has no real roots. So, the trinomial 3x2 + 6x + 9 has 0 roots.

 

The square root of a negative number has no real roots. So, the trinomial x2 − 2x + 7.

 

The trinomial x2 + 10x + 25 has 1 root.

 

The trinomial x2 + 3x − 12 has 2 roots.

 

c. If b2 − 4ac is negative, the equation has no real solutions. If b

2 − 4ac is zero, the equation has one solution. If b2

− 4ac is positive, the equation has 2 solutions. d.

The equation 2x2 − 9x + 15 = 0 has 0 real solutions because b

2 − 4ac is negative.

 

The equation cannot be solved because the square root of a negative number has no real roots.

Trinomial b2 − 4ac Number

of Roots

x2 − 8x + 16 (–8)

2 – 4(1)(16) = 0 1

2x2 − 11x + 3 (–11)

2 – 4(2)(3) = 97  

3x2 + 6x + 9 (6)

2 – 4(3)(9) = −72  

x2 − 2x + 7 (–2)

2 – 4(1)(7) = −24  

x2 + 10x + 25 (10)

2 – 4(1)(25) = 0  

x2 + 3x − 12 (3)

2 – 4(1)(–12) = 57  

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3 97 2

3x2 + 6x + 9 −72 0

x2 − 2x + 7 −24 0

x2 + 10x + 25 0 1

x2 + 3x − 12 57 2

48. CCSS PERSEVERANCE Given y = ax2 + bx + c with a ≠ 0, derive the equation for the axis of symmetry by 

completing the square and rewriting the equation in the form y = a(x − h)2 + k .

SOLUTION:  

Let    and   , then the equation becomes y = (x - h)2 + k .

For an equation in the form y = ax2 + bx + c,  the equation for the axis of symmetry is   .

So, for an equation in the form  y = (x - h)2 + k, the equation for the axis of symmetry becomes x = h.

49. REASONING Determine the number of solutions x2 + bx = c has if . Explain.

SOLUTION:  

None; Sample answer: If you add to each side of the equation and each side of the inequality, you get

. Since the left side of the last equation is a perfect square, it cannot

equal the negative number . So, there are no real solutions.

50. WHICH ONE DOESN’T BELONG? Identify the expression that does not belong with the other three. Explain your reasoning.

 

SOLUTION:  The first 3 trinomials are perfect squares.

The trinomial  is not a perfect square. So, it does not belong.

51. OPEN ENDED Write a quadratic equation for which the only solution is 4.

SOLUTION:  Find a quadratic that has two roots of 4.

Verify that the solution is 4.

So, the quadratic equation x2 − 8x + 16 = 0 has a solution of 4.

52. WRITING IN MATH Compare and contrast the following strategies for solving x2 − 5x − 7 = 0: completing the

square, graphing, and factoring.

SOLUTION:  Sample answer: Because the leading coefficient is 1, solving the equation by completing the square is simpler and yields exact answers for the solutions.   

 

To solve the equation by graphing, use a graphing calculator to graph the related function y = x2 - 5x - 7. Select the

zero option from the 2nd [CALC] menu to determine the roots. 

   

  The roots are given as decimal approximations. So, for this equation the solutions would have to be given as estimations.   There are no factors of –7 that have a sum of –5, so solving by factoring is not possible.

53. The length of a rectangle is 3 times its width. The area of the rectangle is 75 square feet. Find the length of the rectangle in feet. A 25 B 15 C 10 D 5

SOLUTION:  Let x = the width of the rectangle and let 3x = the length of the rectangle.

The length of the rectangle is 3(5) or 15 feet. Choice B is the correct answer.

54. PROBABILITY At a festival, winners of a game draw a token for a prize. There is one token for each prize. The prizes include 9 movie passes, 8 stuffed animals, 5 hats, 10 jump ropes, and 4 glow necklaces. What is the probabilitythat the first person to draw a token will win a movie pass?

F  

G

H

J  

SOLUTION:  There are 9 + 8 + 5 + 10 + 4 or 36 possible outcomes.

P(movie pass) =  or  . Choice J is the correct answer.

55. GRIDDED RESPONSE The population of a town can be modeled by P = 22,000 + 125t, where P represents the population and t represents the number of years from 2000. How many years after 2000 will the population be 26,000?

SOLUTION:  

In 32 years the population of the town will be 26,000.

56. Percy delivers pizzas for Pizza King. He is paid $6 an hour plus $2.50 for each pizza he delivers. Percy earned $280 last week. If he worked a total of 30 hours, how many pizzas did he deliver? A 250 pizzas B 184 pizzas C 40 pizzas D 34 pizzas

SOLUTION:  Let p = the number of pizzas Percy delivered.

Percy delivered 40 pizzas. Choice C is the correct answer.

Describe how the graph of each function is related to the graph of f (x) = x2.

57. g(x) = −12 + x2

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –12, the translation is

down. So, the graph is shifted down 12 units from the parent function. 

58. h(x) = (x + 2)2

SOLUTION:  

The graph of f (x) = (x – c)2 represents a translation left or right from the parent graph. Since c = –2, the translation

is to the left by 2 units. 

 

59. g(x) = 2x2 + 5

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 2 and c = 5. Since 2 > 0 and  > 1, the graph of y = 2x

2 + 5

is the graph of y = x2 vertically stretched and shifted up 5 units.

60. 

SOLUTION:  

The function can be written f (x) = a(x – b)2, where a = and b = 6. Since > 0 and  < 1, the graph of 

is the graph of y = x2 vertically compressed and shifted right 6 units.

61. g(x) = 6 + x2

SOLUTION:  

The function can be written f (x) = ax2 + c, where a =  and c = 6. Since  > 0 and   > 1, the graph of y = 6 +

x2 is the graph of y = x

2 vertically stretched and shifted up 6 units.

62. h(x) = − 1 −   x2

SOLUTION:  

The function can be written f (x) = –ax2 + c, where a =  and c = –1. Since  < 0 and   > 1, the graph of y

= –1 – x2 is the graph of y = x

2 vertically stretched, shifted down 1 unit and reflected across the x-axis.

63. RIDES A popular amusement park ride whisks riders to the top of a 250-foot tower and drops them. A function for

the height of a rider is h = −16t2 + 250, where h is the height and t is the time in seconds. The ride stops the descent

of the rider 40 feet above the ground. Write an equation that models the drop of the rider. How long does it take to fall from 250 feet to 40 feet?

SOLUTION:  

The ride stops the descent of the rider at 40 feet above ground, so h = 40. Then, the equation 40 = −16t2 + 250

models the drop of the rider.

Time cannot be negative. So, it takes about 3.6 seconds to complete the ride.

Simplify. Assume that no denominator is equal to zero.

64. 

SOLUTION:  

65. 

SOLUTION:  

66. 

SOLUTION:  

67. 

SOLUTION:  

68. 

SOLUTION:  

69. b3(m

−3)(b

−6)

SOLUTION:  

Solve each open sentence.

70. |y − 2| > 7

SOLUTION:  

The solution set is {y |y > 9 or y < −5}.

Case 1 y – 2 is positive.

and Case 2 y – 2 is negative.

 

71. |z + 5| < 3

SOLUTION:  

The solution set is {z |−8 < z < −2}.

Case 1 z +5 is positive.

and Case 2 z +5 is negative.

 

72. |2b + 7| ≤ −6

SOLUTION:  

cannot be negative. So,  cannot be less than or equal to –6. The solution set is empty.

73. |3 − 2y | ≥ 8

SOLUTION:  

The solution set is {y |y  ≥ 5.5 or y  ≤ −2.5}.

Case 1 3 – 2y is positive.

and Case 2 3 – 2y is negative.

 

74. |9 − 4m| < −1

SOLUTION:  

cannot be negative. So, cannot be less than or equal to –1. The solution set is empty.

75. |5c − 2| ≤ 13

SOLUTION:  

The solution set is {c|−2.2 ≤ c ≤ 3}.

Case 1 5c – 2 is positive.

and Case 2 5c – 2 is negative.

 

Evaluate  for each set of values. Round to the nearest tenth if necessary.

76. a = 2, b = −5, c = 2

SOLUTION:  

77. a = 1, b = 12, c = 11

SOLUTION:  

78. a = −9, b = 10, c = −1

SOLUTION:  

79. a = 1, b = 7, c = −3

SOLUTION:  

80. a = 2, b = −4, c = −6

SOLUTION:  

81. a = 3, b = 1, c = 2

SOLUTION:  

This value is not a real number.

eSolutions Manual - Powered by Cognero Page 20

9-4 Solving Quadratic Equations by Completing the Square

Find the value of c that makes each trinomial a perfect square.

1. x2 − 18x + c

SOLUTION:  In this trinomial, b = –18.

So, c must be 81 to make the trinomial a perfect square.

2. x2 + 22x + c

SOLUTION:  In this trinomial, b = 22.

So, c must be 121 to make the trinomial a perfect square.

3. x2 + 9x + c

SOLUTION:  In this trinomial, b = 9.

 

So, c must be  to make the trinomial a perfect square.

4. x2 − 7x + c

SOLUTION:  In this trinomial, b = –7.

 

So, c must be  to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

5. x2 + 4x = 6

SOLUTION:  

The solutions are about –5.2 and 1.2.

6. x2 − 8x = −9

SOLUTION:  

The solutions are about 1.4 and 6.6.

7. 4x2 + 9x − 1 = 0

SOLUTION:  

The solutions are about –2.4 and 0.1.

8. −2x2 + 10x + 22 = 4

SOLUTION:  

The solutions are about –1.4 and 6.4.

9. CCSS MODELING Collin is building a deck on the back of his family’s house. He has enough lumber for the deck to be 144 square feet. The length should be 10 feet more than its width. What should the dimensions of the deck be?

SOLUTION:  Let x = the width of the deck and x + 10 = the length of the deck.

The dimensions of the deck can not be negative. So, x = –5 + 13 or 8. The width of the deck is 8 feet and the length of the deck is 8 + 10 or 18 feet.

Find the value of c that makes each trinomial a perfect square.

10. x2 + 26x + c

SOLUTION:  In this trinomial, b = 26.

So, c must be 169 to make the trinomial a perfect square.

11. x2 − 24x + c

SOLUTION:  In this trinomial, b = –24.

So, c must be 144 to make the trinomial a perfect square.

12. x2 − 19x + c

SOLUTION:  In this trinomial, b = –19.

So, c must be  to make the trinomial a perfect square.

13. x2 + 17x + c

SOLUTION:  In this trinomial, b = 17.

So, c must be  to make the trinomial a perfect square.

14. x2 + 5x + c

SOLUTION:  In this trinomial, b = 5.

So, c must be  to make the trinomial a perfect square.

15. x2 − 13x + c

SOLUTION:  In this trinomial, b = –13.

So, c must be  to make the trinomial a perfect square.

16. x2 − 22x + c

SOLUTION:  In this trinomial, b = –22.

So, c must be 121 to make the trinomial a perfect square.

17. x2 − 15x + c

SOLUTION:  In this trinomial, b = –15.

So, c must be  to make the trinomial a perfect square.

18. x2 + 24x + c

SOLUTION:  In this trinomial, b = 24.

So, c must be 144 to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

19. x2 + 6x − 16 = 0

SOLUTION:  

  The solutions are –8 and 2.

20. x2 − 2x − 14 = 0

SOLUTION:  

  The solutions are about –2.9 and 4.9.

21. x2 − 8x − 1 = 8

SOLUTION:  

  The solutions are –1 and 9.

22. x2 + 3x + 21 = 22

SOLUTION:  

  The solutions are about –3.3 and 0.3.

23. x2 − 11x + 3 = 5

SOLUTION:  

  The solutions are about –0.2 and 11.2.

24. 5x2 − 10x = 23

SOLUTION:  

  The solutions are about –1.4 and 3.4.

25. 2x2 − 2x + 7 = 5

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

26. 3x2 + 12x + 81 = 15

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

27. 4x2 + 6x = 12

SOLUTION:  

  The solutions are about –2.6 and 1.1.

28. 4x2 + 5 = 10x

SOLUTION:  

  The solutions are about 0.7 and 1.8.

29. −2x2 + 10x = −14

SOLUTION:  

  The solutions are about –1.1 and 6.1.

30. −3x2 − 12 = 14x

SOLUTION:  

  The solutions are about –3.5 and –1.1.

31. STOCK The price p in dollars for a particular stock can be modeled by the quadratic equation p = 3.5t − 0.05t2, wh

the number of days after the stock is purchased. When is the stock worth $60?

SOLUTION:  Let p = 60.

  The stock is worth $60 on the 30th and 40th days after purchase.

GEOMETRY  Find the value of x for each figure. Round to the nearest tenth if necessary.

32. area = 45 in2

SOLUTION:  

  The height cannot be negative. So, x ≈ 6.3.

33. area = 110 ft2

SOLUTION:  

  The dimensions of the rectangle cannot be negative. So, x ≈ 5.3.

34. NUMBER THEORY The product of two consecutive even integers is 224. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers are 14 and 16 or –16 and –14.

35. CCSS PRECISION The product of two consecutive negative odd integers is 483. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers must be negative. So, they are –23 and –21.

36. GEOMETRY Find the area of the triangle.

SOLUTION:  

  The dimensions of the triangle cannot be negative. So, x = 18. The base of the triangle is 18 meters and the height is 18 + 6 or 24 meters.

The area of the triangle is 216 square meters.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

37. 0.2x2 − 0.2x − 0.4 = 0

SOLUTION:  

  The solutions are –1 and 2.

38. 0.5x2 = 2x − 0.3

SOLUTION:  

  The solutions are about 0.2 and 3.8.

39. 2x2 −

SOLUTION:  

  The solutions are about 0.2 and 0.9.

40. 

SOLUTION:  

  The solutions are –0.5 and 2.5.

41. 

SOLUTION:  

  The solutions are about –8.2 and 0.2.

42. 

SOLUTION:  

  The solutions are about –5.1 and 0.1.

43. ASTRONOMY The height of an object t seconds after it is dropped is given by the equation ,

where h0 is the initial height and g is the acceleration due to gravity. The acceleration due to gravity near the surface

of Mars is 3.73 m/s2, while on Earth it is 9.8 m/s

2. Suppose an object is dropped from an initial height of 120 meters

above the surface of each planet. a. On which planet would the object reach the ground first? b. How long would it take the object to reach the ground on each planet? Round each answer to the nearest tenth. c. Do the times that it takes the object to reach the ground seem reasonable? Explain your reasoning.

SOLUTION:  a. The object on Earth will reach the ground first because it is falling at a faster rate.   b. Mars:

So, t ≈ 8.0 seconds. Earth:

So, t ≈ 4.9 seconds.   c. Sample answer: Yes; the acceleration due to gravity is much greater on Earth than on Mars, so the time to reach the ground should be much less.

44. Find all values of c that make x2 + cx + 100 a perfect square trinomial.

SOLUTION:  

So, c can be –20 or 20 to make the trinomial a perfect square.

45. Find all values of c that make x2 + cx + 225 a perfect square trinomial.

SOLUTION:  

So, c can be –30 or 30 to make the trinomial a perfect square.

46. PAINTING Before she begins painting a picture, Donna stretches her canvas over a wood frame. The frame has a length of 60 inches and a width of 4 inches. She has enough canvas to cover 480 square inches. Donna decides to increase the dimensions of the frame. If the increase in the length is 10 times the increase in the width, what will the dimensions of the frame be?

SOLUTION:  Let x = the increase in the width and let 10x = the increase in the length. So, x + 4 = the new width and 10x + 60 = the new length.  

  The dimensions cannot be negative, so x = 2. The width is 2 + 4 or 6 inches and the length is 10(2) + 60 or 80 inches.

47. MULTIPLE REPRESENTATIONS In this problem, you will investigate a property of quadratic equations. a. TABULAR  Copy the table shown and complete the second column.

b. ALGEBRAIC Set each trinomial equal to zero, and solve the equation by completing the square. Complete the last column of the table with the number of roots of each equation.

c. VERBAL Compare the number of roots of each equation to the result in the b2 − 4ac column. Is there a

relationship between these values? If so, describe it.

d. ANALYTICAL Predict how many solutions 2x2 − 9x + 15 = 0 will have.  Verify your prediction by solving the 

equation.

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3    

3x2 + 6x + 9    

x2 − 2x + 7    

x2 + 10x + 25    

x2 + 3x − 12    

SOLUTION:  a.

b.

The trinomial x2 − 8x + 16 has 1 root.

 

The trinomial 2x

2 − 11x + 3 has 2 roots.

 

The square root of a negative number has no real roots. So, the trinomial 3x2 + 6x + 9 has 0 roots.

 

The square root of a negative number has no real roots. So, the trinomial x2 − 2x + 7.

 

The trinomial x2 + 10x + 25 has 1 root.

 

The trinomial x2 + 3x − 12 has 2 roots.

 

c. If b2 − 4ac is negative, the equation has no real solutions. If b

2 − 4ac is zero, the equation has one solution. If b2

− 4ac is positive, the equation has 2 solutions. d.

The equation 2x2 − 9x + 15 = 0 has 0 real solutions because b

2 − 4ac is negative.

 

The equation cannot be solved because the square root of a negative number has no real roots.

Trinomial b2 − 4ac Number

of Roots

x2 − 8x + 16 (–8)

2 – 4(1)(16) = 0 1

2x2 − 11x + 3 (–11)

2 – 4(2)(3) = 97  

3x2 + 6x + 9 (6)

2 – 4(3)(9) = −72  

x2 − 2x + 7 (–2)

2 – 4(1)(7) = −24  

x2 + 10x + 25 (10)

2 – 4(1)(25) = 0  

x2 + 3x − 12 (3)

2 – 4(1)(–12) = 57  

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3 97 2

3x2 + 6x + 9 −72 0

x2 − 2x + 7 −24 0

x2 + 10x + 25 0 1

x2 + 3x − 12 57 2

48. CCSS PERSEVERANCE Given y = ax2 + bx + c with a ≠ 0, derive the equation for the axis of symmetry by 

completing the square and rewriting the equation in the form y = a(x − h)2 + k .

SOLUTION:  

Let    and   , then the equation becomes y = (x - h)2 + k .

For an equation in the form y = ax2 + bx + c,  the equation for the axis of symmetry is   .

So, for an equation in the form  y = (x - h)2 + k, the equation for the axis of symmetry becomes x = h.

49. REASONING Determine the number of solutions x2 + bx = c has if . Explain.

SOLUTION:  

None; Sample answer: If you add to each side of the equation and each side of the inequality, you get

. Since the left side of the last equation is a perfect square, it cannot

equal the negative number . So, there are no real solutions.

50. WHICH ONE DOESN’T BELONG? Identify the expression that does not belong with the other three. Explain your reasoning.

 

SOLUTION:  The first 3 trinomials are perfect squares.

The trinomial  is not a perfect square. So, it does not belong.

51. OPEN ENDED Write a quadratic equation for which the only solution is 4.

SOLUTION:  Find a quadratic that has two roots of 4.

Verify that the solution is 4.

So, the quadratic equation x2 − 8x + 16 = 0 has a solution of 4.

52. WRITING IN MATH Compare and contrast the following strategies for solving x2 − 5x − 7 = 0: completing the

square, graphing, and factoring.

SOLUTION:  Sample answer: Because the leading coefficient is 1, solving the equation by completing the square is simpler and yields exact answers for the solutions.   

 

To solve the equation by graphing, use a graphing calculator to graph the related function y = x2 - 5x - 7. Select the

zero option from the 2nd [CALC] menu to determine the roots. 

   

  The roots are given as decimal approximations. So, for this equation the solutions would have to be given as estimations.   There are no factors of –7 that have a sum of –5, so solving by factoring is not possible.

53. The length of a rectangle is 3 times its width. The area of the rectangle is 75 square feet. Find the length of the rectangle in feet. A 25 B 15 C 10 D 5

SOLUTION:  Let x = the width of the rectangle and let 3x = the length of the rectangle.

The length of the rectangle is 3(5) or 15 feet. Choice B is the correct answer.

54. PROBABILITY At a festival, winners of a game draw a token for a prize. There is one token for each prize. The prizes include 9 movie passes, 8 stuffed animals, 5 hats, 10 jump ropes, and 4 glow necklaces. What is the probabilitythat the first person to draw a token will win a movie pass?

F  

G

H

J  

SOLUTION:  There are 9 + 8 + 5 + 10 + 4 or 36 possible outcomes.

P(movie pass) =  or  . Choice J is the correct answer.

55. GRIDDED RESPONSE The population of a town can be modeled by P = 22,000 + 125t, where P represents the population and t represents the number of years from 2000. How many years after 2000 will the population be 26,000?

SOLUTION:  

In 32 years the population of the town will be 26,000.

56. Percy delivers pizzas for Pizza King. He is paid $6 an hour plus $2.50 for each pizza he delivers. Percy earned $280 last week. If he worked a total of 30 hours, how many pizzas did he deliver? A 250 pizzas B 184 pizzas C 40 pizzas D 34 pizzas

SOLUTION:  Let p = the number of pizzas Percy delivered.

Percy delivered 40 pizzas. Choice C is the correct answer.

Describe how the graph of each function is related to the graph of f (x) = x2.

57. g(x) = −12 + x2

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –12, the translation is

down. So, the graph is shifted down 12 units from the parent function. 

58. h(x) = (x + 2)2

SOLUTION:  

The graph of f (x) = (x – c)2 represents a translation left or right from the parent graph. Since c = –2, the translation

is to the left by 2 units. 

 

59. g(x) = 2x2 + 5

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 2 and c = 5. Since 2 > 0 and  > 1, the graph of y = 2x

2 + 5

is the graph of y = x2 vertically stretched and shifted up 5 units.

60. 

SOLUTION:  

The function can be written f (x) = a(x – b)2, where a = and b = 6. Since > 0 and  < 1, the graph of 

is the graph of y = x2 vertically compressed and shifted right 6 units.

61. g(x) = 6 + x2

SOLUTION:  

The function can be written f (x) = ax2 + c, where a =  and c = 6. Since  > 0 and   > 1, the graph of y = 6 +

x2 is the graph of y = x

2 vertically stretched and shifted up 6 units.

62. h(x) = − 1 −   x2

SOLUTION:  

The function can be written f (x) = –ax2 + c, where a =  and c = –1. Since  < 0 and   > 1, the graph of y

= –1 – x2 is the graph of y = x

2 vertically stretched, shifted down 1 unit and reflected across the x-axis.

63. RIDES A popular amusement park ride whisks riders to the top of a 250-foot tower and drops them. A function for

the height of a rider is h = −16t2 + 250, where h is the height and t is the time in seconds. The ride stops the descent

of the rider 40 feet above the ground. Write an equation that models the drop of the rider. How long does it take to fall from 250 feet to 40 feet?

SOLUTION:  

The ride stops the descent of the rider at 40 feet above ground, so h = 40. Then, the equation 40 = −16t2 + 250

models the drop of the rider.

Time cannot be negative. So, it takes about 3.6 seconds to complete the ride.

Simplify. Assume that no denominator is equal to zero.

64. 

SOLUTION:  

65. 

SOLUTION:  

66. 

SOLUTION:  

67. 

SOLUTION:  

68. 

SOLUTION:  

69. b3(m

−3)(b

−6)

SOLUTION:  

Solve each open sentence.

70. |y − 2| > 7

SOLUTION:  

The solution set is {y |y > 9 or y < −5}.

Case 1 y – 2 is positive.

and Case 2 y – 2 is negative.

 

71. |z + 5| < 3

SOLUTION:  

The solution set is {z |−8 < z < −2}.

Case 1 z +5 is positive.

and Case 2 z +5 is negative.

 

72. |2b + 7| ≤ −6

SOLUTION:  

cannot be negative. So,  cannot be less than or equal to –6. The solution set is empty.

73. |3 − 2y | ≥ 8

SOLUTION:  

The solution set is {y |y  ≥ 5.5 or y  ≤ −2.5}.

Case 1 3 – 2y is positive.

and Case 2 3 – 2y is negative.

 

74. |9 − 4m| < −1

SOLUTION:  

cannot be negative. So, cannot be less than or equal to –1. The solution set is empty.

75. |5c − 2| ≤ 13

SOLUTION:  

The solution set is {c|−2.2 ≤ c ≤ 3}.

Case 1 5c – 2 is positive.

and Case 2 5c – 2 is negative.

 

Evaluate  for each set of values. Round to the nearest tenth if necessary.

76. a = 2, b = −5, c = 2

SOLUTION:  

77. a = 1, b = 12, c = 11

SOLUTION:  

78. a = −9, b = 10, c = −1

SOLUTION:  

79. a = 1, b = 7, c = −3

SOLUTION:  

80. a = 2, b = −4, c = −6

SOLUTION:  

81. a = 3, b = 1, c = 2

SOLUTION:  

This value is not a real number.

eSolutions Manual - Powered by Cognero Page 21

9-4 Solving Quadratic Equations by Completing the Square

Find the value of c that makes each trinomial a perfect square.

1. x2 − 18x + c

SOLUTION:  In this trinomial, b = –18.

So, c must be 81 to make the trinomial a perfect square.

2. x2 + 22x + c

SOLUTION:  In this trinomial, b = 22.

So, c must be 121 to make the trinomial a perfect square.

3. x2 + 9x + c

SOLUTION:  In this trinomial, b = 9.

 

So, c must be  to make the trinomial a perfect square.

4. x2 − 7x + c

SOLUTION:  In this trinomial, b = –7.

 

So, c must be  to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

5. x2 + 4x = 6

SOLUTION:  

The solutions are about –5.2 and 1.2.

6. x2 − 8x = −9

SOLUTION:  

The solutions are about 1.4 and 6.6.

7. 4x2 + 9x − 1 = 0

SOLUTION:  

The solutions are about –2.4 and 0.1.

8. −2x2 + 10x + 22 = 4

SOLUTION:  

The solutions are about –1.4 and 6.4.

9. CCSS MODELING Collin is building a deck on the back of his family’s house. He has enough lumber for the deck to be 144 square feet. The length should be 10 feet more than its width. What should the dimensions of the deck be?

SOLUTION:  Let x = the width of the deck and x + 10 = the length of the deck.

The dimensions of the deck can not be negative. So, x = –5 + 13 or 8. The width of the deck is 8 feet and the length of the deck is 8 + 10 or 18 feet.

Find the value of c that makes each trinomial a perfect square.

10. x2 + 26x + c

SOLUTION:  In this trinomial, b = 26.

So, c must be 169 to make the trinomial a perfect square.

11. x2 − 24x + c

SOLUTION:  In this trinomial, b = –24.

So, c must be 144 to make the trinomial a perfect square.

12. x2 − 19x + c

SOLUTION:  In this trinomial, b = –19.

So, c must be  to make the trinomial a perfect square.

13. x2 + 17x + c

SOLUTION:  In this trinomial, b = 17.

So, c must be  to make the trinomial a perfect square.

14. x2 + 5x + c

SOLUTION:  In this trinomial, b = 5.

So, c must be  to make the trinomial a perfect square.

15. x2 − 13x + c

SOLUTION:  In this trinomial, b = –13.

So, c must be  to make the trinomial a perfect square.

16. x2 − 22x + c

SOLUTION:  In this trinomial, b = –22.

So, c must be 121 to make the trinomial a perfect square.

17. x2 − 15x + c

SOLUTION:  In this trinomial, b = –15.

So, c must be  to make the trinomial a perfect square.

18. x2 + 24x + c

SOLUTION:  In this trinomial, b = 24.

So, c must be 144 to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

19. x2 + 6x − 16 = 0

SOLUTION:  

  The solutions are –8 and 2.

20. x2 − 2x − 14 = 0

SOLUTION:  

  The solutions are about –2.9 and 4.9.

21. x2 − 8x − 1 = 8

SOLUTION:  

  The solutions are –1 and 9.

22. x2 + 3x + 21 = 22

SOLUTION:  

  The solutions are about –3.3 and 0.3.

23. x2 − 11x + 3 = 5

SOLUTION:  

  The solutions are about –0.2 and 11.2.

24. 5x2 − 10x = 23

SOLUTION:  

  The solutions are about –1.4 and 3.4.

25. 2x2 − 2x + 7 = 5

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

26. 3x2 + 12x + 81 = 15

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

27. 4x2 + 6x = 12

SOLUTION:  

  The solutions are about –2.6 and 1.1.

28. 4x2 + 5 = 10x

SOLUTION:  

  The solutions are about 0.7 and 1.8.

29. −2x2 + 10x = −14

SOLUTION:  

  The solutions are about –1.1 and 6.1.

30. −3x2 − 12 = 14x

SOLUTION:  

  The solutions are about –3.5 and –1.1.

31. STOCK The price p in dollars for a particular stock can be modeled by the quadratic equation p = 3.5t − 0.05t2, wh

the number of days after the stock is purchased. When is the stock worth $60?

SOLUTION:  Let p = 60.

  The stock is worth $60 on the 30th and 40th days after purchase.

GEOMETRY  Find the value of x for each figure. Round to the nearest tenth if necessary.

32. area = 45 in2

SOLUTION:  

  The height cannot be negative. So, x ≈ 6.3.

33. area = 110 ft2

SOLUTION:  

  The dimensions of the rectangle cannot be negative. So, x ≈ 5.3.

34. NUMBER THEORY The product of two consecutive even integers is 224. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers are 14 and 16 or –16 and –14.

35. CCSS PRECISION The product of two consecutive negative odd integers is 483. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers must be negative. So, they are –23 and –21.

36. GEOMETRY Find the area of the triangle.

SOLUTION:  

  The dimensions of the triangle cannot be negative. So, x = 18. The base of the triangle is 18 meters and the height is 18 + 6 or 24 meters.

The area of the triangle is 216 square meters.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

37. 0.2x2 − 0.2x − 0.4 = 0

SOLUTION:  

  The solutions are –1 and 2.

38. 0.5x2 = 2x − 0.3

SOLUTION:  

  The solutions are about 0.2 and 3.8.

39. 2x2 −

SOLUTION:  

  The solutions are about 0.2 and 0.9.

40. 

SOLUTION:  

  The solutions are –0.5 and 2.5.

41. 

SOLUTION:  

  The solutions are about –8.2 and 0.2.

42. 

SOLUTION:  

  The solutions are about –5.1 and 0.1.

43. ASTRONOMY The height of an object t seconds after it is dropped is given by the equation ,

where h0 is the initial height and g is the acceleration due to gravity. The acceleration due to gravity near the surface

of Mars is 3.73 m/s2, while on Earth it is 9.8 m/s

2. Suppose an object is dropped from an initial height of 120 meters

above the surface of each planet. a. On which planet would the object reach the ground first? b. How long would it take the object to reach the ground on each planet? Round each answer to the nearest tenth. c. Do the times that it takes the object to reach the ground seem reasonable? Explain your reasoning.

SOLUTION:  a. The object on Earth will reach the ground first because it is falling at a faster rate.   b. Mars:

So, t ≈ 8.0 seconds. Earth:

So, t ≈ 4.9 seconds.   c. Sample answer: Yes; the acceleration due to gravity is much greater on Earth than on Mars, so the time to reach the ground should be much less.

44. Find all values of c that make x2 + cx + 100 a perfect square trinomial.

SOLUTION:  

So, c can be –20 or 20 to make the trinomial a perfect square.

45. Find all values of c that make x2 + cx + 225 a perfect square trinomial.

SOLUTION:  

So, c can be –30 or 30 to make the trinomial a perfect square.

46. PAINTING Before she begins painting a picture, Donna stretches her canvas over a wood frame. The frame has a length of 60 inches and a width of 4 inches. She has enough canvas to cover 480 square inches. Donna decides to increase the dimensions of the frame. If the increase in the length is 10 times the increase in the width, what will the dimensions of the frame be?

SOLUTION:  Let x = the increase in the width and let 10x = the increase in the length. So, x + 4 = the new width and 10x + 60 = the new length.  

  The dimensions cannot be negative, so x = 2. The width is 2 + 4 or 6 inches and the length is 10(2) + 60 or 80 inches.

47. MULTIPLE REPRESENTATIONS In this problem, you will investigate a property of quadratic equations. a. TABULAR  Copy the table shown and complete the second column.

b. ALGEBRAIC Set each trinomial equal to zero, and solve the equation by completing the square. Complete the last column of the table with the number of roots of each equation.

c. VERBAL Compare the number of roots of each equation to the result in the b2 − 4ac column. Is there a

relationship between these values? If so, describe it.

d. ANALYTICAL Predict how many solutions 2x2 − 9x + 15 = 0 will have.  Verify your prediction by solving the 

equation.

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3    

3x2 + 6x + 9    

x2 − 2x + 7    

x2 + 10x + 25    

x2 + 3x − 12    

SOLUTION:  a.

b.

The trinomial x2 − 8x + 16 has 1 root.

 

The trinomial 2x

2 − 11x + 3 has 2 roots.

 

The square root of a negative number has no real roots. So, the trinomial 3x2 + 6x + 9 has 0 roots.

 

The square root of a negative number has no real roots. So, the trinomial x2 − 2x + 7.

 

The trinomial x2 + 10x + 25 has 1 root.

 

The trinomial x2 + 3x − 12 has 2 roots.

 

c. If b2 − 4ac is negative, the equation has no real solutions. If b

2 − 4ac is zero, the equation has one solution. If b2

− 4ac is positive, the equation has 2 solutions. d.

The equation 2x2 − 9x + 15 = 0 has 0 real solutions because b

2 − 4ac is negative.

 

The equation cannot be solved because the square root of a negative number has no real roots.

Trinomial b2 − 4ac Number

of Roots

x2 − 8x + 16 (–8)

2 – 4(1)(16) = 0 1

2x2 − 11x + 3 (–11)

2 – 4(2)(3) = 97  

3x2 + 6x + 9 (6)

2 – 4(3)(9) = −72  

x2 − 2x + 7 (–2)

2 – 4(1)(7) = −24  

x2 + 10x + 25 (10)

2 – 4(1)(25) = 0  

x2 + 3x − 12 (3)

2 – 4(1)(–12) = 57  

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3 97 2

3x2 + 6x + 9 −72 0

x2 − 2x + 7 −24 0

x2 + 10x + 25 0 1

x2 + 3x − 12 57 2

48. CCSS PERSEVERANCE Given y = ax2 + bx + c with a ≠ 0, derive the equation for the axis of symmetry by 

completing the square and rewriting the equation in the form y = a(x − h)2 + k .

SOLUTION:  

Let    and   , then the equation becomes y = (x - h)2 + k .

For an equation in the form y = ax2 + bx + c,  the equation for the axis of symmetry is   .

So, for an equation in the form  y = (x - h)2 + k, the equation for the axis of symmetry becomes x = h.

49. REASONING Determine the number of solutions x2 + bx = c has if . Explain.

SOLUTION:  

None; Sample answer: If you add to each side of the equation and each side of the inequality, you get

. Since the left side of the last equation is a perfect square, it cannot

equal the negative number . So, there are no real solutions.

50. WHICH ONE DOESN’T BELONG? Identify the expression that does not belong with the other three. Explain your reasoning.

 

SOLUTION:  The first 3 trinomials are perfect squares.

The trinomial  is not a perfect square. So, it does not belong.

51. OPEN ENDED Write a quadratic equation for which the only solution is 4.

SOLUTION:  Find a quadratic that has two roots of 4.

Verify that the solution is 4.

So, the quadratic equation x2 − 8x + 16 = 0 has a solution of 4.

52. WRITING IN MATH Compare and contrast the following strategies for solving x2 − 5x − 7 = 0: completing the

square, graphing, and factoring.

SOLUTION:  Sample answer: Because the leading coefficient is 1, solving the equation by completing the square is simpler and yields exact answers for the solutions.   

 

To solve the equation by graphing, use a graphing calculator to graph the related function y = x2 - 5x - 7. Select the

zero option from the 2nd [CALC] menu to determine the roots. 

   

  The roots are given as decimal approximations. So, for this equation the solutions would have to be given as estimations.   There are no factors of –7 that have a sum of –5, so solving by factoring is not possible.

53. The length of a rectangle is 3 times its width. The area of the rectangle is 75 square feet. Find the length of the rectangle in feet. A 25 B 15 C 10 D 5

SOLUTION:  Let x = the width of the rectangle and let 3x = the length of the rectangle.

The length of the rectangle is 3(5) or 15 feet. Choice B is the correct answer.

54. PROBABILITY At a festival, winners of a game draw a token for a prize. There is one token for each prize. The prizes include 9 movie passes, 8 stuffed animals, 5 hats, 10 jump ropes, and 4 glow necklaces. What is the probabilitythat the first person to draw a token will win a movie pass?

F  

G

H

J  

SOLUTION:  There are 9 + 8 + 5 + 10 + 4 or 36 possible outcomes.

P(movie pass) =  or  . Choice J is the correct answer.

55. GRIDDED RESPONSE The population of a town can be modeled by P = 22,000 + 125t, where P represents the population and t represents the number of years from 2000. How many years after 2000 will the population be 26,000?

SOLUTION:  

In 32 years the population of the town will be 26,000.

56. Percy delivers pizzas for Pizza King. He is paid $6 an hour plus $2.50 for each pizza he delivers. Percy earned $280 last week. If he worked a total of 30 hours, how many pizzas did he deliver? A 250 pizzas B 184 pizzas C 40 pizzas D 34 pizzas

SOLUTION:  Let p = the number of pizzas Percy delivered.

Percy delivered 40 pizzas. Choice C is the correct answer.

Describe how the graph of each function is related to the graph of f (x) = x2.

57. g(x) = −12 + x2

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –12, the translation is

down. So, the graph is shifted down 12 units from the parent function. 

58. h(x) = (x + 2)2

SOLUTION:  

The graph of f (x) = (x – c)2 represents a translation left or right from the parent graph. Since c = –2, the translation

is to the left by 2 units. 

 

59. g(x) = 2x2 + 5

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 2 and c = 5. Since 2 > 0 and  > 1, the graph of y = 2x

2 + 5

is the graph of y = x2 vertically stretched and shifted up 5 units.

60. 

SOLUTION:  

The function can be written f (x) = a(x – b)2, where a = and b = 6. Since > 0 and  < 1, the graph of 

is the graph of y = x2 vertically compressed and shifted right 6 units.

61. g(x) = 6 + x2

SOLUTION:  

The function can be written f (x) = ax2 + c, where a =  and c = 6. Since  > 0 and   > 1, the graph of y = 6 +

x2 is the graph of y = x

2 vertically stretched and shifted up 6 units.

62. h(x) = − 1 −   x2

SOLUTION:  

The function can be written f (x) = –ax2 + c, where a =  and c = –1. Since  < 0 and   > 1, the graph of y

= –1 – x2 is the graph of y = x

2 vertically stretched, shifted down 1 unit and reflected across the x-axis.

63. RIDES A popular amusement park ride whisks riders to the top of a 250-foot tower and drops them. A function for

the height of a rider is h = −16t2 + 250, where h is the height and t is the time in seconds. The ride stops the descent

of the rider 40 feet above the ground. Write an equation that models the drop of the rider. How long does it take to fall from 250 feet to 40 feet?

SOLUTION:  

The ride stops the descent of the rider at 40 feet above ground, so h = 40. Then, the equation 40 = −16t2 + 250

models the drop of the rider.

Time cannot be negative. So, it takes about 3.6 seconds to complete the ride.

Simplify. Assume that no denominator is equal to zero.

64. 

SOLUTION:  

65. 

SOLUTION:  

66. 

SOLUTION:  

67. 

SOLUTION:  

68. 

SOLUTION:  

69. b3(m

−3)(b

−6)

SOLUTION:  

Solve each open sentence.

70. |y − 2| > 7

SOLUTION:  

The solution set is {y |y > 9 or y < −5}.

Case 1 y – 2 is positive.

and Case 2 y – 2 is negative.

 

71. |z + 5| < 3

SOLUTION:  

The solution set is {z |−8 < z < −2}.

Case 1 z +5 is positive.

and Case 2 z +5 is negative.

 

72. |2b + 7| ≤ −6

SOLUTION:  

cannot be negative. So,  cannot be less than or equal to –6. The solution set is empty.

73. |3 − 2y | ≥ 8

SOLUTION:  

The solution set is {y |y  ≥ 5.5 or y  ≤ −2.5}.

Case 1 3 – 2y is positive.

and Case 2 3 – 2y is negative.

 

74. |9 − 4m| < −1

SOLUTION:  

cannot be negative. So, cannot be less than or equal to –1. The solution set is empty.

75. |5c − 2| ≤ 13

SOLUTION:  

The solution set is {c|−2.2 ≤ c ≤ 3}.

Case 1 5c – 2 is positive.

and Case 2 5c – 2 is negative.

 

Evaluate  for each set of values. Round to the nearest tenth if necessary.

76. a = 2, b = −5, c = 2

SOLUTION:  

77. a = 1, b = 12, c = 11

SOLUTION:  

78. a = −9, b = 10, c = −1

SOLUTION:  

79. a = 1, b = 7, c = −3

SOLUTION:  

80. a = 2, b = −4, c = −6

SOLUTION:  

81. a = 3, b = 1, c = 2

SOLUTION:  

This value is not a real number.

eSolutions Manual - Powered by Cognero Page 22

9-4 Solving Quadratic Equations by Completing the Square

Find the value of c that makes each trinomial a perfect square.

1. x2 − 18x + c

SOLUTION:  In this trinomial, b = –18.

So, c must be 81 to make the trinomial a perfect square.

2. x2 + 22x + c

SOLUTION:  In this trinomial, b = 22.

So, c must be 121 to make the trinomial a perfect square.

3. x2 + 9x + c

SOLUTION:  In this trinomial, b = 9.

 

So, c must be  to make the trinomial a perfect square.

4. x2 − 7x + c

SOLUTION:  In this trinomial, b = –7.

 

So, c must be  to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

5. x2 + 4x = 6

SOLUTION:  

The solutions are about –5.2 and 1.2.

6. x2 − 8x = −9

SOLUTION:  

The solutions are about 1.4 and 6.6.

7. 4x2 + 9x − 1 = 0

SOLUTION:  

The solutions are about –2.4 and 0.1.

8. −2x2 + 10x + 22 = 4

SOLUTION:  

The solutions are about –1.4 and 6.4.

9. CCSS MODELING Collin is building a deck on the back of his family’s house. He has enough lumber for the deck to be 144 square feet. The length should be 10 feet more than its width. What should the dimensions of the deck be?

SOLUTION:  Let x = the width of the deck and x + 10 = the length of the deck.

The dimensions of the deck can not be negative. So, x = –5 + 13 or 8. The width of the deck is 8 feet and the length of the deck is 8 + 10 or 18 feet.

Find the value of c that makes each trinomial a perfect square.

10. x2 + 26x + c

SOLUTION:  In this trinomial, b = 26.

So, c must be 169 to make the trinomial a perfect square.

11. x2 − 24x + c

SOLUTION:  In this trinomial, b = –24.

So, c must be 144 to make the trinomial a perfect square.

12. x2 − 19x + c

SOLUTION:  In this trinomial, b = –19.

So, c must be  to make the trinomial a perfect square.

13. x2 + 17x + c

SOLUTION:  In this trinomial, b = 17.

So, c must be  to make the trinomial a perfect square.

14. x2 + 5x + c

SOLUTION:  In this trinomial, b = 5.

So, c must be  to make the trinomial a perfect square.

15. x2 − 13x + c

SOLUTION:  In this trinomial, b = –13.

So, c must be  to make the trinomial a perfect square.

16. x2 − 22x + c

SOLUTION:  In this trinomial, b = –22.

So, c must be 121 to make the trinomial a perfect square.

17. x2 − 15x + c

SOLUTION:  In this trinomial, b = –15.

So, c must be  to make the trinomial a perfect square.

18. x2 + 24x + c

SOLUTION:  In this trinomial, b = 24.

So, c must be 144 to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

19. x2 + 6x − 16 = 0

SOLUTION:  

  The solutions are –8 and 2.

20. x2 − 2x − 14 = 0

SOLUTION:  

  The solutions are about –2.9 and 4.9.

21. x2 − 8x − 1 = 8

SOLUTION:  

  The solutions are –1 and 9.

22. x2 + 3x + 21 = 22

SOLUTION:  

  The solutions are about –3.3 and 0.3.

23. x2 − 11x + 3 = 5

SOLUTION:  

  The solutions are about –0.2 and 11.2.

24. 5x2 − 10x = 23

SOLUTION:  

  The solutions are about –1.4 and 3.4.

25. 2x2 − 2x + 7 = 5

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

26. 3x2 + 12x + 81 = 15

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

27. 4x2 + 6x = 12

SOLUTION:  

  The solutions are about –2.6 and 1.1.

28. 4x2 + 5 = 10x

SOLUTION:  

  The solutions are about 0.7 and 1.8.

29. −2x2 + 10x = −14

SOLUTION:  

  The solutions are about –1.1 and 6.1.

30. −3x2 − 12 = 14x

SOLUTION:  

  The solutions are about –3.5 and –1.1.

31. STOCK The price p in dollars for a particular stock can be modeled by the quadratic equation p = 3.5t − 0.05t2, wh

the number of days after the stock is purchased. When is the stock worth $60?

SOLUTION:  Let p = 60.

  The stock is worth $60 on the 30th and 40th days after purchase.

GEOMETRY  Find the value of x for each figure. Round to the nearest tenth if necessary.

32. area = 45 in2

SOLUTION:  

  The height cannot be negative. So, x ≈ 6.3.

33. area = 110 ft2

SOLUTION:  

  The dimensions of the rectangle cannot be negative. So, x ≈ 5.3.

34. NUMBER THEORY The product of two consecutive even integers is 224. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers are 14 and 16 or –16 and –14.

35. CCSS PRECISION The product of two consecutive negative odd integers is 483. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers must be negative. So, they are –23 and –21.

36. GEOMETRY Find the area of the triangle.

SOLUTION:  

  The dimensions of the triangle cannot be negative. So, x = 18. The base of the triangle is 18 meters and the height is 18 + 6 or 24 meters.

The area of the triangle is 216 square meters.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

37. 0.2x2 − 0.2x − 0.4 = 0

SOLUTION:  

  The solutions are –1 and 2.

38. 0.5x2 = 2x − 0.3

SOLUTION:  

  The solutions are about 0.2 and 3.8.

39. 2x2 −

SOLUTION:  

  The solutions are about 0.2 and 0.9.

40. 

SOLUTION:  

  The solutions are –0.5 and 2.5.

41. 

SOLUTION:  

  The solutions are about –8.2 and 0.2.

42. 

SOLUTION:  

  The solutions are about –5.1 and 0.1.

43. ASTRONOMY The height of an object t seconds after it is dropped is given by the equation ,

where h0 is the initial height and g is the acceleration due to gravity. The acceleration due to gravity near the surface

of Mars is 3.73 m/s2, while on Earth it is 9.8 m/s

2. Suppose an object is dropped from an initial height of 120 meters

above the surface of each planet. a. On which planet would the object reach the ground first? b. How long would it take the object to reach the ground on each planet? Round each answer to the nearest tenth. c. Do the times that it takes the object to reach the ground seem reasonable? Explain your reasoning.

SOLUTION:  a. The object on Earth will reach the ground first because it is falling at a faster rate.   b. Mars:

So, t ≈ 8.0 seconds. Earth:

So, t ≈ 4.9 seconds.   c. Sample answer: Yes; the acceleration due to gravity is much greater on Earth than on Mars, so the time to reach the ground should be much less.

44. Find all values of c that make x2 + cx + 100 a perfect square trinomial.

SOLUTION:  

So, c can be –20 or 20 to make the trinomial a perfect square.

45. Find all values of c that make x2 + cx + 225 a perfect square trinomial.

SOLUTION:  

So, c can be –30 or 30 to make the trinomial a perfect square.

46. PAINTING Before she begins painting a picture, Donna stretches her canvas over a wood frame. The frame has a length of 60 inches and a width of 4 inches. She has enough canvas to cover 480 square inches. Donna decides to increase the dimensions of the frame. If the increase in the length is 10 times the increase in the width, what will the dimensions of the frame be?

SOLUTION:  Let x = the increase in the width and let 10x = the increase in the length. So, x + 4 = the new width and 10x + 60 = the new length.  

  The dimensions cannot be negative, so x = 2. The width is 2 + 4 or 6 inches and the length is 10(2) + 60 or 80 inches.

47. MULTIPLE REPRESENTATIONS In this problem, you will investigate a property of quadratic equations. a. TABULAR  Copy the table shown and complete the second column.

b. ALGEBRAIC Set each trinomial equal to zero, and solve the equation by completing the square. Complete the last column of the table with the number of roots of each equation.

c. VERBAL Compare the number of roots of each equation to the result in the b2 − 4ac column. Is there a

relationship between these values? If so, describe it.

d. ANALYTICAL Predict how many solutions 2x2 − 9x + 15 = 0 will have.  Verify your prediction by solving the 

equation.

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3    

3x2 + 6x + 9    

x2 − 2x + 7    

x2 + 10x + 25    

x2 + 3x − 12    

SOLUTION:  a.

b.

The trinomial x2 − 8x + 16 has 1 root.

 

The trinomial 2x

2 − 11x + 3 has 2 roots.

 

The square root of a negative number has no real roots. So, the trinomial 3x2 + 6x + 9 has 0 roots.

 

The square root of a negative number has no real roots. So, the trinomial x2 − 2x + 7.

 

The trinomial x2 + 10x + 25 has 1 root.

 

The trinomial x2 + 3x − 12 has 2 roots.

 

c. If b2 − 4ac is negative, the equation has no real solutions. If b

2 − 4ac is zero, the equation has one solution. If b2

− 4ac is positive, the equation has 2 solutions. d.

The equation 2x2 − 9x + 15 = 0 has 0 real solutions because b

2 − 4ac is negative.

 

The equation cannot be solved because the square root of a negative number has no real roots.

Trinomial b2 − 4ac Number

of Roots

x2 − 8x + 16 (–8)

2 – 4(1)(16) = 0 1

2x2 − 11x + 3 (–11)

2 – 4(2)(3) = 97  

3x2 + 6x + 9 (6)

2 – 4(3)(9) = −72  

x2 − 2x + 7 (–2)

2 – 4(1)(7) = −24  

x2 + 10x + 25 (10)

2 – 4(1)(25) = 0  

x2 + 3x − 12 (3)

2 – 4(1)(–12) = 57  

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3 97 2

3x2 + 6x + 9 −72 0

x2 − 2x + 7 −24 0

x2 + 10x + 25 0 1

x2 + 3x − 12 57 2

48. CCSS PERSEVERANCE Given y = ax2 + bx + c with a ≠ 0, derive the equation for the axis of symmetry by 

completing the square and rewriting the equation in the form y = a(x − h)2 + k .

SOLUTION:  

Let    and   , then the equation becomes y = (x - h)2 + k .

For an equation in the form y = ax2 + bx + c,  the equation for the axis of symmetry is   .

So, for an equation in the form  y = (x - h)2 + k, the equation for the axis of symmetry becomes x = h.

49. REASONING Determine the number of solutions x2 + bx = c has if . Explain.

SOLUTION:  

None; Sample answer: If you add to each side of the equation and each side of the inequality, you get

. Since the left side of the last equation is a perfect square, it cannot

equal the negative number . So, there are no real solutions.

50. WHICH ONE DOESN’T BELONG? Identify the expression that does not belong with the other three. Explain your reasoning.

 

SOLUTION:  The first 3 trinomials are perfect squares.

The trinomial  is not a perfect square. So, it does not belong.

51. OPEN ENDED Write a quadratic equation for which the only solution is 4.

SOLUTION:  Find a quadratic that has two roots of 4.

Verify that the solution is 4.

So, the quadratic equation x2 − 8x + 16 = 0 has a solution of 4.

52. WRITING IN MATH Compare and contrast the following strategies for solving x2 − 5x − 7 = 0: completing the

square, graphing, and factoring.

SOLUTION:  Sample answer: Because the leading coefficient is 1, solving the equation by completing the square is simpler and yields exact answers for the solutions.   

 

To solve the equation by graphing, use a graphing calculator to graph the related function y = x2 - 5x - 7. Select the

zero option from the 2nd [CALC] menu to determine the roots. 

   

  The roots are given as decimal approximations. So, for this equation the solutions would have to be given as estimations.   There are no factors of –7 that have a sum of –5, so solving by factoring is not possible.

53. The length of a rectangle is 3 times its width. The area of the rectangle is 75 square feet. Find the length of the rectangle in feet. A 25 B 15 C 10 D 5

SOLUTION:  Let x = the width of the rectangle and let 3x = the length of the rectangle.

The length of the rectangle is 3(5) or 15 feet. Choice B is the correct answer.

54. PROBABILITY At a festival, winners of a game draw a token for a prize. There is one token for each prize. The prizes include 9 movie passes, 8 stuffed animals, 5 hats, 10 jump ropes, and 4 glow necklaces. What is the probabilitythat the first person to draw a token will win a movie pass?

F  

G

H

J  

SOLUTION:  There are 9 + 8 + 5 + 10 + 4 or 36 possible outcomes.

P(movie pass) =  or  . Choice J is the correct answer.

55. GRIDDED RESPONSE The population of a town can be modeled by P = 22,000 + 125t, where P represents the population and t represents the number of years from 2000. How many years after 2000 will the population be 26,000?

SOLUTION:  

In 32 years the population of the town will be 26,000.

56. Percy delivers pizzas for Pizza King. He is paid $6 an hour plus $2.50 for each pizza he delivers. Percy earned $280 last week. If he worked a total of 30 hours, how many pizzas did he deliver? A 250 pizzas B 184 pizzas C 40 pizzas D 34 pizzas

SOLUTION:  Let p = the number of pizzas Percy delivered.

Percy delivered 40 pizzas. Choice C is the correct answer.

Describe how the graph of each function is related to the graph of f (x) = x2.

57. g(x) = −12 + x2

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –12, the translation is

down. So, the graph is shifted down 12 units from the parent function. 

58. h(x) = (x + 2)2

SOLUTION:  

The graph of f (x) = (x – c)2 represents a translation left or right from the parent graph. Since c = –2, the translation

is to the left by 2 units. 

 

59. g(x) = 2x2 + 5

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 2 and c = 5. Since 2 > 0 and  > 1, the graph of y = 2x

2 + 5

is the graph of y = x2 vertically stretched and shifted up 5 units.

60. 

SOLUTION:  

The function can be written f (x) = a(x – b)2, where a = and b = 6. Since > 0 and  < 1, the graph of 

is the graph of y = x2 vertically compressed and shifted right 6 units.

61. g(x) = 6 + x2

SOLUTION:  

The function can be written f (x) = ax2 + c, where a =  and c = 6. Since  > 0 and   > 1, the graph of y = 6 +

x2 is the graph of y = x

2 vertically stretched and shifted up 6 units.

62. h(x) = − 1 −   x2

SOLUTION:  

The function can be written f (x) = –ax2 + c, where a =  and c = –1. Since  < 0 and   > 1, the graph of y

= –1 – x2 is the graph of y = x

2 vertically stretched, shifted down 1 unit and reflected across the x-axis.

63. RIDES A popular amusement park ride whisks riders to the top of a 250-foot tower and drops them. A function for

the height of a rider is h = −16t2 + 250, where h is the height and t is the time in seconds. The ride stops the descent

of the rider 40 feet above the ground. Write an equation that models the drop of the rider. How long does it take to fall from 250 feet to 40 feet?

SOLUTION:  

The ride stops the descent of the rider at 40 feet above ground, so h = 40. Then, the equation 40 = −16t2 + 250

models the drop of the rider.

Time cannot be negative. So, it takes about 3.6 seconds to complete the ride.

Simplify. Assume that no denominator is equal to zero.

64. 

SOLUTION:  

65. 

SOLUTION:  

66. 

SOLUTION:  

67. 

SOLUTION:  

68. 

SOLUTION:  

69. b3(m

−3)(b

−6)

SOLUTION:  

Solve each open sentence.

70. |y − 2| > 7

SOLUTION:  

The solution set is {y |y > 9 or y < −5}.

Case 1 y – 2 is positive.

and Case 2 y – 2 is negative.

 

71. |z + 5| < 3

SOLUTION:  

The solution set is {z |−8 < z < −2}.

Case 1 z +5 is positive.

and Case 2 z +5 is negative.

 

72. |2b + 7| ≤ −6

SOLUTION:  

cannot be negative. So,  cannot be less than or equal to –6. The solution set is empty.

73. |3 − 2y | ≥ 8

SOLUTION:  

The solution set is {y |y  ≥ 5.5 or y  ≤ −2.5}.

Case 1 3 – 2y is positive.

and Case 2 3 – 2y is negative.

 

74. |9 − 4m| < −1

SOLUTION:  

cannot be negative. So, cannot be less than or equal to –1. The solution set is empty.

75. |5c − 2| ≤ 13

SOLUTION:  

The solution set is {c|−2.2 ≤ c ≤ 3}.

Case 1 5c – 2 is positive.

and Case 2 5c – 2 is negative.

 

Evaluate  for each set of values. Round to the nearest tenth if necessary.

76. a = 2, b = −5, c = 2

SOLUTION:  

77. a = 1, b = 12, c = 11

SOLUTION:  

78. a = −9, b = 10, c = −1

SOLUTION:  

79. a = 1, b = 7, c = −3

SOLUTION:  

80. a = 2, b = −4, c = −6

SOLUTION:  

81. a = 3, b = 1, c = 2

SOLUTION:  

This value is not a real number.

eSolutions Manual - Powered by Cognero Page 23

9-4 Solving Quadratic Equations by Completing the Square

Find the value of c that makes each trinomial a perfect square.

1. x2 − 18x + c

SOLUTION:  In this trinomial, b = –18.

So, c must be 81 to make the trinomial a perfect square.

2. x2 + 22x + c

SOLUTION:  In this trinomial, b = 22.

So, c must be 121 to make the trinomial a perfect square.

3. x2 + 9x + c

SOLUTION:  In this trinomial, b = 9.

 

So, c must be  to make the trinomial a perfect square.

4. x2 − 7x + c

SOLUTION:  In this trinomial, b = –7.

 

So, c must be  to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

5. x2 + 4x = 6

SOLUTION:  

The solutions are about –5.2 and 1.2.

6. x2 − 8x = −9

SOLUTION:  

The solutions are about 1.4 and 6.6.

7. 4x2 + 9x − 1 = 0

SOLUTION:  

The solutions are about –2.4 and 0.1.

8. −2x2 + 10x + 22 = 4

SOLUTION:  

The solutions are about –1.4 and 6.4.

9. CCSS MODELING Collin is building a deck on the back of his family’s house. He has enough lumber for the deck to be 144 square feet. The length should be 10 feet more than its width. What should the dimensions of the deck be?

SOLUTION:  Let x = the width of the deck and x + 10 = the length of the deck.

The dimensions of the deck can not be negative. So, x = –5 + 13 or 8. The width of the deck is 8 feet and the length of the deck is 8 + 10 or 18 feet.

Find the value of c that makes each trinomial a perfect square.

10. x2 + 26x + c

SOLUTION:  In this trinomial, b = 26.

So, c must be 169 to make the trinomial a perfect square.

11. x2 − 24x + c

SOLUTION:  In this trinomial, b = –24.

So, c must be 144 to make the trinomial a perfect square.

12. x2 − 19x + c

SOLUTION:  In this trinomial, b = –19.

So, c must be  to make the trinomial a perfect square.

13. x2 + 17x + c

SOLUTION:  In this trinomial, b = 17.

So, c must be  to make the trinomial a perfect square.

14. x2 + 5x + c

SOLUTION:  In this trinomial, b = 5.

So, c must be  to make the trinomial a perfect square.

15. x2 − 13x + c

SOLUTION:  In this trinomial, b = –13.

So, c must be  to make the trinomial a perfect square.

16. x2 − 22x + c

SOLUTION:  In this trinomial, b = –22.

So, c must be 121 to make the trinomial a perfect square.

17. x2 − 15x + c

SOLUTION:  In this trinomial, b = –15.

So, c must be  to make the trinomial a perfect square.

18. x2 + 24x + c

SOLUTION:  In this trinomial, b = 24.

So, c must be 144 to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

19. x2 + 6x − 16 = 0

SOLUTION:  

  The solutions are –8 and 2.

20. x2 − 2x − 14 = 0

SOLUTION:  

  The solutions are about –2.9 and 4.9.

21. x2 − 8x − 1 = 8

SOLUTION:  

  The solutions are –1 and 9.

22. x2 + 3x + 21 = 22

SOLUTION:  

  The solutions are about –3.3 and 0.3.

23. x2 − 11x + 3 = 5

SOLUTION:  

  The solutions are about –0.2 and 11.2.

24. 5x2 − 10x = 23

SOLUTION:  

  The solutions are about –1.4 and 3.4.

25. 2x2 − 2x + 7 = 5

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

26. 3x2 + 12x + 81 = 15

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

27. 4x2 + 6x = 12

SOLUTION:  

  The solutions are about –2.6 and 1.1.

28. 4x2 + 5 = 10x

SOLUTION:  

  The solutions are about 0.7 and 1.8.

29. −2x2 + 10x = −14

SOLUTION:  

  The solutions are about –1.1 and 6.1.

30. −3x2 − 12 = 14x

SOLUTION:  

  The solutions are about –3.5 and –1.1.

31. STOCK The price p in dollars for a particular stock can be modeled by the quadratic equation p = 3.5t − 0.05t2, wh

the number of days after the stock is purchased. When is the stock worth $60?

SOLUTION:  Let p = 60.

  The stock is worth $60 on the 30th and 40th days after purchase.

GEOMETRY  Find the value of x for each figure. Round to the nearest tenth if necessary.

32. area = 45 in2

SOLUTION:  

  The height cannot be negative. So, x ≈ 6.3.

33. area = 110 ft2

SOLUTION:  

  The dimensions of the rectangle cannot be negative. So, x ≈ 5.3.

34. NUMBER THEORY The product of two consecutive even integers is 224. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers are 14 and 16 or –16 and –14.

35. CCSS PRECISION The product of two consecutive negative odd integers is 483. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers must be negative. So, they are –23 and –21.

36. GEOMETRY Find the area of the triangle.

SOLUTION:  

  The dimensions of the triangle cannot be negative. So, x = 18. The base of the triangle is 18 meters and the height is 18 + 6 or 24 meters.

The area of the triangle is 216 square meters.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

37. 0.2x2 − 0.2x − 0.4 = 0

SOLUTION:  

  The solutions are –1 and 2.

38. 0.5x2 = 2x − 0.3

SOLUTION:  

  The solutions are about 0.2 and 3.8.

39. 2x2 −

SOLUTION:  

  The solutions are about 0.2 and 0.9.

40. 

SOLUTION:  

  The solutions are –0.5 and 2.5.

41. 

SOLUTION:  

  The solutions are about –8.2 and 0.2.

42. 

SOLUTION:  

  The solutions are about –5.1 and 0.1.

43. ASTRONOMY The height of an object t seconds after it is dropped is given by the equation ,

where h0 is the initial height and g is the acceleration due to gravity. The acceleration due to gravity near the surface

of Mars is 3.73 m/s2, while on Earth it is 9.8 m/s

2. Suppose an object is dropped from an initial height of 120 meters

above the surface of each planet. a. On which planet would the object reach the ground first? b. How long would it take the object to reach the ground on each planet? Round each answer to the nearest tenth. c. Do the times that it takes the object to reach the ground seem reasonable? Explain your reasoning.

SOLUTION:  a. The object on Earth will reach the ground first because it is falling at a faster rate.   b. Mars:

So, t ≈ 8.0 seconds. Earth:

So, t ≈ 4.9 seconds.   c. Sample answer: Yes; the acceleration due to gravity is much greater on Earth than on Mars, so the time to reach the ground should be much less.

44. Find all values of c that make x2 + cx + 100 a perfect square trinomial.

SOLUTION:  

So, c can be –20 or 20 to make the trinomial a perfect square.

45. Find all values of c that make x2 + cx + 225 a perfect square trinomial.

SOLUTION:  

So, c can be –30 or 30 to make the trinomial a perfect square.

46. PAINTING Before she begins painting a picture, Donna stretches her canvas over a wood frame. The frame has a length of 60 inches and a width of 4 inches. She has enough canvas to cover 480 square inches. Donna decides to increase the dimensions of the frame. If the increase in the length is 10 times the increase in the width, what will the dimensions of the frame be?

SOLUTION:  Let x = the increase in the width and let 10x = the increase in the length. So, x + 4 = the new width and 10x + 60 = the new length.  

  The dimensions cannot be negative, so x = 2. The width is 2 + 4 or 6 inches and the length is 10(2) + 60 or 80 inches.

47. MULTIPLE REPRESENTATIONS In this problem, you will investigate a property of quadratic equations. a. TABULAR  Copy the table shown and complete the second column.

b. ALGEBRAIC Set each trinomial equal to zero, and solve the equation by completing the square. Complete the last column of the table with the number of roots of each equation.

c. VERBAL Compare the number of roots of each equation to the result in the b2 − 4ac column. Is there a

relationship between these values? If so, describe it.

d. ANALYTICAL Predict how many solutions 2x2 − 9x + 15 = 0 will have.  Verify your prediction by solving the 

equation.

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3    

3x2 + 6x + 9    

x2 − 2x + 7    

x2 + 10x + 25    

x2 + 3x − 12    

SOLUTION:  a.

b.

The trinomial x2 − 8x + 16 has 1 root.

 

The trinomial 2x

2 − 11x + 3 has 2 roots.

 

The square root of a negative number has no real roots. So, the trinomial 3x2 + 6x + 9 has 0 roots.

 

The square root of a negative number has no real roots. So, the trinomial x2 − 2x + 7.

 

The trinomial x2 + 10x + 25 has 1 root.

 

The trinomial x2 + 3x − 12 has 2 roots.

 

c. If b2 − 4ac is negative, the equation has no real solutions. If b

2 − 4ac is zero, the equation has one solution. If b2

− 4ac is positive, the equation has 2 solutions. d.

The equation 2x2 − 9x + 15 = 0 has 0 real solutions because b

2 − 4ac is negative.

 

The equation cannot be solved because the square root of a negative number has no real roots.

Trinomial b2 − 4ac Number

of Roots

x2 − 8x + 16 (–8)

2 – 4(1)(16) = 0 1

2x2 − 11x + 3 (–11)

2 – 4(2)(3) = 97  

3x2 + 6x + 9 (6)

2 – 4(3)(9) = −72  

x2 − 2x + 7 (–2)

2 – 4(1)(7) = −24  

x2 + 10x + 25 (10)

2 – 4(1)(25) = 0  

x2 + 3x − 12 (3)

2 – 4(1)(–12) = 57  

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3 97 2

3x2 + 6x + 9 −72 0

x2 − 2x + 7 −24 0

x2 + 10x + 25 0 1

x2 + 3x − 12 57 2

48. CCSS PERSEVERANCE Given y = ax2 + bx + c with a ≠ 0, derive the equation for the axis of symmetry by 

completing the square and rewriting the equation in the form y = a(x − h)2 + k .

SOLUTION:  

Let    and   , then the equation becomes y = (x - h)2 + k .

For an equation in the form y = ax2 + bx + c,  the equation for the axis of symmetry is   .

So, for an equation in the form  y = (x - h)2 + k, the equation for the axis of symmetry becomes x = h.

49. REASONING Determine the number of solutions x2 + bx = c has if . Explain.

SOLUTION:  

None; Sample answer: If you add to each side of the equation and each side of the inequality, you get

. Since the left side of the last equation is a perfect square, it cannot

equal the negative number . So, there are no real solutions.

50. WHICH ONE DOESN’T BELONG? Identify the expression that does not belong with the other three. Explain your reasoning.

 

SOLUTION:  The first 3 trinomials are perfect squares.

The trinomial  is not a perfect square. So, it does not belong.

51. OPEN ENDED Write a quadratic equation for which the only solution is 4.

SOLUTION:  Find a quadratic that has two roots of 4.

Verify that the solution is 4.

So, the quadratic equation x2 − 8x + 16 = 0 has a solution of 4.

52. WRITING IN MATH Compare and contrast the following strategies for solving x2 − 5x − 7 = 0: completing the

square, graphing, and factoring.

SOLUTION:  Sample answer: Because the leading coefficient is 1, solving the equation by completing the square is simpler and yields exact answers for the solutions.   

 

To solve the equation by graphing, use a graphing calculator to graph the related function y = x2 - 5x - 7. Select the

zero option from the 2nd [CALC] menu to determine the roots. 

   

  The roots are given as decimal approximations. So, for this equation the solutions would have to be given as estimations.   There are no factors of –7 that have a sum of –5, so solving by factoring is not possible.

53. The length of a rectangle is 3 times its width. The area of the rectangle is 75 square feet. Find the length of the rectangle in feet. A 25 B 15 C 10 D 5

SOLUTION:  Let x = the width of the rectangle and let 3x = the length of the rectangle.

The length of the rectangle is 3(5) or 15 feet. Choice B is the correct answer.

54. PROBABILITY At a festival, winners of a game draw a token for a prize. There is one token for each prize. The prizes include 9 movie passes, 8 stuffed animals, 5 hats, 10 jump ropes, and 4 glow necklaces. What is the probabilitythat the first person to draw a token will win a movie pass?

F  

G

H

J  

SOLUTION:  There are 9 + 8 + 5 + 10 + 4 or 36 possible outcomes.

P(movie pass) =  or  . Choice J is the correct answer.

55. GRIDDED RESPONSE The population of a town can be modeled by P = 22,000 + 125t, where P represents the population and t represents the number of years from 2000. How many years after 2000 will the population be 26,000?

SOLUTION:  

In 32 years the population of the town will be 26,000.

56. Percy delivers pizzas for Pizza King. He is paid $6 an hour plus $2.50 for each pizza he delivers. Percy earned $280 last week. If he worked a total of 30 hours, how many pizzas did he deliver? A 250 pizzas B 184 pizzas C 40 pizzas D 34 pizzas

SOLUTION:  Let p = the number of pizzas Percy delivered.

Percy delivered 40 pizzas. Choice C is the correct answer.

Describe how the graph of each function is related to the graph of f (x) = x2.

57. g(x) = −12 + x2

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –12, the translation is

down. So, the graph is shifted down 12 units from the parent function. 

58. h(x) = (x + 2)2

SOLUTION:  

The graph of f (x) = (x – c)2 represents a translation left or right from the parent graph. Since c = –2, the translation

is to the left by 2 units. 

 

59. g(x) = 2x2 + 5

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 2 and c = 5. Since 2 > 0 and  > 1, the graph of y = 2x

2 + 5

is the graph of y = x2 vertically stretched and shifted up 5 units.

60. 

SOLUTION:  

The function can be written f (x) = a(x – b)2, where a = and b = 6. Since > 0 and  < 1, the graph of 

is the graph of y = x2 vertically compressed and shifted right 6 units.

61. g(x) = 6 + x2

SOLUTION:  

The function can be written f (x) = ax2 + c, where a =  and c = 6. Since  > 0 and   > 1, the graph of y = 6 +

x2 is the graph of y = x

2 vertically stretched and shifted up 6 units.

62. h(x) = − 1 −   x2

SOLUTION:  

The function can be written f (x) = –ax2 + c, where a =  and c = –1. Since  < 0 and   > 1, the graph of y

= –1 – x2 is the graph of y = x

2 vertically stretched, shifted down 1 unit and reflected across the x-axis.

63. RIDES A popular amusement park ride whisks riders to the top of a 250-foot tower and drops them. A function for

the height of a rider is h = −16t2 + 250, where h is the height and t is the time in seconds. The ride stops the descent

of the rider 40 feet above the ground. Write an equation that models the drop of the rider. How long does it take to fall from 250 feet to 40 feet?

SOLUTION:  

The ride stops the descent of the rider at 40 feet above ground, so h = 40. Then, the equation 40 = −16t2 + 250

models the drop of the rider.

Time cannot be negative. So, it takes about 3.6 seconds to complete the ride.

Simplify. Assume that no denominator is equal to zero.

64. 

SOLUTION:  

65. 

SOLUTION:  

66. 

SOLUTION:  

67. 

SOLUTION:  

68. 

SOLUTION:  

69. b3(m

−3)(b

−6)

SOLUTION:  

Solve each open sentence.

70. |y − 2| > 7

SOLUTION:  

The solution set is {y |y > 9 or y < −5}.

Case 1 y – 2 is positive.

and Case 2 y – 2 is negative.

 

71. |z + 5| < 3

SOLUTION:  

The solution set is {z |−8 < z < −2}.

Case 1 z +5 is positive.

and Case 2 z +5 is negative.

 

72. |2b + 7| ≤ −6

SOLUTION:  

cannot be negative. So,  cannot be less than or equal to –6. The solution set is empty.

73. |3 − 2y | ≥ 8

SOLUTION:  

The solution set is {y |y  ≥ 5.5 or y  ≤ −2.5}.

Case 1 3 – 2y is positive.

and Case 2 3 – 2y is negative.

 

74. |9 − 4m| < −1

SOLUTION:  

cannot be negative. So, cannot be less than or equal to –1. The solution set is empty.

75. |5c − 2| ≤ 13

SOLUTION:  

The solution set is {c|−2.2 ≤ c ≤ 3}.

Case 1 5c – 2 is positive.

and Case 2 5c – 2 is negative.

 

Evaluate  for each set of values. Round to the nearest tenth if necessary.

76. a = 2, b = −5, c = 2

SOLUTION:  

77. a = 1, b = 12, c = 11

SOLUTION:  

78. a = −9, b = 10, c = −1

SOLUTION:  

79. a = 1, b = 7, c = −3

SOLUTION:  

80. a = 2, b = −4, c = −6

SOLUTION:  

81. a = 3, b = 1, c = 2

SOLUTION:  

This value is not a real number.

eSolutions Manual - Powered by Cognero Page 24

9-4 Solving Quadratic Equations by Completing the Square

Find the value of c that makes each trinomial a perfect square.

1. x2 − 18x + c

SOLUTION:  In this trinomial, b = –18.

So, c must be 81 to make the trinomial a perfect square.

2. x2 + 22x + c

SOLUTION:  In this trinomial, b = 22.

So, c must be 121 to make the trinomial a perfect square.

3. x2 + 9x + c

SOLUTION:  In this trinomial, b = 9.

 

So, c must be  to make the trinomial a perfect square.

4. x2 − 7x + c

SOLUTION:  In this trinomial, b = –7.

 

So, c must be  to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

5. x2 + 4x = 6

SOLUTION:  

The solutions are about –5.2 and 1.2.

6. x2 − 8x = −9

SOLUTION:  

The solutions are about 1.4 and 6.6.

7. 4x2 + 9x − 1 = 0

SOLUTION:  

The solutions are about –2.4 and 0.1.

8. −2x2 + 10x + 22 = 4

SOLUTION:  

The solutions are about –1.4 and 6.4.

9. CCSS MODELING Collin is building a deck on the back of his family’s house. He has enough lumber for the deck to be 144 square feet. The length should be 10 feet more than its width. What should the dimensions of the deck be?

SOLUTION:  Let x = the width of the deck and x + 10 = the length of the deck.

The dimensions of the deck can not be negative. So, x = –5 + 13 or 8. The width of the deck is 8 feet and the length of the deck is 8 + 10 or 18 feet.

Find the value of c that makes each trinomial a perfect square.

10. x2 + 26x + c

SOLUTION:  In this trinomial, b = 26.

So, c must be 169 to make the trinomial a perfect square.

11. x2 − 24x + c

SOLUTION:  In this trinomial, b = –24.

So, c must be 144 to make the trinomial a perfect square.

12. x2 − 19x + c

SOLUTION:  In this trinomial, b = –19.

So, c must be  to make the trinomial a perfect square.

13. x2 + 17x + c

SOLUTION:  In this trinomial, b = 17.

So, c must be  to make the trinomial a perfect square.

14. x2 + 5x + c

SOLUTION:  In this trinomial, b = 5.

So, c must be  to make the trinomial a perfect square.

15. x2 − 13x + c

SOLUTION:  In this trinomial, b = –13.

So, c must be  to make the trinomial a perfect square.

16. x2 − 22x + c

SOLUTION:  In this trinomial, b = –22.

So, c must be 121 to make the trinomial a perfect square.

17. x2 − 15x + c

SOLUTION:  In this trinomial, b = –15.

So, c must be  to make the trinomial a perfect square.

18. x2 + 24x + c

SOLUTION:  In this trinomial, b = 24.

So, c must be 144 to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

19. x2 + 6x − 16 = 0

SOLUTION:  

  The solutions are –8 and 2.

20. x2 − 2x − 14 = 0

SOLUTION:  

  The solutions are about –2.9 and 4.9.

21. x2 − 8x − 1 = 8

SOLUTION:  

  The solutions are –1 and 9.

22. x2 + 3x + 21 = 22

SOLUTION:  

  The solutions are about –3.3 and 0.3.

23. x2 − 11x + 3 = 5

SOLUTION:  

  The solutions are about –0.2 and 11.2.

24. 5x2 − 10x = 23

SOLUTION:  

  The solutions are about –1.4 and 3.4.

25. 2x2 − 2x + 7 = 5

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

26. 3x2 + 12x + 81 = 15

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

27. 4x2 + 6x = 12

SOLUTION:  

  The solutions are about –2.6 and 1.1.

28. 4x2 + 5 = 10x

SOLUTION:  

  The solutions are about 0.7 and 1.8.

29. −2x2 + 10x = −14

SOLUTION:  

  The solutions are about –1.1 and 6.1.

30. −3x2 − 12 = 14x

SOLUTION:  

  The solutions are about –3.5 and –1.1.

31. STOCK The price p in dollars for a particular stock can be modeled by the quadratic equation p = 3.5t − 0.05t2, wh

the number of days after the stock is purchased. When is the stock worth $60?

SOLUTION:  Let p = 60.

  The stock is worth $60 on the 30th and 40th days after purchase.

GEOMETRY  Find the value of x for each figure. Round to the nearest tenth if necessary.

32. area = 45 in2

SOLUTION:  

  The height cannot be negative. So, x ≈ 6.3.

33. area = 110 ft2

SOLUTION:  

  The dimensions of the rectangle cannot be negative. So, x ≈ 5.3.

34. NUMBER THEORY The product of two consecutive even integers is 224. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers are 14 and 16 or –16 and –14.

35. CCSS PRECISION The product of two consecutive negative odd integers is 483. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers must be negative. So, they are –23 and –21.

36. GEOMETRY Find the area of the triangle.

SOLUTION:  

  The dimensions of the triangle cannot be negative. So, x = 18. The base of the triangle is 18 meters and the height is 18 + 6 or 24 meters.

The area of the triangle is 216 square meters.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

37. 0.2x2 − 0.2x − 0.4 = 0

SOLUTION:  

  The solutions are –1 and 2.

38. 0.5x2 = 2x − 0.3

SOLUTION:  

  The solutions are about 0.2 and 3.8.

39. 2x2 −

SOLUTION:  

  The solutions are about 0.2 and 0.9.

40. 

SOLUTION:  

  The solutions are –0.5 and 2.5.

41. 

SOLUTION:  

  The solutions are about –8.2 and 0.2.

42. 

SOLUTION:  

  The solutions are about –5.1 and 0.1.

43. ASTRONOMY The height of an object t seconds after it is dropped is given by the equation ,

where h0 is the initial height and g is the acceleration due to gravity. The acceleration due to gravity near the surface

of Mars is 3.73 m/s2, while on Earth it is 9.8 m/s

2. Suppose an object is dropped from an initial height of 120 meters

above the surface of each planet. a. On which planet would the object reach the ground first? b. How long would it take the object to reach the ground on each planet? Round each answer to the nearest tenth. c. Do the times that it takes the object to reach the ground seem reasonable? Explain your reasoning.

SOLUTION:  a. The object on Earth will reach the ground first because it is falling at a faster rate.   b. Mars:

So, t ≈ 8.0 seconds. Earth:

So, t ≈ 4.9 seconds.   c. Sample answer: Yes; the acceleration due to gravity is much greater on Earth than on Mars, so the time to reach the ground should be much less.

44. Find all values of c that make x2 + cx + 100 a perfect square trinomial.

SOLUTION:  

So, c can be –20 or 20 to make the trinomial a perfect square.

45. Find all values of c that make x2 + cx + 225 a perfect square trinomial.

SOLUTION:  

So, c can be –30 or 30 to make the trinomial a perfect square.

46. PAINTING Before she begins painting a picture, Donna stretches her canvas over a wood frame. The frame has a length of 60 inches and a width of 4 inches. She has enough canvas to cover 480 square inches. Donna decides to increase the dimensions of the frame. If the increase in the length is 10 times the increase in the width, what will the dimensions of the frame be?

SOLUTION:  Let x = the increase in the width and let 10x = the increase in the length. So, x + 4 = the new width and 10x + 60 = the new length.  

  The dimensions cannot be negative, so x = 2. The width is 2 + 4 or 6 inches and the length is 10(2) + 60 or 80 inches.

47. MULTIPLE REPRESENTATIONS In this problem, you will investigate a property of quadratic equations. a. TABULAR  Copy the table shown and complete the second column.

b. ALGEBRAIC Set each trinomial equal to zero, and solve the equation by completing the square. Complete the last column of the table with the number of roots of each equation.

c. VERBAL Compare the number of roots of each equation to the result in the b2 − 4ac column. Is there a

relationship between these values? If so, describe it.

d. ANALYTICAL Predict how many solutions 2x2 − 9x + 15 = 0 will have.  Verify your prediction by solving the 

equation.

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3    

3x2 + 6x + 9    

x2 − 2x + 7    

x2 + 10x + 25    

x2 + 3x − 12    

SOLUTION:  a.

b.

The trinomial x2 − 8x + 16 has 1 root.

 

The trinomial 2x

2 − 11x + 3 has 2 roots.

 

The square root of a negative number has no real roots. So, the trinomial 3x2 + 6x + 9 has 0 roots.

 

The square root of a negative number has no real roots. So, the trinomial x2 − 2x + 7.

 

The trinomial x2 + 10x + 25 has 1 root.

 

The trinomial x2 + 3x − 12 has 2 roots.

 

c. If b2 − 4ac is negative, the equation has no real solutions. If b

2 − 4ac is zero, the equation has one solution. If b2

− 4ac is positive, the equation has 2 solutions. d.

The equation 2x2 − 9x + 15 = 0 has 0 real solutions because b

2 − 4ac is negative.

 

The equation cannot be solved because the square root of a negative number has no real roots.

Trinomial b2 − 4ac Number

of Roots

x2 − 8x + 16 (–8)

2 – 4(1)(16) = 0 1

2x2 − 11x + 3 (–11)

2 – 4(2)(3) = 97  

3x2 + 6x + 9 (6)

2 – 4(3)(9) = −72  

x2 − 2x + 7 (–2)

2 – 4(1)(7) = −24  

x2 + 10x + 25 (10)

2 – 4(1)(25) = 0  

x2 + 3x − 12 (3)

2 – 4(1)(–12) = 57  

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3 97 2

3x2 + 6x + 9 −72 0

x2 − 2x + 7 −24 0

x2 + 10x + 25 0 1

x2 + 3x − 12 57 2

48. CCSS PERSEVERANCE Given y = ax2 + bx + c with a ≠ 0, derive the equation for the axis of symmetry by 

completing the square and rewriting the equation in the form y = a(x − h)2 + k .

SOLUTION:  

Let    and   , then the equation becomes y = (x - h)2 + k .

For an equation in the form y = ax2 + bx + c,  the equation for the axis of symmetry is   .

So, for an equation in the form  y = (x - h)2 + k, the equation for the axis of symmetry becomes x = h.

49. REASONING Determine the number of solutions x2 + bx = c has if . Explain.

SOLUTION:  

None; Sample answer: If you add to each side of the equation and each side of the inequality, you get

. Since the left side of the last equation is a perfect square, it cannot

equal the negative number . So, there are no real solutions.

50. WHICH ONE DOESN’T BELONG? Identify the expression that does not belong with the other three. Explain your reasoning.

 

SOLUTION:  The first 3 trinomials are perfect squares.

The trinomial  is not a perfect square. So, it does not belong.

51. OPEN ENDED Write a quadratic equation for which the only solution is 4.

SOLUTION:  Find a quadratic that has two roots of 4.

Verify that the solution is 4.

So, the quadratic equation x2 − 8x + 16 = 0 has a solution of 4.

52. WRITING IN MATH Compare and contrast the following strategies for solving x2 − 5x − 7 = 0: completing the

square, graphing, and factoring.

SOLUTION:  Sample answer: Because the leading coefficient is 1, solving the equation by completing the square is simpler and yields exact answers for the solutions.   

 

To solve the equation by graphing, use a graphing calculator to graph the related function y = x2 - 5x - 7. Select the

zero option from the 2nd [CALC] menu to determine the roots. 

   

  The roots are given as decimal approximations. So, for this equation the solutions would have to be given as estimations.   There are no factors of –7 that have a sum of –5, so solving by factoring is not possible.

53. The length of a rectangle is 3 times its width. The area of the rectangle is 75 square feet. Find the length of the rectangle in feet. A 25 B 15 C 10 D 5

SOLUTION:  Let x = the width of the rectangle and let 3x = the length of the rectangle.

The length of the rectangle is 3(5) or 15 feet. Choice B is the correct answer.

54. PROBABILITY At a festival, winners of a game draw a token for a prize. There is one token for each prize. The prizes include 9 movie passes, 8 stuffed animals, 5 hats, 10 jump ropes, and 4 glow necklaces. What is the probabilitythat the first person to draw a token will win a movie pass?

F  

G

H

J  

SOLUTION:  There are 9 + 8 + 5 + 10 + 4 or 36 possible outcomes.

P(movie pass) =  or  . Choice J is the correct answer.

55. GRIDDED RESPONSE The population of a town can be modeled by P = 22,000 + 125t, where P represents the population and t represents the number of years from 2000. How many years after 2000 will the population be 26,000?

SOLUTION:  

In 32 years the population of the town will be 26,000.

56. Percy delivers pizzas for Pizza King. He is paid $6 an hour plus $2.50 for each pizza he delivers. Percy earned $280 last week. If he worked a total of 30 hours, how many pizzas did he deliver? A 250 pizzas B 184 pizzas C 40 pizzas D 34 pizzas

SOLUTION:  Let p = the number of pizzas Percy delivered.

Percy delivered 40 pizzas. Choice C is the correct answer.

Describe how the graph of each function is related to the graph of f (x) = x2.

57. g(x) = −12 + x2

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –12, the translation is

down. So, the graph is shifted down 12 units from the parent function. 

58. h(x) = (x + 2)2

SOLUTION:  

The graph of f (x) = (x – c)2 represents a translation left or right from the parent graph. Since c = –2, the translation

is to the left by 2 units. 

 

59. g(x) = 2x2 + 5

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 2 and c = 5. Since 2 > 0 and  > 1, the graph of y = 2x

2 + 5

is the graph of y = x2 vertically stretched and shifted up 5 units.

60. 

SOLUTION:  

The function can be written f (x) = a(x – b)2, where a = and b = 6. Since > 0 and  < 1, the graph of 

is the graph of y = x2 vertically compressed and shifted right 6 units.

61. g(x) = 6 + x2

SOLUTION:  

The function can be written f (x) = ax2 + c, where a =  and c = 6. Since  > 0 and   > 1, the graph of y = 6 +

x2 is the graph of y = x

2 vertically stretched and shifted up 6 units.

62. h(x) = − 1 −   x2

SOLUTION:  

The function can be written f (x) = –ax2 + c, where a =  and c = –1. Since  < 0 and   > 1, the graph of y

= –1 – x2 is the graph of y = x

2 vertically stretched, shifted down 1 unit and reflected across the x-axis.

63. RIDES A popular amusement park ride whisks riders to the top of a 250-foot tower and drops them. A function for

the height of a rider is h = −16t2 + 250, where h is the height and t is the time in seconds. The ride stops the descent

of the rider 40 feet above the ground. Write an equation that models the drop of the rider. How long does it take to fall from 250 feet to 40 feet?

SOLUTION:  

The ride stops the descent of the rider at 40 feet above ground, so h = 40. Then, the equation 40 = −16t2 + 250

models the drop of the rider.

Time cannot be negative. So, it takes about 3.6 seconds to complete the ride.

Simplify. Assume that no denominator is equal to zero.

64. 

SOLUTION:  

65. 

SOLUTION:  

66. 

SOLUTION:  

67. 

SOLUTION:  

68. 

SOLUTION:  

69. b3(m

−3)(b

−6)

SOLUTION:  

Solve each open sentence.

70. |y − 2| > 7

SOLUTION:  

The solution set is {y |y > 9 or y < −5}.

Case 1 y – 2 is positive.

and Case 2 y – 2 is negative.

 

71. |z + 5| < 3

SOLUTION:  

The solution set is {z |−8 < z < −2}.

Case 1 z +5 is positive.

and Case 2 z +5 is negative.

 

72. |2b + 7| ≤ −6

SOLUTION:  

cannot be negative. So,  cannot be less than or equal to –6. The solution set is empty.

73. |3 − 2y | ≥ 8

SOLUTION:  

The solution set is {y |y  ≥ 5.5 or y  ≤ −2.5}.

Case 1 3 – 2y is positive.

and Case 2 3 – 2y is negative.

 

74. |9 − 4m| < −1

SOLUTION:  

cannot be negative. So, cannot be less than or equal to –1. The solution set is empty.

75. |5c − 2| ≤ 13

SOLUTION:  

The solution set is {c|−2.2 ≤ c ≤ 3}.

Case 1 5c – 2 is positive.

and Case 2 5c – 2 is negative.

 

Evaluate  for each set of values. Round to the nearest tenth if necessary.

76. a = 2, b = −5, c = 2

SOLUTION:  

77. a = 1, b = 12, c = 11

SOLUTION:  

78. a = −9, b = 10, c = −1

SOLUTION:  

79. a = 1, b = 7, c = −3

SOLUTION:  

80. a = 2, b = −4, c = −6

SOLUTION:  

81. a = 3, b = 1, c = 2

SOLUTION:  

This value is not a real number.

eSolutions Manual - Powered by Cognero Page 25

9-4 Solving Quadratic Equations by Completing the Square

Find the value of c that makes each trinomial a perfect square.

1. x2 − 18x + c

SOLUTION:  In this trinomial, b = –18.

So, c must be 81 to make the trinomial a perfect square.

2. x2 + 22x + c

SOLUTION:  In this trinomial, b = 22.

So, c must be 121 to make the trinomial a perfect square.

3. x2 + 9x + c

SOLUTION:  In this trinomial, b = 9.

 

So, c must be  to make the trinomial a perfect square.

4. x2 − 7x + c

SOLUTION:  In this trinomial, b = –7.

 

So, c must be  to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

5. x2 + 4x = 6

SOLUTION:  

The solutions are about –5.2 and 1.2.

6. x2 − 8x = −9

SOLUTION:  

The solutions are about 1.4 and 6.6.

7. 4x2 + 9x − 1 = 0

SOLUTION:  

The solutions are about –2.4 and 0.1.

8. −2x2 + 10x + 22 = 4

SOLUTION:  

The solutions are about –1.4 and 6.4.

9. CCSS MODELING Collin is building a deck on the back of his family’s house. He has enough lumber for the deck to be 144 square feet. The length should be 10 feet more than its width. What should the dimensions of the deck be?

SOLUTION:  Let x = the width of the deck and x + 10 = the length of the deck.

The dimensions of the deck can not be negative. So, x = –5 + 13 or 8. The width of the deck is 8 feet and the length of the deck is 8 + 10 or 18 feet.

Find the value of c that makes each trinomial a perfect square.

10. x2 + 26x + c

SOLUTION:  In this trinomial, b = 26.

So, c must be 169 to make the trinomial a perfect square.

11. x2 − 24x + c

SOLUTION:  In this trinomial, b = –24.

So, c must be 144 to make the trinomial a perfect square.

12. x2 − 19x + c

SOLUTION:  In this trinomial, b = –19.

So, c must be  to make the trinomial a perfect square.

13. x2 + 17x + c

SOLUTION:  In this trinomial, b = 17.

So, c must be  to make the trinomial a perfect square.

14. x2 + 5x + c

SOLUTION:  In this trinomial, b = 5.

So, c must be  to make the trinomial a perfect square.

15. x2 − 13x + c

SOLUTION:  In this trinomial, b = –13.

So, c must be  to make the trinomial a perfect square.

16. x2 − 22x + c

SOLUTION:  In this trinomial, b = –22.

So, c must be 121 to make the trinomial a perfect square.

17. x2 − 15x + c

SOLUTION:  In this trinomial, b = –15.

So, c must be  to make the trinomial a perfect square.

18. x2 + 24x + c

SOLUTION:  In this trinomial, b = 24.

So, c must be 144 to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

19. x2 + 6x − 16 = 0

SOLUTION:  

  The solutions are –8 and 2.

20. x2 − 2x − 14 = 0

SOLUTION:  

  The solutions are about –2.9 and 4.9.

21. x2 − 8x − 1 = 8

SOLUTION:  

  The solutions are –1 and 9.

22. x2 + 3x + 21 = 22

SOLUTION:  

  The solutions are about –3.3 and 0.3.

23. x2 − 11x + 3 = 5

SOLUTION:  

  The solutions are about –0.2 and 11.2.

24. 5x2 − 10x = 23

SOLUTION:  

  The solutions are about –1.4 and 3.4.

25. 2x2 − 2x + 7 = 5

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

26. 3x2 + 12x + 81 = 15

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

27. 4x2 + 6x = 12

SOLUTION:  

  The solutions are about –2.6 and 1.1.

28. 4x2 + 5 = 10x

SOLUTION:  

  The solutions are about 0.7 and 1.8.

29. −2x2 + 10x = −14

SOLUTION:  

  The solutions are about –1.1 and 6.1.

30. −3x2 − 12 = 14x

SOLUTION:  

  The solutions are about –3.5 and –1.1.

31. STOCK The price p in dollars for a particular stock can be modeled by the quadratic equation p = 3.5t − 0.05t2, wh

the number of days after the stock is purchased. When is the stock worth $60?

SOLUTION:  Let p = 60.

  The stock is worth $60 on the 30th and 40th days after purchase.

GEOMETRY  Find the value of x for each figure. Round to the nearest tenth if necessary.

32. area = 45 in2

SOLUTION:  

  The height cannot be negative. So, x ≈ 6.3.

33. area = 110 ft2

SOLUTION:  

  The dimensions of the rectangle cannot be negative. So, x ≈ 5.3.

34. NUMBER THEORY The product of two consecutive even integers is 224. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers are 14 and 16 or –16 and –14.

35. CCSS PRECISION The product of two consecutive negative odd integers is 483. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers must be negative. So, they are –23 and –21.

36. GEOMETRY Find the area of the triangle.

SOLUTION:  

  The dimensions of the triangle cannot be negative. So, x = 18. The base of the triangle is 18 meters and the height is 18 + 6 or 24 meters.

The area of the triangle is 216 square meters.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

37. 0.2x2 − 0.2x − 0.4 = 0

SOLUTION:  

  The solutions are –1 and 2.

38. 0.5x2 = 2x − 0.3

SOLUTION:  

  The solutions are about 0.2 and 3.8.

39. 2x2 −

SOLUTION:  

  The solutions are about 0.2 and 0.9.

40. 

SOLUTION:  

  The solutions are –0.5 and 2.5.

41. 

SOLUTION:  

  The solutions are about –8.2 and 0.2.

42. 

SOLUTION:  

  The solutions are about –5.1 and 0.1.

43. ASTRONOMY The height of an object t seconds after it is dropped is given by the equation ,

where h0 is the initial height and g is the acceleration due to gravity. The acceleration due to gravity near the surface

of Mars is 3.73 m/s2, while on Earth it is 9.8 m/s

2. Suppose an object is dropped from an initial height of 120 meters

above the surface of each planet. a. On which planet would the object reach the ground first? b. How long would it take the object to reach the ground on each planet? Round each answer to the nearest tenth. c. Do the times that it takes the object to reach the ground seem reasonable? Explain your reasoning.

SOLUTION:  a. The object on Earth will reach the ground first because it is falling at a faster rate.   b. Mars:

So, t ≈ 8.0 seconds. Earth:

So, t ≈ 4.9 seconds.   c. Sample answer: Yes; the acceleration due to gravity is much greater on Earth than on Mars, so the time to reach the ground should be much less.

44. Find all values of c that make x2 + cx + 100 a perfect square trinomial.

SOLUTION:  

So, c can be –20 or 20 to make the trinomial a perfect square.

45. Find all values of c that make x2 + cx + 225 a perfect square trinomial.

SOLUTION:  

So, c can be –30 or 30 to make the trinomial a perfect square.

46. PAINTING Before she begins painting a picture, Donna stretches her canvas over a wood frame. The frame has a length of 60 inches and a width of 4 inches. She has enough canvas to cover 480 square inches. Donna decides to increase the dimensions of the frame. If the increase in the length is 10 times the increase in the width, what will the dimensions of the frame be?

SOLUTION:  Let x = the increase in the width and let 10x = the increase in the length. So, x + 4 = the new width and 10x + 60 = the new length.  

  The dimensions cannot be negative, so x = 2. The width is 2 + 4 or 6 inches and the length is 10(2) + 60 or 80 inches.

47. MULTIPLE REPRESENTATIONS In this problem, you will investigate a property of quadratic equations. a. TABULAR  Copy the table shown and complete the second column.

b. ALGEBRAIC Set each trinomial equal to zero, and solve the equation by completing the square. Complete the last column of the table with the number of roots of each equation.

c. VERBAL Compare the number of roots of each equation to the result in the b2 − 4ac column. Is there a

relationship between these values? If so, describe it.

d. ANALYTICAL Predict how many solutions 2x2 − 9x + 15 = 0 will have.  Verify your prediction by solving the 

equation.

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3    

3x2 + 6x + 9    

x2 − 2x + 7    

x2 + 10x + 25    

x2 + 3x − 12    

SOLUTION:  a.

b.

The trinomial x2 − 8x + 16 has 1 root.

 

The trinomial 2x

2 − 11x + 3 has 2 roots.

 

The square root of a negative number has no real roots. So, the trinomial 3x2 + 6x + 9 has 0 roots.

 

The square root of a negative number has no real roots. So, the trinomial x2 − 2x + 7.

 

The trinomial x2 + 10x + 25 has 1 root.

 

The trinomial x2 + 3x − 12 has 2 roots.

 

c. If b2 − 4ac is negative, the equation has no real solutions. If b

2 − 4ac is zero, the equation has one solution. If b2

− 4ac is positive, the equation has 2 solutions. d.

The equation 2x2 − 9x + 15 = 0 has 0 real solutions because b

2 − 4ac is negative.

 

The equation cannot be solved because the square root of a negative number has no real roots.

Trinomial b2 − 4ac Number

of Roots

x2 − 8x + 16 (–8)

2 – 4(1)(16) = 0 1

2x2 − 11x + 3 (–11)

2 – 4(2)(3) = 97  

3x2 + 6x + 9 (6)

2 – 4(3)(9) = −72  

x2 − 2x + 7 (–2)

2 – 4(1)(7) = −24  

x2 + 10x + 25 (10)

2 – 4(1)(25) = 0  

x2 + 3x − 12 (3)

2 – 4(1)(–12) = 57  

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3 97 2

3x2 + 6x + 9 −72 0

x2 − 2x + 7 −24 0

x2 + 10x + 25 0 1

x2 + 3x − 12 57 2

48. CCSS PERSEVERANCE Given y = ax2 + bx + c with a ≠ 0, derive the equation for the axis of symmetry by 

completing the square and rewriting the equation in the form y = a(x − h)2 + k .

SOLUTION:  

Let    and   , then the equation becomes y = (x - h)2 + k .

For an equation in the form y = ax2 + bx + c,  the equation for the axis of symmetry is   .

So, for an equation in the form  y = (x - h)2 + k, the equation for the axis of symmetry becomes x = h.

49. REASONING Determine the number of solutions x2 + bx = c has if . Explain.

SOLUTION:  

None; Sample answer: If you add to each side of the equation and each side of the inequality, you get

. Since the left side of the last equation is a perfect square, it cannot

equal the negative number . So, there are no real solutions.

50. WHICH ONE DOESN’T BELONG? Identify the expression that does not belong with the other three. Explain your reasoning.

 

SOLUTION:  The first 3 trinomials are perfect squares.

The trinomial  is not a perfect square. So, it does not belong.

51. OPEN ENDED Write a quadratic equation for which the only solution is 4.

SOLUTION:  Find a quadratic that has two roots of 4.

Verify that the solution is 4.

So, the quadratic equation x2 − 8x + 16 = 0 has a solution of 4.

52. WRITING IN MATH Compare and contrast the following strategies for solving x2 − 5x − 7 = 0: completing the

square, graphing, and factoring.

SOLUTION:  Sample answer: Because the leading coefficient is 1, solving the equation by completing the square is simpler and yields exact answers for the solutions.   

 

To solve the equation by graphing, use a graphing calculator to graph the related function y = x2 - 5x - 7. Select the

zero option from the 2nd [CALC] menu to determine the roots. 

   

  The roots are given as decimal approximations. So, for this equation the solutions would have to be given as estimations.   There are no factors of –7 that have a sum of –5, so solving by factoring is not possible.

53. The length of a rectangle is 3 times its width. The area of the rectangle is 75 square feet. Find the length of the rectangle in feet. A 25 B 15 C 10 D 5

SOLUTION:  Let x = the width of the rectangle and let 3x = the length of the rectangle.

The length of the rectangle is 3(5) or 15 feet. Choice B is the correct answer.

54. PROBABILITY At a festival, winners of a game draw a token for a prize. There is one token for each prize. The prizes include 9 movie passes, 8 stuffed animals, 5 hats, 10 jump ropes, and 4 glow necklaces. What is the probabilitythat the first person to draw a token will win a movie pass?

F  

G

H

J  

SOLUTION:  There are 9 + 8 + 5 + 10 + 4 or 36 possible outcomes.

P(movie pass) =  or  . Choice J is the correct answer.

55. GRIDDED RESPONSE The population of a town can be modeled by P = 22,000 + 125t, where P represents the population and t represents the number of years from 2000. How many years after 2000 will the population be 26,000?

SOLUTION:  

In 32 years the population of the town will be 26,000.

56. Percy delivers pizzas for Pizza King. He is paid $6 an hour plus $2.50 for each pizza he delivers. Percy earned $280 last week. If he worked a total of 30 hours, how many pizzas did he deliver? A 250 pizzas B 184 pizzas C 40 pizzas D 34 pizzas

SOLUTION:  Let p = the number of pizzas Percy delivered.

Percy delivered 40 pizzas. Choice C is the correct answer.

Describe how the graph of each function is related to the graph of f (x) = x2.

57. g(x) = −12 + x2

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –12, the translation is

down. So, the graph is shifted down 12 units from the parent function. 

58. h(x) = (x + 2)2

SOLUTION:  

The graph of f (x) = (x – c)2 represents a translation left or right from the parent graph. Since c = –2, the translation

is to the left by 2 units. 

 

59. g(x) = 2x2 + 5

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 2 and c = 5. Since 2 > 0 and  > 1, the graph of y = 2x

2 + 5

is the graph of y = x2 vertically stretched and shifted up 5 units.

60. 

SOLUTION:  

The function can be written f (x) = a(x – b)2, where a = and b = 6. Since > 0 and  < 1, the graph of 

is the graph of y = x2 vertically compressed and shifted right 6 units.

61. g(x) = 6 + x2

SOLUTION:  

The function can be written f (x) = ax2 + c, where a =  and c = 6. Since  > 0 and   > 1, the graph of y = 6 +

x2 is the graph of y = x

2 vertically stretched and shifted up 6 units.

62. h(x) = − 1 −   x2

SOLUTION:  

The function can be written f (x) = –ax2 + c, where a =  and c = –1. Since  < 0 and   > 1, the graph of y

= –1 – x2 is the graph of y = x

2 vertically stretched, shifted down 1 unit and reflected across the x-axis.

63. RIDES A popular amusement park ride whisks riders to the top of a 250-foot tower and drops them. A function for

the height of a rider is h = −16t2 + 250, where h is the height and t is the time in seconds. The ride stops the descent

of the rider 40 feet above the ground. Write an equation that models the drop of the rider. How long does it take to fall from 250 feet to 40 feet?

SOLUTION:  

The ride stops the descent of the rider at 40 feet above ground, so h = 40. Then, the equation 40 = −16t2 + 250

models the drop of the rider.

Time cannot be negative. So, it takes about 3.6 seconds to complete the ride.

Simplify. Assume that no denominator is equal to zero.

64. 

SOLUTION:  

65. 

SOLUTION:  

66. 

SOLUTION:  

67. 

SOLUTION:  

68. 

SOLUTION:  

69. b3(m

−3)(b

−6)

SOLUTION:  

Solve each open sentence.

70. |y − 2| > 7

SOLUTION:  

The solution set is {y |y > 9 or y < −5}.

Case 1 y – 2 is positive.

and Case 2 y – 2 is negative.

 

71. |z + 5| < 3

SOLUTION:  

The solution set is {z |−8 < z < −2}.

Case 1 z +5 is positive.

and Case 2 z +5 is negative.

 

72. |2b + 7| ≤ −6

SOLUTION:  

cannot be negative. So,  cannot be less than or equal to –6. The solution set is empty.

73. |3 − 2y | ≥ 8

SOLUTION:  

The solution set is {y |y  ≥ 5.5 or y  ≤ −2.5}.

Case 1 3 – 2y is positive.

and Case 2 3 – 2y is negative.

 

74. |9 − 4m| < −1

SOLUTION:  

cannot be negative. So, cannot be less than or equal to –1. The solution set is empty.

75. |5c − 2| ≤ 13

SOLUTION:  

The solution set is {c|−2.2 ≤ c ≤ 3}.

Case 1 5c – 2 is positive.

and Case 2 5c – 2 is negative.

 

Evaluate  for each set of values. Round to the nearest tenth if necessary.

76. a = 2, b = −5, c = 2

SOLUTION:  

77. a = 1, b = 12, c = 11

SOLUTION:  

78. a = −9, b = 10, c = −1

SOLUTION:  

79. a = 1, b = 7, c = −3

SOLUTION:  

80. a = 2, b = −4, c = −6

SOLUTION:  

81. a = 3, b = 1, c = 2

SOLUTION:  

This value is not a real number.

eSolutions Manual - Powered by Cognero Page 26

9-4 Solving Quadratic Equations by Completing the Square

Find the value of c that makes each trinomial a perfect square.

1. x2 − 18x + c

SOLUTION:  In this trinomial, b = –18.

So, c must be 81 to make the trinomial a perfect square.

2. x2 + 22x + c

SOLUTION:  In this trinomial, b = 22.

So, c must be 121 to make the trinomial a perfect square.

3. x2 + 9x + c

SOLUTION:  In this trinomial, b = 9.

 

So, c must be  to make the trinomial a perfect square.

4. x2 − 7x + c

SOLUTION:  In this trinomial, b = –7.

 

So, c must be  to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

5. x2 + 4x = 6

SOLUTION:  

The solutions are about –5.2 and 1.2.

6. x2 − 8x = −9

SOLUTION:  

The solutions are about 1.4 and 6.6.

7. 4x2 + 9x − 1 = 0

SOLUTION:  

The solutions are about –2.4 and 0.1.

8. −2x2 + 10x + 22 = 4

SOLUTION:  

The solutions are about –1.4 and 6.4.

9. CCSS MODELING Collin is building a deck on the back of his family’s house. He has enough lumber for the deck to be 144 square feet. The length should be 10 feet more than its width. What should the dimensions of the deck be?

SOLUTION:  Let x = the width of the deck and x + 10 = the length of the deck.

The dimensions of the deck can not be negative. So, x = –5 + 13 or 8. The width of the deck is 8 feet and the length of the deck is 8 + 10 or 18 feet.

Find the value of c that makes each trinomial a perfect square.

10. x2 + 26x + c

SOLUTION:  In this trinomial, b = 26.

So, c must be 169 to make the trinomial a perfect square.

11. x2 − 24x + c

SOLUTION:  In this trinomial, b = –24.

So, c must be 144 to make the trinomial a perfect square.

12. x2 − 19x + c

SOLUTION:  In this trinomial, b = –19.

So, c must be  to make the trinomial a perfect square.

13. x2 + 17x + c

SOLUTION:  In this trinomial, b = 17.

So, c must be  to make the trinomial a perfect square.

14. x2 + 5x + c

SOLUTION:  In this trinomial, b = 5.

So, c must be  to make the trinomial a perfect square.

15. x2 − 13x + c

SOLUTION:  In this trinomial, b = –13.

So, c must be  to make the trinomial a perfect square.

16. x2 − 22x + c

SOLUTION:  In this trinomial, b = –22.

So, c must be 121 to make the trinomial a perfect square.

17. x2 − 15x + c

SOLUTION:  In this trinomial, b = –15.

So, c must be  to make the trinomial a perfect square.

18. x2 + 24x + c

SOLUTION:  In this trinomial, b = 24.

So, c must be 144 to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

19. x2 + 6x − 16 = 0

SOLUTION:  

  The solutions are –8 and 2.

20. x2 − 2x − 14 = 0

SOLUTION:  

  The solutions are about –2.9 and 4.9.

21. x2 − 8x − 1 = 8

SOLUTION:  

  The solutions are –1 and 9.

22. x2 + 3x + 21 = 22

SOLUTION:  

  The solutions are about –3.3 and 0.3.

23. x2 − 11x + 3 = 5

SOLUTION:  

  The solutions are about –0.2 and 11.2.

24. 5x2 − 10x = 23

SOLUTION:  

  The solutions are about –1.4 and 3.4.

25. 2x2 − 2x + 7 = 5

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

26. 3x2 + 12x + 81 = 15

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

27. 4x2 + 6x = 12

SOLUTION:  

  The solutions are about –2.6 and 1.1.

28. 4x2 + 5 = 10x

SOLUTION:  

  The solutions are about 0.7 and 1.8.

29. −2x2 + 10x = −14

SOLUTION:  

  The solutions are about –1.1 and 6.1.

30. −3x2 − 12 = 14x

SOLUTION:  

  The solutions are about –3.5 and –1.1.

31. STOCK The price p in dollars for a particular stock can be modeled by the quadratic equation p = 3.5t − 0.05t2, wh

the number of days after the stock is purchased. When is the stock worth $60?

SOLUTION:  Let p = 60.

  The stock is worth $60 on the 30th and 40th days after purchase.

GEOMETRY  Find the value of x for each figure. Round to the nearest tenth if necessary.

32. area = 45 in2

SOLUTION:  

  The height cannot be negative. So, x ≈ 6.3.

33. area = 110 ft2

SOLUTION:  

  The dimensions of the rectangle cannot be negative. So, x ≈ 5.3.

34. NUMBER THEORY The product of two consecutive even integers is 224. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers are 14 and 16 or –16 and –14.

35. CCSS PRECISION The product of two consecutive negative odd integers is 483. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers must be negative. So, they are –23 and –21.

36. GEOMETRY Find the area of the triangle.

SOLUTION:  

  The dimensions of the triangle cannot be negative. So, x = 18. The base of the triangle is 18 meters and the height is 18 + 6 or 24 meters.

The area of the triangle is 216 square meters.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

37. 0.2x2 − 0.2x − 0.4 = 0

SOLUTION:  

  The solutions are –1 and 2.

38. 0.5x2 = 2x − 0.3

SOLUTION:  

  The solutions are about 0.2 and 3.8.

39. 2x2 −

SOLUTION:  

  The solutions are about 0.2 and 0.9.

40. 

SOLUTION:  

  The solutions are –0.5 and 2.5.

41. 

SOLUTION:  

  The solutions are about –8.2 and 0.2.

42. 

SOLUTION:  

  The solutions are about –5.1 and 0.1.

43. ASTRONOMY The height of an object t seconds after it is dropped is given by the equation ,

where h0 is the initial height and g is the acceleration due to gravity. The acceleration due to gravity near the surface

of Mars is 3.73 m/s2, while on Earth it is 9.8 m/s

2. Suppose an object is dropped from an initial height of 120 meters

above the surface of each planet. a. On which planet would the object reach the ground first? b. How long would it take the object to reach the ground on each planet? Round each answer to the nearest tenth. c. Do the times that it takes the object to reach the ground seem reasonable? Explain your reasoning.

SOLUTION:  a. The object on Earth will reach the ground first because it is falling at a faster rate.   b. Mars:

So, t ≈ 8.0 seconds. Earth:

So, t ≈ 4.9 seconds.   c. Sample answer: Yes; the acceleration due to gravity is much greater on Earth than on Mars, so the time to reach the ground should be much less.

44. Find all values of c that make x2 + cx + 100 a perfect square trinomial.

SOLUTION:  

So, c can be –20 or 20 to make the trinomial a perfect square.

45. Find all values of c that make x2 + cx + 225 a perfect square trinomial.

SOLUTION:  

So, c can be –30 or 30 to make the trinomial a perfect square.

46. PAINTING Before she begins painting a picture, Donna stretches her canvas over a wood frame. The frame has a length of 60 inches and a width of 4 inches. She has enough canvas to cover 480 square inches. Donna decides to increase the dimensions of the frame. If the increase in the length is 10 times the increase in the width, what will the dimensions of the frame be?

SOLUTION:  Let x = the increase in the width and let 10x = the increase in the length. So, x + 4 = the new width and 10x + 60 = the new length.  

  The dimensions cannot be negative, so x = 2. The width is 2 + 4 or 6 inches and the length is 10(2) + 60 or 80 inches.

47. MULTIPLE REPRESENTATIONS In this problem, you will investigate a property of quadratic equations. a. TABULAR  Copy the table shown and complete the second column.

b. ALGEBRAIC Set each trinomial equal to zero, and solve the equation by completing the square. Complete the last column of the table with the number of roots of each equation.

c. VERBAL Compare the number of roots of each equation to the result in the b2 − 4ac column. Is there a

relationship between these values? If so, describe it.

d. ANALYTICAL Predict how many solutions 2x2 − 9x + 15 = 0 will have.  Verify your prediction by solving the 

equation.

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3    

3x2 + 6x + 9    

x2 − 2x + 7    

x2 + 10x + 25    

x2 + 3x − 12    

SOLUTION:  a.

b.

The trinomial x2 − 8x + 16 has 1 root.

 

The trinomial 2x

2 − 11x + 3 has 2 roots.

 

The square root of a negative number has no real roots. So, the trinomial 3x2 + 6x + 9 has 0 roots.

 

The square root of a negative number has no real roots. So, the trinomial x2 − 2x + 7.

 

The trinomial x2 + 10x + 25 has 1 root.

 

The trinomial x2 + 3x − 12 has 2 roots.

 

c. If b2 − 4ac is negative, the equation has no real solutions. If b

2 − 4ac is zero, the equation has one solution. If b2

− 4ac is positive, the equation has 2 solutions. d.

The equation 2x2 − 9x + 15 = 0 has 0 real solutions because b

2 − 4ac is negative.

 

The equation cannot be solved because the square root of a negative number has no real roots.

Trinomial b2 − 4ac Number

of Roots

x2 − 8x + 16 (–8)

2 – 4(1)(16) = 0 1

2x2 − 11x + 3 (–11)

2 – 4(2)(3) = 97  

3x2 + 6x + 9 (6)

2 – 4(3)(9) = −72  

x2 − 2x + 7 (–2)

2 – 4(1)(7) = −24  

x2 + 10x + 25 (10)

2 – 4(1)(25) = 0  

x2 + 3x − 12 (3)

2 – 4(1)(–12) = 57  

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3 97 2

3x2 + 6x + 9 −72 0

x2 − 2x + 7 −24 0

x2 + 10x + 25 0 1

x2 + 3x − 12 57 2

48. CCSS PERSEVERANCE Given y = ax2 + bx + c with a ≠ 0, derive the equation for the axis of symmetry by 

completing the square and rewriting the equation in the form y = a(x − h)2 + k .

SOLUTION:  

Let    and   , then the equation becomes y = (x - h)2 + k .

For an equation in the form y = ax2 + bx + c,  the equation for the axis of symmetry is   .

So, for an equation in the form  y = (x - h)2 + k, the equation for the axis of symmetry becomes x = h.

49. REASONING Determine the number of solutions x2 + bx = c has if . Explain.

SOLUTION:  

None; Sample answer: If you add to each side of the equation and each side of the inequality, you get

. Since the left side of the last equation is a perfect square, it cannot

equal the negative number . So, there are no real solutions.

50. WHICH ONE DOESN’T BELONG? Identify the expression that does not belong with the other three. Explain your reasoning.

 

SOLUTION:  The first 3 trinomials are perfect squares.

The trinomial  is not a perfect square. So, it does not belong.

51. OPEN ENDED Write a quadratic equation for which the only solution is 4.

SOLUTION:  Find a quadratic that has two roots of 4.

Verify that the solution is 4.

So, the quadratic equation x2 − 8x + 16 = 0 has a solution of 4.

52. WRITING IN MATH Compare and contrast the following strategies for solving x2 − 5x − 7 = 0: completing the

square, graphing, and factoring.

SOLUTION:  Sample answer: Because the leading coefficient is 1, solving the equation by completing the square is simpler and yields exact answers for the solutions.   

 

To solve the equation by graphing, use a graphing calculator to graph the related function y = x2 - 5x - 7. Select the

zero option from the 2nd [CALC] menu to determine the roots. 

   

  The roots are given as decimal approximations. So, for this equation the solutions would have to be given as estimations.   There are no factors of –7 that have a sum of –5, so solving by factoring is not possible.

53. The length of a rectangle is 3 times its width. The area of the rectangle is 75 square feet. Find the length of the rectangle in feet. A 25 B 15 C 10 D 5

SOLUTION:  Let x = the width of the rectangle and let 3x = the length of the rectangle.

The length of the rectangle is 3(5) or 15 feet. Choice B is the correct answer.

54. PROBABILITY At a festival, winners of a game draw a token for a prize. There is one token for each prize. The prizes include 9 movie passes, 8 stuffed animals, 5 hats, 10 jump ropes, and 4 glow necklaces. What is the probabilitythat the first person to draw a token will win a movie pass?

F  

G

H

J  

SOLUTION:  There are 9 + 8 + 5 + 10 + 4 or 36 possible outcomes.

P(movie pass) =  or  . Choice J is the correct answer.

55. GRIDDED RESPONSE The population of a town can be modeled by P = 22,000 + 125t, where P represents the population and t represents the number of years from 2000. How many years after 2000 will the population be 26,000?

SOLUTION:  

In 32 years the population of the town will be 26,000.

56. Percy delivers pizzas for Pizza King. He is paid $6 an hour plus $2.50 for each pizza he delivers. Percy earned $280 last week. If he worked a total of 30 hours, how many pizzas did he deliver? A 250 pizzas B 184 pizzas C 40 pizzas D 34 pizzas

SOLUTION:  Let p = the number of pizzas Percy delivered.

Percy delivered 40 pizzas. Choice C is the correct answer.

Describe how the graph of each function is related to the graph of f (x) = x2.

57. g(x) = −12 + x2

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –12, the translation is

down. So, the graph is shifted down 12 units from the parent function. 

58. h(x) = (x + 2)2

SOLUTION:  

The graph of f (x) = (x – c)2 represents a translation left or right from the parent graph. Since c = –2, the translation

is to the left by 2 units. 

 

59. g(x) = 2x2 + 5

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 2 and c = 5. Since 2 > 0 and  > 1, the graph of y = 2x

2 + 5

is the graph of y = x2 vertically stretched and shifted up 5 units.

60. 

SOLUTION:  

The function can be written f (x) = a(x – b)2, where a = and b = 6. Since > 0 and  < 1, the graph of 

is the graph of y = x2 vertically compressed and shifted right 6 units.

61. g(x) = 6 + x2

SOLUTION:  

The function can be written f (x) = ax2 + c, where a =  and c = 6. Since  > 0 and   > 1, the graph of y = 6 +

x2 is the graph of y = x

2 vertically stretched and shifted up 6 units.

62. h(x) = − 1 −   x2

SOLUTION:  

The function can be written f (x) = –ax2 + c, where a =  and c = –1. Since  < 0 and   > 1, the graph of y

= –1 – x2 is the graph of y = x

2 vertically stretched, shifted down 1 unit and reflected across the x-axis.

63. RIDES A popular amusement park ride whisks riders to the top of a 250-foot tower and drops them. A function for

the height of a rider is h = −16t2 + 250, where h is the height and t is the time in seconds. The ride stops the descent

of the rider 40 feet above the ground. Write an equation that models the drop of the rider. How long does it take to fall from 250 feet to 40 feet?

SOLUTION:  

The ride stops the descent of the rider at 40 feet above ground, so h = 40. Then, the equation 40 = −16t2 + 250

models the drop of the rider.

Time cannot be negative. So, it takes about 3.6 seconds to complete the ride.

Simplify. Assume that no denominator is equal to zero.

64. 

SOLUTION:  

65. 

SOLUTION:  

66. 

SOLUTION:  

67. 

SOLUTION:  

68. 

SOLUTION:  

69. b3(m

−3)(b

−6)

SOLUTION:  

Solve each open sentence.

70. |y − 2| > 7

SOLUTION:  

The solution set is {y |y > 9 or y < −5}.

Case 1 y – 2 is positive.

and Case 2 y – 2 is negative.

 

71. |z + 5| < 3

SOLUTION:  

The solution set is {z |−8 < z < −2}.

Case 1 z +5 is positive.

and Case 2 z +5 is negative.

 

72. |2b + 7| ≤ −6

SOLUTION:  

cannot be negative. So,  cannot be less than or equal to –6. The solution set is empty.

73. |3 − 2y | ≥ 8

SOLUTION:  

The solution set is {y |y  ≥ 5.5 or y  ≤ −2.5}.

Case 1 3 – 2y is positive.

and Case 2 3 – 2y is negative.

 

74. |9 − 4m| < −1

SOLUTION:  

cannot be negative. So, cannot be less than or equal to –1. The solution set is empty.

75. |5c − 2| ≤ 13

SOLUTION:  

The solution set is {c|−2.2 ≤ c ≤ 3}.

Case 1 5c – 2 is positive.

and Case 2 5c – 2 is negative.

 

Evaluate  for each set of values. Round to the nearest tenth if necessary.

76. a = 2, b = −5, c = 2

SOLUTION:  

77. a = 1, b = 12, c = 11

SOLUTION:  

78. a = −9, b = 10, c = −1

SOLUTION:  

79. a = 1, b = 7, c = −3

SOLUTION:  

80. a = 2, b = −4, c = −6

SOLUTION:  

81. a = 3, b = 1, c = 2

SOLUTION:  

This value is not a real number.

eSolutions Manual - Powered by Cognero Page 27

9-4 Solving Quadratic Equations by Completing the Square

Find the value of c that makes each trinomial a perfect square.

1. x2 − 18x + c

SOLUTION:  In this trinomial, b = –18.

So, c must be 81 to make the trinomial a perfect square.

2. x2 + 22x + c

SOLUTION:  In this trinomial, b = 22.

So, c must be 121 to make the trinomial a perfect square.

3. x2 + 9x + c

SOLUTION:  In this trinomial, b = 9.

 

So, c must be  to make the trinomial a perfect square.

4. x2 − 7x + c

SOLUTION:  In this trinomial, b = –7.

 

So, c must be  to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

5. x2 + 4x = 6

SOLUTION:  

The solutions are about –5.2 and 1.2.

6. x2 − 8x = −9

SOLUTION:  

The solutions are about 1.4 and 6.6.

7. 4x2 + 9x − 1 = 0

SOLUTION:  

The solutions are about –2.4 and 0.1.

8. −2x2 + 10x + 22 = 4

SOLUTION:  

The solutions are about –1.4 and 6.4.

9. CCSS MODELING Collin is building a deck on the back of his family’s house. He has enough lumber for the deck to be 144 square feet. The length should be 10 feet more than its width. What should the dimensions of the deck be?

SOLUTION:  Let x = the width of the deck and x + 10 = the length of the deck.

The dimensions of the deck can not be negative. So, x = –5 + 13 or 8. The width of the deck is 8 feet and the length of the deck is 8 + 10 or 18 feet.

Find the value of c that makes each trinomial a perfect square.

10. x2 + 26x + c

SOLUTION:  In this trinomial, b = 26.

So, c must be 169 to make the trinomial a perfect square.

11. x2 − 24x + c

SOLUTION:  In this trinomial, b = –24.

So, c must be 144 to make the trinomial a perfect square.

12. x2 − 19x + c

SOLUTION:  In this trinomial, b = –19.

So, c must be  to make the trinomial a perfect square.

13. x2 + 17x + c

SOLUTION:  In this trinomial, b = 17.

So, c must be  to make the trinomial a perfect square.

14. x2 + 5x + c

SOLUTION:  In this trinomial, b = 5.

So, c must be  to make the trinomial a perfect square.

15. x2 − 13x + c

SOLUTION:  In this trinomial, b = –13.

So, c must be  to make the trinomial a perfect square.

16. x2 − 22x + c

SOLUTION:  In this trinomial, b = –22.

So, c must be 121 to make the trinomial a perfect square.

17. x2 − 15x + c

SOLUTION:  In this trinomial, b = –15.

So, c must be  to make the trinomial a perfect square.

18. x2 + 24x + c

SOLUTION:  In this trinomial, b = 24.

So, c must be 144 to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

19. x2 + 6x − 16 = 0

SOLUTION:  

  The solutions are –8 and 2.

20. x2 − 2x − 14 = 0

SOLUTION:  

  The solutions are about –2.9 and 4.9.

21. x2 − 8x − 1 = 8

SOLUTION:  

  The solutions are –1 and 9.

22. x2 + 3x + 21 = 22

SOLUTION:  

  The solutions are about –3.3 and 0.3.

23. x2 − 11x + 3 = 5

SOLUTION:  

  The solutions are about –0.2 and 11.2.

24. 5x2 − 10x = 23

SOLUTION:  

  The solutions are about –1.4 and 3.4.

25. 2x2 − 2x + 7 = 5

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

26. 3x2 + 12x + 81 = 15

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

27. 4x2 + 6x = 12

SOLUTION:  

  The solutions are about –2.6 and 1.1.

28. 4x2 + 5 = 10x

SOLUTION:  

  The solutions are about 0.7 and 1.8.

29. −2x2 + 10x = −14

SOLUTION:  

  The solutions are about –1.1 and 6.1.

30. −3x2 − 12 = 14x

SOLUTION:  

  The solutions are about –3.5 and –1.1.

31. STOCK The price p in dollars for a particular stock can be modeled by the quadratic equation p = 3.5t − 0.05t2, wh

the number of days after the stock is purchased. When is the stock worth $60?

SOLUTION:  Let p = 60.

  The stock is worth $60 on the 30th and 40th days after purchase.

GEOMETRY  Find the value of x for each figure. Round to the nearest tenth if necessary.

32. area = 45 in2

SOLUTION:  

  The height cannot be negative. So, x ≈ 6.3.

33. area = 110 ft2

SOLUTION:  

  The dimensions of the rectangle cannot be negative. So, x ≈ 5.3.

34. NUMBER THEORY The product of two consecutive even integers is 224. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers are 14 and 16 or –16 and –14.

35. CCSS PRECISION The product of two consecutive negative odd integers is 483. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers must be negative. So, they are –23 and –21.

36. GEOMETRY Find the area of the triangle.

SOLUTION:  

  The dimensions of the triangle cannot be negative. So, x = 18. The base of the triangle is 18 meters and the height is 18 + 6 or 24 meters.

The area of the triangle is 216 square meters.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

37. 0.2x2 − 0.2x − 0.4 = 0

SOLUTION:  

  The solutions are –1 and 2.

38. 0.5x2 = 2x − 0.3

SOLUTION:  

  The solutions are about 0.2 and 3.8.

39. 2x2 −

SOLUTION:  

  The solutions are about 0.2 and 0.9.

40. 

SOLUTION:  

  The solutions are –0.5 and 2.5.

41. 

SOLUTION:  

  The solutions are about –8.2 and 0.2.

42. 

SOLUTION:  

  The solutions are about –5.1 and 0.1.

43. ASTRONOMY The height of an object t seconds after it is dropped is given by the equation ,

where h0 is the initial height and g is the acceleration due to gravity. The acceleration due to gravity near the surface

of Mars is 3.73 m/s2, while on Earth it is 9.8 m/s

2. Suppose an object is dropped from an initial height of 120 meters

above the surface of each planet. a. On which planet would the object reach the ground first? b. How long would it take the object to reach the ground on each planet? Round each answer to the nearest tenth. c. Do the times that it takes the object to reach the ground seem reasonable? Explain your reasoning.

SOLUTION:  a. The object on Earth will reach the ground first because it is falling at a faster rate.   b. Mars:

So, t ≈ 8.0 seconds. Earth:

So, t ≈ 4.9 seconds.   c. Sample answer: Yes; the acceleration due to gravity is much greater on Earth than on Mars, so the time to reach the ground should be much less.

44. Find all values of c that make x2 + cx + 100 a perfect square trinomial.

SOLUTION:  

So, c can be –20 or 20 to make the trinomial a perfect square.

45. Find all values of c that make x2 + cx + 225 a perfect square trinomial.

SOLUTION:  

So, c can be –30 or 30 to make the trinomial a perfect square.

46. PAINTING Before she begins painting a picture, Donna stretches her canvas over a wood frame. The frame has a length of 60 inches and a width of 4 inches. She has enough canvas to cover 480 square inches. Donna decides to increase the dimensions of the frame. If the increase in the length is 10 times the increase in the width, what will the dimensions of the frame be?

SOLUTION:  Let x = the increase in the width and let 10x = the increase in the length. So, x + 4 = the new width and 10x + 60 = the new length.  

  The dimensions cannot be negative, so x = 2. The width is 2 + 4 or 6 inches and the length is 10(2) + 60 or 80 inches.

47. MULTIPLE REPRESENTATIONS In this problem, you will investigate a property of quadratic equations. a. TABULAR  Copy the table shown and complete the second column.

b. ALGEBRAIC Set each trinomial equal to zero, and solve the equation by completing the square. Complete the last column of the table with the number of roots of each equation.

c. VERBAL Compare the number of roots of each equation to the result in the b2 − 4ac column. Is there a

relationship between these values? If so, describe it.

d. ANALYTICAL Predict how many solutions 2x2 − 9x + 15 = 0 will have.  Verify your prediction by solving the 

equation.

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3    

3x2 + 6x + 9    

x2 − 2x + 7    

x2 + 10x + 25    

x2 + 3x − 12    

SOLUTION:  a.

b.

The trinomial x2 − 8x + 16 has 1 root.

 

The trinomial 2x

2 − 11x + 3 has 2 roots.

 

The square root of a negative number has no real roots. So, the trinomial 3x2 + 6x + 9 has 0 roots.

 

The square root of a negative number has no real roots. So, the trinomial x2 − 2x + 7.

 

The trinomial x2 + 10x + 25 has 1 root.

 

The trinomial x2 + 3x − 12 has 2 roots.

 

c. If b2 − 4ac is negative, the equation has no real solutions. If b

2 − 4ac is zero, the equation has one solution. If b2

− 4ac is positive, the equation has 2 solutions. d.

The equation 2x2 − 9x + 15 = 0 has 0 real solutions because b

2 − 4ac is negative.

 

The equation cannot be solved because the square root of a negative number has no real roots.

Trinomial b2 − 4ac Number

of Roots

x2 − 8x + 16 (–8)

2 – 4(1)(16) = 0 1

2x2 − 11x + 3 (–11)

2 – 4(2)(3) = 97  

3x2 + 6x + 9 (6)

2 – 4(3)(9) = −72  

x2 − 2x + 7 (–2)

2 – 4(1)(7) = −24  

x2 + 10x + 25 (10)

2 – 4(1)(25) = 0  

x2 + 3x − 12 (3)

2 – 4(1)(–12) = 57  

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3 97 2

3x2 + 6x + 9 −72 0

x2 − 2x + 7 −24 0

x2 + 10x + 25 0 1

x2 + 3x − 12 57 2

48. CCSS PERSEVERANCE Given y = ax2 + bx + c with a ≠ 0, derive the equation for the axis of symmetry by 

completing the square and rewriting the equation in the form y = a(x − h)2 + k .

SOLUTION:  

Let    and   , then the equation becomes y = (x - h)2 + k .

For an equation in the form y = ax2 + bx + c,  the equation for the axis of symmetry is   .

So, for an equation in the form  y = (x - h)2 + k, the equation for the axis of symmetry becomes x = h.

49. REASONING Determine the number of solutions x2 + bx = c has if . Explain.

SOLUTION:  

None; Sample answer: If you add to each side of the equation and each side of the inequality, you get

. Since the left side of the last equation is a perfect square, it cannot

equal the negative number . So, there are no real solutions.

50. WHICH ONE DOESN’T BELONG? Identify the expression that does not belong with the other three. Explain your reasoning.

 

SOLUTION:  The first 3 trinomials are perfect squares.

The trinomial  is not a perfect square. So, it does not belong.

51. OPEN ENDED Write a quadratic equation for which the only solution is 4.

SOLUTION:  Find a quadratic that has two roots of 4.

Verify that the solution is 4.

So, the quadratic equation x2 − 8x + 16 = 0 has a solution of 4.

52. WRITING IN MATH Compare and contrast the following strategies for solving x2 − 5x − 7 = 0: completing the

square, graphing, and factoring.

SOLUTION:  Sample answer: Because the leading coefficient is 1, solving the equation by completing the square is simpler and yields exact answers for the solutions.   

 

To solve the equation by graphing, use a graphing calculator to graph the related function y = x2 - 5x - 7. Select the

zero option from the 2nd [CALC] menu to determine the roots. 

   

  The roots are given as decimal approximations. So, for this equation the solutions would have to be given as estimations.   There are no factors of –7 that have a sum of –5, so solving by factoring is not possible.

53. The length of a rectangle is 3 times its width. The area of the rectangle is 75 square feet. Find the length of the rectangle in feet. A 25 B 15 C 10 D 5

SOLUTION:  Let x = the width of the rectangle and let 3x = the length of the rectangle.

The length of the rectangle is 3(5) or 15 feet. Choice B is the correct answer.

54. PROBABILITY At a festival, winners of a game draw a token for a prize. There is one token for each prize. The prizes include 9 movie passes, 8 stuffed animals, 5 hats, 10 jump ropes, and 4 glow necklaces. What is the probabilitythat the first person to draw a token will win a movie pass?

F  

G

H

J  

SOLUTION:  There are 9 + 8 + 5 + 10 + 4 or 36 possible outcomes.

P(movie pass) =  or  . Choice J is the correct answer.

55. GRIDDED RESPONSE The population of a town can be modeled by P = 22,000 + 125t, where P represents the population and t represents the number of years from 2000. How many years after 2000 will the population be 26,000?

SOLUTION:  

In 32 years the population of the town will be 26,000.

56. Percy delivers pizzas for Pizza King. He is paid $6 an hour plus $2.50 for each pizza he delivers. Percy earned $280 last week. If he worked a total of 30 hours, how many pizzas did he deliver? A 250 pizzas B 184 pizzas C 40 pizzas D 34 pizzas

SOLUTION:  Let p = the number of pizzas Percy delivered.

Percy delivered 40 pizzas. Choice C is the correct answer.

Describe how the graph of each function is related to the graph of f (x) = x2.

57. g(x) = −12 + x2

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –12, the translation is

down. So, the graph is shifted down 12 units from the parent function. 

58. h(x) = (x + 2)2

SOLUTION:  

The graph of f (x) = (x – c)2 represents a translation left or right from the parent graph. Since c = –2, the translation

is to the left by 2 units. 

 

59. g(x) = 2x2 + 5

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 2 and c = 5. Since 2 > 0 and  > 1, the graph of y = 2x

2 + 5

is the graph of y = x2 vertically stretched and shifted up 5 units.

60. 

SOLUTION:  

The function can be written f (x) = a(x – b)2, where a = and b = 6. Since > 0 and  < 1, the graph of 

is the graph of y = x2 vertically compressed and shifted right 6 units.

61. g(x) = 6 + x2

SOLUTION:  

The function can be written f (x) = ax2 + c, where a =  and c = 6. Since  > 0 and   > 1, the graph of y = 6 +

x2 is the graph of y = x

2 vertically stretched and shifted up 6 units.

62. h(x) = − 1 −   x2

SOLUTION:  

The function can be written f (x) = –ax2 + c, where a =  and c = –1. Since  < 0 and   > 1, the graph of y

= –1 – x2 is the graph of y = x

2 vertically stretched, shifted down 1 unit and reflected across the x-axis.

63. RIDES A popular amusement park ride whisks riders to the top of a 250-foot tower and drops them. A function for

the height of a rider is h = −16t2 + 250, where h is the height and t is the time in seconds. The ride stops the descent

of the rider 40 feet above the ground. Write an equation that models the drop of the rider. How long does it take to fall from 250 feet to 40 feet?

SOLUTION:  

The ride stops the descent of the rider at 40 feet above ground, so h = 40. Then, the equation 40 = −16t2 + 250

models the drop of the rider.

Time cannot be negative. So, it takes about 3.6 seconds to complete the ride.

Simplify. Assume that no denominator is equal to zero.

64. 

SOLUTION:  

65. 

SOLUTION:  

66. 

SOLUTION:  

67. 

SOLUTION:  

68. 

SOLUTION:  

69. b3(m

−3)(b

−6)

SOLUTION:  

Solve each open sentence.

70. |y − 2| > 7

SOLUTION:  

The solution set is {y |y > 9 or y < −5}.

Case 1 y – 2 is positive.

and Case 2 y – 2 is negative.

 

71. |z + 5| < 3

SOLUTION:  

The solution set is {z |−8 < z < −2}.

Case 1 z +5 is positive.

and Case 2 z +5 is negative.

 

72. |2b + 7| ≤ −6

SOLUTION:  

cannot be negative. So,  cannot be less than or equal to –6. The solution set is empty.

73. |3 − 2y | ≥ 8

SOLUTION:  

The solution set is {y |y  ≥ 5.5 or y  ≤ −2.5}.

Case 1 3 – 2y is positive.

and Case 2 3 – 2y is negative.

 

74. |9 − 4m| < −1

SOLUTION:  

cannot be negative. So, cannot be less than or equal to –1. The solution set is empty.

75. |5c − 2| ≤ 13

SOLUTION:  

The solution set is {c|−2.2 ≤ c ≤ 3}.

Case 1 5c – 2 is positive.

and Case 2 5c – 2 is negative.

 

Evaluate  for each set of values. Round to the nearest tenth if necessary.

76. a = 2, b = −5, c = 2

SOLUTION:  

77. a = 1, b = 12, c = 11

SOLUTION:  

78. a = −9, b = 10, c = −1

SOLUTION:  

79. a = 1, b = 7, c = −3

SOLUTION:  

80. a = 2, b = −4, c = −6

SOLUTION:  

81. a = 3, b = 1, c = 2

SOLUTION:  

This value is not a real number.

eSolutions Manual - Powered by Cognero Page 28

9-4 Solving Quadratic Equations by Completing the Square

Find the value of c that makes each trinomial a perfect square.

1. x2 − 18x + c

SOLUTION:  In this trinomial, b = –18.

So, c must be 81 to make the trinomial a perfect square.

2. x2 + 22x + c

SOLUTION:  In this trinomial, b = 22.

So, c must be 121 to make the trinomial a perfect square.

3. x2 + 9x + c

SOLUTION:  In this trinomial, b = 9.

 

So, c must be  to make the trinomial a perfect square.

4. x2 − 7x + c

SOLUTION:  In this trinomial, b = –7.

 

So, c must be  to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

5. x2 + 4x = 6

SOLUTION:  

The solutions are about –5.2 and 1.2.

6. x2 − 8x = −9

SOLUTION:  

The solutions are about 1.4 and 6.6.

7. 4x2 + 9x − 1 = 0

SOLUTION:  

The solutions are about –2.4 and 0.1.

8. −2x2 + 10x + 22 = 4

SOLUTION:  

The solutions are about –1.4 and 6.4.

9. CCSS MODELING Collin is building a deck on the back of his family’s house. He has enough lumber for the deck to be 144 square feet. The length should be 10 feet more than its width. What should the dimensions of the deck be?

SOLUTION:  Let x = the width of the deck and x + 10 = the length of the deck.

The dimensions of the deck can not be negative. So, x = –5 + 13 or 8. The width of the deck is 8 feet and the length of the deck is 8 + 10 or 18 feet.

Find the value of c that makes each trinomial a perfect square.

10. x2 + 26x + c

SOLUTION:  In this trinomial, b = 26.

So, c must be 169 to make the trinomial a perfect square.

11. x2 − 24x + c

SOLUTION:  In this trinomial, b = –24.

So, c must be 144 to make the trinomial a perfect square.

12. x2 − 19x + c

SOLUTION:  In this trinomial, b = –19.

So, c must be  to make the trinomial a perfect square.

13. x2 + 17x + c

SOLUTION:  In this trinomial, b = 17.

So, c must be  to make the trinomial a perfect square.

14. x2 + 5x + c

SOLUTION:  In this trinomial, b = 5.

So, c must be  to make the trinomial a perfect square.

15. x2 − 13x + c

SOLUTION:  In this trinomial, b = –13.

So, c must be  to make the trinomial a perfect square.

16. x2 − 22x + c

SOLUTION:  In this trinomial, b = –22.

So, c must be 121 to make the trinomial a perfect square.

17. x2 − 15x + c

SOLUTION:  In this trinomial, b = –15.

So, c must be  to make the trinomial a perfect square.

18. x2 + 24x + c

SOLUTION:  In this trinomial, b = 24.

So, c must be 144 to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

19. x2 + 6x − 16 = 0

SOLUTION:  

  The solutions are –8 and 2.

20. x2 − 2x − 14 = 0

SOLUTION:  

  The solutions are about –2.9 and 4.9.

21. x2 − 8x − 1 = 8

SOLUTION:  

  The solutions are –1 and 9.

22. x2 + 3x + 21 = 22

SOLUTION:  

  The solutions are about –3.3 and 0.3.

23. x2 − 11x + 3 = 5

SOLUTION:  

  The solutions are about –0.2 and 11.2.

24. 5x2 − 10x = 23

SOLUTION:  

  The solutions are about –1.4 and 3.4.

25. 2x2 − 2x + 7 = 5

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

26. 3x2 + 12x + 81 = 15

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

27. 4x2 + 6x = 12

SOLUTION:  

  The solutions are about –2.6 and 1.1.

28. 4x2 + 5 = 10x

SOLUTION:  

  The solutions are about 0.7 and 1.8.

29. −2x2 + 10x = −14

SOLUTION:  

  The solutions are about –1.1 and 6.1.

30. −3x2 − 12 = 14x

SOLUTION:  

  The solutions are about –3.5 and –1.1.

31. STOCK The price p in dollars for a particular stock can be modeled by the quadratic equation p = 3.5t − 0.05t2, wh

the number of days after the stock is purchased. When is the stock worth $60?

SOLUTION:  Let p = 60.

  The stock is worth $60 on the 30th and 40th days after purchase.

GEOMETRY  Find the value of x for each figure. Round to the nearest tenth if necessary.

32. area = 45 in2

SOLUTION:  

  The height cannot be negative. So, x ≈ 6.3.

33. area = 110 ft2

SOLUTION:  

  The dimensions of the rectangle cannot be negative. So, x ≈ 5.3.

34. NUMBER THEORY The product of two consecutive even integers is 224. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers are 14 and 16 or –16 and –14.

35. CCSS PRECISION The product of two consecutive negative odd integers is 483. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers must be negative. So, they are –23 and –21.

36. GEOMETRY Find the area of the triangle.

SOLUTION:  

  The dimensions of the triangle cannot be negative. So, x = 18. The base of the triangle is 18 meters and the height is 18 + 6 or 24 meters.

The area of the triangle is 216 square meters.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

37. 0.2x2 − 0.2x − 0.4 = 0

SOLUTION:  

  The solutions are –1 and 2.

38. 0.5x2 = 2x − 0.3

SOLUTION:  

  The solutions are about 0.2 and 3.8.

39. 2x2 −

SOLUTION:  

  The solutions are about 0.2 and 0.9.

40. 

SOLUTION:  

  The solutions are –0.5 and 2.5.

41. 

SOLUTION:  

  The solutions are about –8.2 and 0.2.

42. 

SOLUTION:  

  The solutions are about –5.1 and 0.1.

43. ASTRONOMY The height of an object t seconds after it is dropped is given by the equation ,

where h0 is the initial height and g is the acceleration due to gravity. The acceleration due to gravity near the surface

of Mars is 3.73 m/s2, while on Earth it is 9.8 m/s

2. Suppose an object is dropped from an initial height of 120 meters

above the surface of each planet. a. On which planet would the object reach the ground first? b. How long would it take the object to reach the ground on each planet? Round each answer to the nearest tenth. c. Do the times that it takes the object to reach the ground seem reasonable? Explain your reasoning.

SOLUTION:  a. The object on Earth will reach the ground first because it is falling at a faster rate.   b. Mars:

So, t ≈ 8.0 seconds. Earth:

So, t ≈ 4.9 seconds.   c. Sample answer: Yes; the acceleration due to gravity is much greater on Earth than on Mars, so the time to reach the ground should be much less.

44. Find all values of c that make x2 + cx + 100 a perfect square trinomial.

SOLUTION:  

So, c can be –20 or 20 to make the trinomial a perfect square.

45. Find all values of c that make x2 + cx + 225 a perfect square trinomial.

SOLUTION:  

So, c can be –30 or 30 to make the trinomial a perfect square.

46. PAINTING Before she begins painting a picture, Donna stretches her canvas over a wood frame. The frame has a length of 60 inches and a width of 4 inches. She has enough canvas to cover 480 square inches. Donna decides to increase the dimensions of the frame. If the increase in the length is 10 times the increase in the width, what will the dimensions of the frame be?

SOLUTION:  Let x = the increase in the width and let 10x = the increase in the length. So, x + 4 = the new width and 10x + 60 = the new length.  

  The dimensions cannot be negative, so x = 2. The width is 2 + 4 or 6 inches and the length is 10(2) + 60 or 80 inches.

47. MULTIPLE REPRESENTATIONS In this problem, you will investigate a property of quadratic equations. a. TABULAR  Copy the table shown and complete the second column.

b. ALGEBRAIC Set each trinomial equal to zero, and solve the equation by completing the square. Complete the last column of the table with the number of roots of each equation.

c. VERBAL Compare the number of roots of each equation to the result in the b2 − 4ac column. Is there a

relationship between these values? If so, describe it.

d. ANALYTICAL Predict how many solutions 2x2 − 9x + 15 = 0 will have.  Verify your prediction by solving the 

equation.

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3    

3x2 + 6x + 9    

x2 − 2x + 7    

x2 + 10x + 25    

x2 + 3x − 12    

SOLUTION:  a.

b.

The trinomial x2 − 8x + 16 has 1 root.

 

The trinomial 2x

2 − 11x + 3 has 2 roots.

 

The square root of a negative number has no real roots. So, the trinomial 3x2 + 6x + 9 has 0 roots.

 

The square root of a negative number has no real roots. So, the trinomial x2 − 2x + 7.

 

The trinomial x2 + 10x + 25 has 1 root.

 

The trinomial x2 + 3x − 12 has 2 roots.

 

c. If b2 − 4ac is negative, the equation has no real solutions. If b

2 − 4ac is zero, the equation has one solution. If b2

− 4ac is positive, the equation has 2 solutions. d.

The equation 2x2 − 9x + 15 = 0 has 0 real solutions because b

2 − 4ac is negative.

 

The equation cannot be solved because the square root of a negative number has no real roots.

Trinomial b2 − 4ac Number

of Roots

x2 − 8x + 16 (–8)

2 – 4(1)(16) = 0 1

2x2 − 11x + 3 (–11)

2 – 4(2)(3) = 97  

3x2 + 6x + 9 (6)

2 – 4(3)(9) = −72  

x2 − 2x + 7 (–2)

2 – 4(1)(7) = −24  

x2 + 10x + 25 (10)

2 – 4(1)(25) = 0  

x2 + 3x − 12 (3)

2 – 4(1)(–12) = 57  

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3 97 2

3x2 + 6x + 9 −72 0

x2 − 2x + 7 −24 0

x2 + 10x + 25 0 1

x2 + 3x − 12 57 2

48. CCSS PERSEVERANCE Given y = ax2 + bx + c with a ≠ 0, derive the equation for the axis of symmetry by 

completing the square and rewriting the equation in the form y = a(x − h)2 + k .

SOLUTION:  

Let    and   , then the equation becomes y = (x - h)2 + k .

For an equation in the form y = ax2 + bx + c,  the equation for the axis of symmetry is   .

So, for an equation in the form  y = (x - h)2 + k, the equation for the axis of symmetry becomes x = h.

49. REASONING Determine the number of solutions x2 + bx = c has if . Explain.

SOLUTION:  

None; Sample answer: If you add to each side of the equation and each side of the inequality, you get

. Since the left side of the last equation is a perfect square, it cannot

equal the negative number . So, there are no real solutions.

50. WHICH ONE DOESN’T BELONG? Identify the expression that does not belong with the other three. Explain your reasoning.

 

SOLUTION:  The first 3 trinomials are perfect squares.

The trinomial  is not a perfect square. So, it does not belong.

51. OPEN ENDED Write a quadratic equation for which the only solution is 4.

SOLUTION:  Find a quadratic that has two roots of 4.

Verify that the solution is 4.

So, the quadratic equation x2 − 8x + 16 = 0 has a solution of 4.

52. WRITING IN MATH Compare and contrast the following strategies for solving x2 − 5x − 7 = 0: completing the

square, graphing, and factoring.

SOLUTION:  Sample answer: Because the leading coefficient is 1, solving the equation by completing the square is simpler and yields exact answers for the solutions.   

 

To solve the equation by graphing, use a graphing calculator to graph the related function y = x2 - 5x - 7. Select the

zero option from the 2nd [CALC] menu to determine the roots. 

   

  The roots are given as decimal approximations. So, for this equation the solutions would have to be given as estimations.   There are no factors of –7 that have a sum of –5, so solving by factoring is not possible.

53. The length of a rectangle is 3 times its width. The area of the rectangle is 75 square feet. Find the length of the rectangle in feet. A 25 B 15 C 10 D 5

SOLUTION:  Let x = the width of the rectangle and let 3x = the length of the rectangle.

The length of the rectangle is 3(5) or 15 feet. Choice B is the correct answer.

54. PROBABILITY At a festival, winners of a game draw a token for a prize. There is one token for each prize. The prizes include 9 movie passes, 8 stuffed animals, 5 hats, 10 jump ropes, and 4 glow necklaces. What is the probabilitythat the first person to draw a token will win a movie pass?

F  

G

H

J  

SOLUTION:  There are 9 + 8 + 5 + 10 + 4 or 36 possible outcomes.

P(movie pass) =  or  . Choice J is the correct answer.

55. GRIDDED RESPONSE The population of a town can be modeled by P = 22,000 + 125t, where P represents the population and t represents the number of years from 2000. How many years after 2000 will the population be 26,000?

SOLUTION:  

In 32 years the population of the town will be 26,000.

56. Percy delivers pizzas for Pizza King. He is paid $6 an hour plus $2.50 for each pizza he delivers. Percy earned $280 last week. If he worked a total of 30 hours, how many pizzas did he deliver? A 250 pizzas B 184 pizzas C 40 pizzas D 34 pizzas

SOLUTION:  Let p = the number of pizzas Percy delivered.

Percy delivered 40 pizzas. Choice C is the correct answer.

Describe how the graph of each function is related to the graph of f (x) = x2.

57. g(x) = −12 + x2

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –12, the translation is

down. So, the graph is shifted down 12 units from the parent function. 

58. h(x) = (x + 2)2

SOLUTION:  

The graph of f (x) = (x – c)2 represents a translation left or right from the parent graph. Since c = –2, the translation

is to the left by 2 units. 

 

59. g(x) = 2x2 + 5

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 2 and c = 5. Since 2 > 0 and  > 1, the graph of y = 2x

2 + 5

is the graph of y = x2 vertically stretched and shifted up 5 units.

60. 

SOLUTION:  

The function can be written f (x) = a(x – b)2, where a = and b = 6. Since > 0 and  < 1, the graph of 

is the graph of y = x2 vertically compressed and shifted right 6 units.

61. g(x) = 6 + x2

SOLUTION:  

The function can be written f (x) = ax2 + c, where a =  and c = 6. Since  > 0 and   > 1, the graph of y = 6 +

x2 is the graph of y = x

2 vertically stretched and shifted up 6 units.

62. h(x) = − 1 −   x2

SOLUTION:  

The function can be written f (x) = –ax2 + c, where a =  and c = –1. Since  < 0 and   > 1, the graph of y

= –1 – x2 is the graph of y = x

2 vertically stretched, shifted down 1 unit and reflected across the x-axis.

63. RIDES A popular amusement park ride whisks riders to the top of a 250-foot tower and drops them. A function for

the height of a rider is h = −16t2 + 250, where h is the height and t is the time in seconds. The ride stops the descent

of the rider 40 feet above the ground. Write an equation that models the drop of the rider. How long does it take to fall from 250 feet to 40 feet?

SOLUTION:  

The ride stops the descent of the rider at 40 feet above ground, so h = 40. Then, the equation 40 = −16t2 + 250

models the drop of the rider.

Time cannot be negative. So, it takes about 3.6 seconds to complete the ride.

Simplify. Assume that no denominator is equal to zero.

64. 

SOLUTION:  

65. 

SOLUTION:  

66. 

SOLUTION:  

67. 

SOLUTION:  

68. 

SOLUTION:  

69. b3(m

−3)(b

−6)

SOLUTION:  

Solve each open sentence.

70. |y − 2| > 7

SOLUTION:  

The solution set is {y |y > 9 or y < −5}.

Case 1 y – 2 is positive.

and Case 2 y – 2 is negative.

 

71. |z + 5| < 3

SOLUTION:  

The solution set is {z |−8 < z < −2}.

Case 1 z +5 is positive.

and Case 2 z +5 is negative.

 

72. |2b + 7| ≤ −6

SOLUTION:  

cannot be negative. So,  cannot be less than or equal to –6. The solution set is empty.

73. |3 − 2y | ≥ 8

SOLUTION:  

The solution set is {y |y  ≥ 5.5 or y  ≤ −2.5}.

Case 1 3 – 2y is positive.

and Case 2 3 – 2y is negative.

 

74. |9 − 4m| < −1

SOLUTION:  

cannot be negative. So, cannot be less than or equal to –1. The solution set is empty.

75. |5c − 2| ≤ 13

SOLUTION:  

The solution set is {c|−2.2 ≤ c ≤ 3}.

Case 1 5c – 2 is positive.

and Case 2 5c – 2 is negative.

 

Evaluate  for each set of values. Round to the nearest tenth if necessary.

76. a = 2, b = −5, c = 2

SOLUTION:  

77. a = 1, b = 12, c = 11

SOLUTION:  

78. a = −9, b = 10, c = −1

SOLUTION:  

79. a = 1, b = 7, c = −3

SOLUTION:  

80. a = 2, b = −4, c = −6

SOLUTION:  

81. a = 3, b = 1, c = 2

SOLUTION:  

This value is not a real number.

eSolutions Manual - Powered by Cognero Page 29

9-4 Solving Quadratic Equations by Completing the Square

Find the value of c that makes each trinomial a perfect square.

1. x2 − 18x + c

SOLUTION:  In this trinomial, b = –18.

So, c must be 81 to make the trinomial a perfect square.

2. x2 + 22x + c

SOLUTION:  In this trinomial, b = 22.

So, c must be 121 to make the trinomial a perfect square.

3. x2 + 9x + c

SOLUTION:  In this trinomial, b = 9.

 

So, c must be  to make the trinomial a perfect square.

4. x2 − 7x + c

SOLUTION:  In this trinomial, b = –7.

 

So, c must be  to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

5. x2 + 4x = 6

SOLUTION:  

The solutions are about –5.2 and 1.2.

6. x2 − 8x = −9

SOLUTION:  

The solutions are about 1.4 and 6.6.

7. 4x2 + 9x − 1 = 0

SOLUTION:  

The solutions are about –2.4 and 0.1.

8. −2x2 + 10x + 22 = 4

SOLUTION:  

The solutions are about –1.4 and 6.4.

9. CCSS MODELING Collin is building a deck on the back of his family’s house. He has enough lumber for the deck to be 144 square feet. The length should be 10 feet more than its width. What should the dimensions of the deck be?

SOLUTION:  Let x = the width of the deck and x + 10 = the length of the deck.

The dimensions of the deck can not be negative. So, x = –5 + 13 or 8. The width of the deck is 8 feet and the length of the deck is 8 + 10 or 18 feet.

Find the value of c that makes each trinomial a perfect square.

10. x2 + 26x + c

SOLUTION:  In this trinomial, b = 26.

So, c must be 169 to make the trinomial a perfect square.

11. x2 − 24x + c

SOLUTION:  In this trinomial, b = –24.

So, c must be 144 to make the trinomial a perfect square.

12. x2 − 19x + c

SOLUTION:  In this trinomial, b = –19.

So, c must be  to make the trinomial a perfect square.

13. x2 + 17x + c

SOLUTION:  In this trinomial, b = 17.

So, c must be  to make the trinomial a perfect square.

14. x2 + 5x + c

SOLUTION:  In this trinomial, b = 5.

So, c must be  to make the trinomial a perfect square.

15. x2 − 13x + c

SOLUTION:  In this trinomial, b = –13.

So, c must be  to make the trinomial a perfect square.

16. x2 − 22x + c

SOLUTION:  In this trinomial, b = –22.

So, c must be 121 to make the trinomial a perfect square.

17. x2 − 15x + c

SOLUTION:  In this trinomial, b = –15.

So, c must be  to make the trinomial a perfect square.

18. x2 + 24x + c

SOLUTION:  In this trinomial, b = 24.

So, c must be 144 to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

19. x2 + 6x − 16 = 0

SOLUTION:  

  The solutions are –8 and 2.

20. x2 − 2x − 14 = 0

SOLUTION:  

  The solutions are about –2.9 and 4.9.

21. x2 − 8x − 1 = 8

SOLUTION:  

  The solutions are –1 and 9.

22. x2 + 3x + 21 = 22

SOLUTION:  

  The solutions are about –3.3 and 0.3.

23. x2 − 11x + 3 = 5

SOLUTION:  

  The solutions are about –0.2 and 11.2.

24. 5x2 − 10x = 23

SOLUTION:  

  The solutions are about –1.4 and 3.4.

25. 2x2 − 2x + 7 = 5

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

26. 3x2 + 12x + 81 = 15

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

27. 4x2 + 6x = 12

SOLUTION:  

  The solutions are about –2.6 and 1.1.

28. 4x2 + 5 = 10x

SOLUTION:  

  The solutions are about 0.7 and 1.8.

29. −2x2 + 10x = −14

SOLUTION:  

  The solutions are about –1.1 and 6.1.

30. −3x2 − 12 = 14x

SOLUTION:  

  The solutions are about –3.5 and –1.1.

31. STOCK The price p in dollars for a particular stock can be modeled by the quadratic equation p = 3.5t − 0.05t2, wh

the number of days after the stock is purchased. When is the stock worth $60?

SOLUTION:  Let p = 60.

  The stock is worth $60 on the 30th and 40th days after purchase.

GEOMETRY  Find the value of x for each figure. Round to the nearest tenth if necessary.

32. area = 45 in2

SOLUTION:  

  The height cannot be negative. So, x ≈ 6.3.

33. area = 110 ft2

SOLUTION:  

  The dimensions of the rectangle cannot be negative. So, x ≈ 5.3.

34. NUMBER THEORY The product of two consecutive even integers is 224. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers are 14 and 16 or –16 and –14.

35. CCSS PRECISION The product of two consecutive negative odd integers is 483. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers must be negative. So, they are –23 and –21.

36. GEOMETRY Find the area of the triangle.

SOLUTION:  

  The dimensions of the triangle cannot be negative. So, x = 18. The base of the triangle is 18 meters and the height is 18 + 6 or 24 meters.

The area of the triangle is 216 square meters.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

37. 0.2x2 − 0.2x − 0.4 = 0

SOLUTION:  

  The solutions are –1 and 2.

38. 0.5x2 = 2x − 0.3

SOLUTION:  

  The solutions are about 0.2 and 3.8.

39. 2x2 −

SOLUTION:  

  The solutions are about 0.2 and 0.9.

40. 

SOLUTION:  

  The solutions are –0.5 and 2.5.

41. 

SOLUTION:  

  The solutions are about –8.2 and 0.2.

42. 

SOLUTION:  

  The solutions are about –5.1 and 0.1.

43. ASTRONOMY The height of an object t seconds after it is dropped is given by the equation ,

where h0 is the initial height and g is the acceleration due to gravity. The acceleration due to gravity near the surface

of Mars is 3.73 m/s2, while on Earth it is 9.8 m/s

2. Suppose an object is dropped from an initial height of 120 meters

above the surface of each planet. a. On which planet would the object reach the ground first? b. How long would it take the object to reach the ground on each planet? Round each answer to the nearest tenth. c. Do the times that it takes the object to reach the ground seem reasonable? Explain your reasoning.

SOLUTION:  a. The object on Earth will reach the ground first because it is falling at a faster rate.   b. Mars:

So, t ≈ 8.0 seconds. Earth:

So, t ≈ 4.9 seconds.   c. Sample answer: Yes; the acceleration due to gravity is much greater on Earth than on Mars, so the time to reach the ground should be much less.

44. Find all values of c that make x2 + cx + 100 a perfect square trinomial.

SOLUTION:  

So, c can be –20 or 20 to make the trinomial a perfect square.

45. Find all values of c that make x2 + cx + 225 a perfect square trinomial.

SOLUTION:  

So, c can be –30 or 30 to make the trinomial a perfect square.

46. PAINTING Before she begins painting a picture, Donna stretches her canvas over a wood frame. The frame has a length of 60 inches and a width of 4 inches. She has enough canvas to cover 480 square inches. Donna decides to increase the dimensions of the frame. If the increase in the length is 10 times the increase in the width, what will the dimensions of the frame be?

SOLUTION:  Let x = the increase in the width and let 10x = the increase in the length. So, x + 4 = the new width and 10x + 60 = the new length.  

  The dimensions cannot be negative, so x = 2. The width is 2 + 4 or 6 inches and the length is 10(2) + 60 or 80 inches.

47. MULTIPLE REPRESENTATIONS In this problem, you will investigate a property of quadratic equations. a. TABULAR  Copy the table shown and complete the second column.

b. ALGEBRAIC Set each trinomial equal to zero, and solve the equation by completing the square. Complete the last column of the table with the number of roots of each equation.

c. VERBAL Compare the number of roots of each equation to the result in the b2 − 4ac column. Is there a

relationship between these values? If so, describe it.

d. ANALYTICAL Predict how many solutions 2x2 − 9x + 15 = 0 will have.  Verify your prediction by solving the 

equation.

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3    

3x2 + 6x + 9    

x2 − 2x + 7    

x2 + 10x + 25    

x2 + 3x − 12    

SOLUTION:  a.

b.

The trinomial x2 − 8x + 16 has 1 root.

 

The trinomial 2x

2 − 11x + 3 has 2 roots.

 

The square root of a negative number has no real roots. So, the trinomial 3x2 + 6x + 9 has 0 roots.

 

The square root of a negative number has no real roots. So, the trinomial x2 − 2x + 7.

 

The trinomial x2 + 10x + 25 has 1 root.

 

The trinomial x2 + 3x − 12 has 2 roots.

 

c. If b2 − 4ac is negative, the equation has no real solutions. If b

2 − 4ac is zero, the equation has one solution. If b2

− 4ac is positive, the equation has 2 solutions. d.

The equation 2x2 − 9x + 15 = 0 has 0 real solutions because b

2 − 4ac is negative.

 

The equation cannot be solved because the square root of a negative number has no real roots.

Trinomial b2 − 4ac Number

of Roots

x2 − 8x + 16 (–8)

2 – 4(1)(16) = 0 1

2x2 − 11x + 3 (–11)

2 – 4(2)(3) = 97  

3x2 + 6x + 9 (6)

2 – 4(3)(9) = −72  

x2 − 2x + 7 (–2)

2 – 4(1)(7) = −24  

x2 + 10x + 25 (10)

2 – 4(1)(25) = 0  

x2 + 3x − 12 (3)

2 – 4(1)(–12) = 57  

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3 97 2

3x2 + 6x + 9 −72 0

x2 − 2x + 7 −24 0

x2 + 10x + 25 0 1

x2 + 3x − 12 57 2

48. CCSS PERSEVERANCE Given y = ax2 + bx + c with a ≠ 0, derive the equation for the axis of symmetry by 

completing the square and rewriting the equation in the form y = a(x − h)2 + k .

SOLUTION:  

Let    and   , then the equation becomes y = (x - h)2 + k .

For an equation in the form y = ax2 + bx + c,  the equation for the axis of symmetry is   .

So, for an equation in the form  y = (x - h)2 + k, the equation for the axis of symmetry becomes x = h.

49. REASONING Determine the number of solutions x2 + bx = c has if . Explain.

SOLUTION:  

None; Sample answer: If you add to each side of the equation and each side of the inequality, you get

. Since the left side of the last equation is a perfect square, it cannot

equal the negative number . So, there are no real solutions.

50. WHICH ONE DOESN’T BELONG? Identify the expression that does not belong with the other three. Explain your reasoning.

 

SOLUTION:  The first 3 trinomials are perfect squares.

The trinomial  is not a perfect square. So, it does not belong.

51. OPEN ENDED Write a quadratic equation for which the only solution is 4.

SOLUTION:  Find a quadratic that has two roots of 4.

Verify that the solution is 4.

So, the quadratic equation x2 − 8x + 16 = 0 has a solution of 4.

52. WRITING IN MATH Compare and contrast the following strategies for solving x2 − 5x − 7 = 0: completing the

square, graphing, and factoring.

SOLUTION:  Sample answer: Because the leading coefficient is 1, solving the equation by completing the square is simpler and yields exact answers for the solutions.   

 

To solve the equation by graphing, use a graphing calculator to graph the related function y = x2 - 5x - 7. Select the

zero option from the 2nd [CALC] menu to determine the roots. 

   

  The roots are given as decimal approximations. So, for this equation the solutions would have to be given as estimations.   There are no factors of –7 that have a sum of –5, so solving by factoring is not possible.

53. The length of a rectangle is 3 times its width. The area of the rectangle is 75 square feet. Find the length of the rectangle in feet. A 25 B 15 C 10 D 5

SOLUTION:  Let x = the width of the rectangle and let 3x = the length of the rectangle.

The length of the rectangle is 3(5) or 15 feet. Choice B is the correct answer.

54. PROBABILITY At a festival, winners of a game draw a token for a prize. There is one token for each prize. The prizes include 9 movie passes, 8 stuffed animals, 5 hats, 10 jump ropes, and 4 glow necklaces. What is the probabilitythat the first person to draw a token will win a movie pass?

F  

G

H

J  

SOLUTION:  There are 9 + 8 + 5 + 10 + 4 or 36 possible outcomes.

P(movie pass) =  or  . Choice J is the correct answer.

55. GRIDDED RESPONSE The population of a town can be modeled by P = 22,000 + 125t, where P represents the population and t represents the number of years from 2000. How many years after 2000 will the population be 26,000?

SOLUTION:  

In 32 years the population of the town will be 26,000.

56. Percy delivers pizzas for Pizza King. He is paid $6 an hour plus $2.50 for each pizza he delivers. Percy earned $280 last week. If he worked a total of 30 hours, how many pizzas did he deliver? A 250 pizzas B 184 pizzas C 40 pizzas D 34 pizzas

SOLUTION:  Let p = the number of pizzas Percy delivered.

Percy delivered 40 pizzas. Choice C is the correct answer.

Describe how the graph of each function is related to the graph of f (x) = x2.

57. g(x) = −12 + x2

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –12, the translation is

down. So, the graph is shifted down 12 units from the parent function. 

58. h(x) = (x + 2)2

SOLUTION:  

The graph of f (x) = (x – c)2 represents a translation left or right from the parent graph. Since c = –2, the translation

is to the left by 2 units. 

 

59. g(x) = 2x2 + 5

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 2 and c = 5. Since 2 > 0 and  > 1, the graph of y = 2x

2 + 5

is the graph of y = x2 vertically stretched and shifted up 5 units.

60. 

SOLUTION:  

The function can be written f (x) = a(x – b)2, where a = and b = 6. Since > 0 and  < 1, the graph of 

is the graph of y = x2 vertically compressed and shifted right 6 units.

61. g(x) = 6 + x2

SOLUTION:  

The function can be written f (x) = ax2 + c, where a =  and c = 6. Since  > 0 and   > 1, the graph of y = 6 +

x2 is the graph of y = x

2 vertically stretched and shifted up 6 units.

62. h(x) = − 1 −   x2

SOLUTION:  

The function can be written f (x) = –ax2 + c, where a =  and c = –1. Since  < 0 and   > 1, the graph of y

= –1 – x2 is the graph of y = x

2 vertically stretched, shifted down 1 unit and reflected across the x-axis.

63. RIDES A popular amusement park ride whisks riders to the top of a 250-foot tower and drops them. A function for

the height of a rider is h = −16t2 + 250, where h is the height and t is the time in seconds. The ride stops the descent

of the rider 40 feet above the ground. Write an equation that models the drop of the rider. How long does it take to fall from 250 feet to 40 feet?

SOLUTION:  

The ride stops the descent of the rider at 40 feet above ground, so h = 40. Then, the equation 40 = −16t2 + 250

models the drop of the rider.

Time cannot be negative. So, it takes about 3.6 seconds to complete the ride.

Simplify. Assume that no denominator is equal to zero.

64. 

SOLUTION:  

65. 

SOLUTION:  

66. 

SOLUTION:  

67. 

SOLUTION:  

68. 

SOLUTION:  

69. b3(m

−3)(b

−6)

SOLUTION:  

Solve each open sentence.

70. |y − 2| > 7

SOLUTION:  

The solution set is {y |y > 9 or y < −5}.

Case 1 y – 2 is positive.

and Case 2 y – 2 is negative.

 

71. |z + 5| < 3

SOLUTION:  

The solution set is {z |−8 < z < −2}.

Case 1 z +5 is positive.

and Case 2 z +5 is negative.

 

72. |2b + 7| ≤ −6

SOLUTION:  

cannot be negative. So,  cannot be less than or equal to –6. The solution set is empty.

73. |3 − 2y | ≥ 8

SOLUTION:  

The solution set is {y |y  ≥ 5.5 or y  ≤ −2.5}.

Case 1 3 – 2y is positive.

and Case 2 3 – 2y is negative.

 

74. |9 − 4m| < −1

SOLUTION:  

cannot be negative. So, cannot be less than or equal to –1. The solution set is empty.

75. |5c − 2| ≤ 13

SOLUTION:  

The solution set is {c|−2.2 ≤ c ≤ 3}.

Case 1 5c – 2 is positive.

and Case 2 5c – 2 is negative.

 

Evaluate  for each set of values. Round to the nearest tenth if necessary.

76. a = 2, b = −5, c = 2

SOLUTION:  

77. a = 1, b = 12, c = 11

SOLUTION:  

78. a = −9, b = 10, c = −1

SOLUTION:  

79. a = 1, b = 7, c = −3

SOLUTION:  

80. a = 2, b = −4, c = −6

SOLUTION:  

81. a = 3, b = 1, c = 2

SOLUTION:  

This value is not a real number.

eSolutions Manual - Powered by Cognero Page 30

9-4 Solving Quadratic Equations by Completing the Square

Find the value of c that makes each trinomial a perfect square.

1. x2 − 18x + c

SOLUTION:  In this trinomial, b = –18.

So, c must be 81 to make the trinomial a perfect square.

2. x2 + 22x + c

SOLUTION:  In this trinomial, b = 22.

So, c must be 121 to make the trinomial a perfect square.

3. x2 + 9x + c

SOLUTION:  In this trinomial, b = 9.

 

So, c must be  to make the trinomial a perfect square.

4. x2 − 7x + c

SOLUTION:  In this trinomial, b = –7.

 

So, c must be  to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

5. x2 + 4x = 6

SOLUTION:  

The solutions are about –5.2 and 1.2.

6. x2 − 8x = −9

SOLUTION:  

The solutions are about 1.4 and 6.6.

7. 4x2 + 9x − 1 = 0

SOLUTION:  

The solutions are about –2.4 and 0.1.

8. −2x2 + 10x + 22 = 4

SOLUTION:  

The solutions are about –1.4 and 6.4.

9. CCSS MODELING Collin is building a deck on the back of his family’s house. He has enough lumber for the deck to be 144 square feet. The length should be 10 feet more than its width. What should the dimensions of the deck be?

SOLUTION:  Let x = the width of the deck and x + 10 = the length of the deck.

The dimensions of the deck can not be negative. So, x = –5 + 13 or 8. The width of the deck is 8 feet and the length of the deck is 8 + 10 or 18 feet.

Find the value of c that makes each trinomial a perfect square.

10. x2 + 26x + c

SOLUTION:  In this trinomial, b = 26.

So, c must be 169 to make the trinomial a perfect square.

11. x2 − 24x + c

SOLUTION:  In this trinomial, b = –24.

So, c must be 144 to make the trinomial a perfect square.

12. x2 − 19x + c

SOLUTION:  In this trinomial, b = –19.

So, c must be  to make the trinomial a perfect square.

13. x2 + 17x + c

SOLUTION:  In this trinomial, b = 17.

So, c must be  to make the trinomial a perfect square.

14. x2 + 5x + c

SOLUTION:  In this trinomial, b = 5.

So, c must be  to make the trinomial a perfect square.

15. x2 − 13x + c

SOLUTION:  In this trinomial, b = –13.

So, c must be  to make the trinomial a perfect square.

16. x2 − 22x + c

SOLUTION:  In this trinomial, b = –22.

So, c must be 121 to make the trinomial a perfect square.

17. x2 − 15x + c

SOLUTION:  In this trinomial, b = –15.

So, c must be  to make the trinomial a perfect square.

18. x2 + 24x + c

SOLUTION:  In this trinomial, b = 24.

So, c must be 144 to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

19. x2 + 6x − 16 = 0

SOLUTION:  

  The solutions are –8 and 2.

20. x2 − 2x − 14 = 0

SOLUTION:  

  The solutions are about –2.9 and 4.9.

21. x2 − 8x − 1 = 8

SOLUTION:  

  The solutions are –1 and 9.

22. x2 + 3x + 21 = 22

SOLUTION:  

  The solutions are about –3.3 and 0.3.

23. x2 − 11x + 3 = 5

SOLUTION:  

  The solutions are about –0.2 and 11.2.

24. 5x2 − 10x = 23

SOLUTION:  

  The solutions are about –1.4 and 3.4.

25. 2x2 − 2x + 7 = 5

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

26. 3x2 + 12x + 81 = 15

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

27. 4x2 + 6x = 12

SOLUTION:  

  The solutions are about –2.6 and 1.1.

28. 4x2 + 5 = 10x

SOLUTION:  

  The solutions are about 0.7 and 1.8.

29. −2x2 + 10x = −14

SOLUTION:  

  The solutions are about –1.1 and 6.1.

30. −3x2 − 12 = 14x

SOLUTION:  

  The solutions are about –3.5 and –1.1.

31. STOCK The price p in dollars for a particular stock can be modeled by the quadratic equation p = 3.5t − 0.05t2, wh

the number of days after the stock is purchased. When is the stock worth $60?

SOLUTION:  Let p = 60.

  The stock is worth $60 on the 30th and 40th days after purchase.

GEOMETRY  Find the value of x for each figure. Round to the nearest tenth if necessary.

32. area = 45 in2

SOLUTION:  

  The height cannot be negative. So, x ≈ 6.3.

33. area = 110 ft2

SOLUTION:  

  The dimensions of the rectangle cannot be negative. So, x ≈ 5.3.

34. NUMBER THEORY The product of two consecutive even integers is 224. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers are 14 and 16 or –16 and –14.

35. CCSS PRECISION The product of two consecutive negative odd integers is 483. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers must be negative. So, they are –23 and –21.

36. GEOMETRY Find the area of the triangle.

SOLUTION:  

  The dimensions of the triangle cannot be negative. So, x = 18. The base of the triangle is 18 meters and the height is 18 + 6 or 24 meters.

The area of the triangle is 216 square meters.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

37. 0.2x2 − 0.2x − 0.4 = 0

SOLUTION:  

  The solutions are –1 and 2.

38. 0.5x2 = 2x − 0.3

SOLUTION:  

  The solutions are about 0.2 and 3.8.

39. 2x2 −

SOLUTION:  

  The solutions are about 0.2 and 0.9.

40. 

SOLUTION:  

  The solutions are –0.5 and 2.5.

41. 

SOLUTION:  

  The solutions are about –8.2 and 0.2.

42. 

SOLUTION:  

  The solutions are about –5.1 and 0.1.

43. ASTRONOMY The height of an object t seconds after it is dropped is given by the equation ,

where h0 is the initial height and g is the acceleration due to gravity. The acceleration due to gravity near the surface

of Mars is 3.73 m/s2, while on Earth it is 9.8 m/s

2. Suppose an object is dropped from an initial height of 120 meters

above the surface of each planet. a. On which planet would the object reach the ground first? b. How long would it take the object to reach the ground on each planet? Round each answer to the nearest tenth. c. Do the times that it takes the object to reach the ground seem reasonable? Explain your reasoning.

SOLUTION:  a. The object on Earth will reach the ground first because it is falling at a faster rate.   b. Mars:

So, t ≈ 8.0 seconds. Earth:

So, t ≈ 4.9 seconds.   c. Sample answer: Yes; the acceleration due to gravity is much greater on Earth than on Mars, so the time to reach the ground should be much less.

44. Find all values of c that make x2 + cx + 100 a perfect square trinomial.

SOLUTION:  

So, c can be –20 or 20 to make the trinomial a perfect square.

45. Find all values of c that make x2 + cx + 225 a perfect square trinomial.

SOLUTION:  

So, c can be –30 or 30 to make the trinomial a perfect square.

46. PAINTING Before she begins painting a picture, Donna stretches her canvas over a wood frame. The frame has a length of 60 inches and a width of 4 inches. She has enough canvas to cover 480 square inches. Donna decides to increase the dimensions of the frame. If the increase in the length is 10 times the increase in the width, what will the dimensions of the frame be?

SOLUTION:  Let x = the increase in the width and let 10x = the increase in the length. So, x + 4 = the new width and 10x + 60 = the new length.  

  The dimensions cannot be negative, so x = 2. The width is 2 + 4 or 6 inches and the length is 10(2) + 60 or 80 inches.

47. MULTIPLE REPRESENTATIONS In this problem, you will investigate a property of quadratic equations. a. TABULAR  Copy the table shown and complete the second column.

b. ALGEBRAIC Set each trinomial equal to zero, and solve the equation by completing the square. Complete the last column of the table with the number of roots of each equation.

c. VERBAL Compare the number of roots of each equation to the result in the b2 − 4ac column. Is there a

relationship between these values? If so, describe it.

d. ANALYTICAL Predict how many solutions 2x2 − 9x + 15 = 0 will have.  Verify your prediction by solving the 

equation.

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3    

3x2 + 6x + 9    

x2 − 2x + 7    

x2 + 10x + 25    

x2 + 3x − 12    

SOLUTION:  a.

b.

The trinomial x2 − 8x + 16 has 1 root.

 

The trinomial 2x

2 − 11x + 3 has 2 roots.

 

The square root of a negative number has no real roots. So, the trinomial 3x2 + 6x + 9 has 0 roots.

 

The square root of a negative number has no real roots. So, the trinomial x2 − 2x + 7.

 

The trinomial x2 + 10x + 25 has 1 root.

 

The trinomial x2 + 3x − 12 has 2 roots.

 

c. If b2 − 4ac is negative, the equation has no real solutions. If b

2 − 4ac is zero, the equation has one solution. If b2

− 4ac is positive, the equation has 2 solutions. d.

The equation 2x2 − 9x + 15 = 0 has 0 real solutions because b

2 − 4ac is negative.

 

The equation cannot be solved because the square root of a negative number has no real roots.

Trinomial b2 − 4ac Number

of Roots

x2 − 8x + 16 (–8)

2 – 4(1)(16) = 0 1

2x2 − 11x + 3 (–11)

2 – 4(2)(3) = 97  

3x2 + 6x + 9 (6)

2 – 4(3)(9) = −72  

x2 − 2x + 7 (–2)

2 – 4(1)(7) = −24  

x2 + 10x + 25 (10)

2 – 4(1)(25) = 0  

x2 + 3x − 12 (3)

2 – 4(1)(–12) = 57  

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3 97 2

3x2 + 6x + 9 −72 0

x2 − 2x + 7 −24 0

x2 + 10x + 25 0 1

x2 + 3x − 12 57 2

48. CCSS PERSEVERANCE Given y = ax2 + bx + c with a ≠ 0, derive the equation for the axis of symmetry by 

completing the square and rewriting the equation in the form y = a(x − h)2 + k .

SOLUTION:  

Let    and   , then the equation becomes y = (x - h)2 + k .

For an equation in the form y = ax2 + bx + c,  the equation for the axis of symmetry is   .

So, for an equation in the form  y = (x - h)2 + k, the equation for the axis of symmetry becomes x = h.

49. REASONING Determine the number of solutions x2 + bx = c has if . Explain.

SOLUTION:  

None; Sample answer: If you add to each side of the equation and each side of the inequality, you get

. Since the left side of the last equation is a perfect square, it cannot

equal the negative number . So, there are no real solutions.

50. WHICH ONE DOESN’T BELONG? Identify the expression that does not belong with the other three. Explain your reasoning.

 

SOLUTION:  The first 3 trinomials are perfect squares.

The trinomial  is not a perfect square. So, it does not belong.

51. OPEN ENDED Write a quadratic equation for which the only solution is 4.

SOLUTION:  Find a quadratic that has two roots of 4.

Verify that the solution is 4.

So, the quadratic equation x2 − 8x + 16 = 0 has a solution of 4.

52. WRITING IN MATH Compare and contrast the following strategies for solving x2 − 5x − 7 = 0: completing the

square, graphing, and factoring.

SOLUTION:  Sample answer: Because the leading coefficient is 1, solving the equation by completing the square is simpler and yields exact answers for the solutions.   

 

To solve the equation by graphing, use a graphing calculator to graph the related function y = x2 - 5x - 7. Select the

zero option from the 2nd [CALC] menu to determine the roots. 

   

  The roots are given as decimal approximations. So, for this equation the solutions would have to be given as estimations.   There are no factors of –7 that have a sum of –5, so solving by factoring is not possible.

53. The length of a rectangle is 3 times its width. The area of the rectangle is 75 square feet. Find the length of the rectangle in feet. A 25 B 15 C 10 D 5

SOLUTION:  Let x = the width of the rectangle and let 3x = the length of the rectangle.

The length of the rectangle is 3(5) or 15 feet. Choice B is the correct answer.

54. PROBABILITY At a festival, winners of a game draw a token for a prize. There is one token for each prize. The prizes include 9 movie passes, 8 stuffed animals, 5 hats, 10 jump ropes, and 4 glow necklaces. What is the probabilitythat the first person to draw a token will win a movie pass?

F  

G

H

J  

SOLUTION:  There are 9 + 8 + 5 + 10 + 4 or 36 possible outcomes.

P(movie pass) =  or  . Choice J is the correct answer.

55. GRIDDED RESPONSE The population of a town can be modeled by P = 22,000 + 125t, where P represents the population and t represents the number of years from 2000. How many years after 2000 will the population be 26,000?

SOLUTION:  

In 32 years the population of the town will be 26,000.

56. Percy delivers pizzas for Pizza King. He is paid $6 an hour plus $2.50 for each pizza he delivers. Percy earned $280 last week. If he worked a total of 30 hours, how many pizzas did he deliver? A 250 pizzas B 184 pizzas C 40 pizzas D 34 pizzas

SOLUTION:  Let p = the number of pizzas Percy delivered.

Percy delivered 40 pizzas. Choice C is the correct answer.

Describe how the graph of each function is related to the graph of f (x) = x2.

57. g(x) = −12 + x2

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –12, the translation is

down. So, the graph is shifted down 12 units from the parent function. 

58. h(x) = (x + 2)2

SOLUTION:  

The graph of f (x) = (x – c)2 represents a translation left or right from the parent graph. Since c = –2, the translation

is to the left by 2 units. 

 

59. g(x) = 2x2 + 5

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 2 and c = 5. Since 2 > 0 and  > 1, the graph of y = 2x

2 + 5

is the graph of y = x2 vertically stretched and shifted up 5 units.

60. 

SOLUTION:  

The function can be written f (x) = a(x – b)2, where a = and b = 6. Since > 0 and  < 1, the graph of 

is the graph of y = x2 vertically compressed and shifted right 6 units.

61. g(x) = 6 + x2

SOLUTION:  

The function can be written f (x) = ax2 + c, where a =  and c = 6. Since  > 0 and   > 1, the graph of y = 6 +

x2 is the graph of y = x

2 vertically stretched and shifted up 6 units.

62. h(x) = − 1 −   x2

SOLUTION:  

The function can be written f (x) = –ax2 + c, where a =  and c = –1. Since  < 0 and   > 1, the graph of y

= –1 – x2 is the graph of y = x

2 vertically stretched, shifted down 1 unit and reflected across the x-axis.

63. RIDES A popular amusement park ride whisks riders to the top of a 250-foot tower and drops them. A function for

the height of a rider is h = −16t2 + 250, where h is the height and t is the time in seconds. The ride stops the descent

of the rider 40 feet above the ground. Write an equation that models the drop of the rider. How long does it take to fall from 250 feet to 40 feet?

SOLUTION:  

The ride stops the descent of the rider at 40 feet above ground, so h = 40. Then, the equation 40 = −16t2 + 250

models the drop of the rider.

Time cannot be negative. So, it takes about 3.6 seconds to complete the ride.

Simplify. Assume that no denominator is equal to zero.

64. 

SOLUTION:  

65. 

SOLUTION:  

66. 

SOLUTION:  

67. 

SOLUTION:  

68. 

SOLUTION:  

69. b3(m

−3)(b

−6)

SOLUTION:  

Solve each open sentence.

70. |y − 2| > 7

SOLUTION:  

The solution set is {y |y > 9 or y < −5}.

Case 1 y – 2 is positive.

and Case 2 y – 2 is negative.

 

71. |z + 5| < 3

SOLUTION:  

The solution set is {z |−8 < z < −2}.

Case 1 z +5 is positive.

and Case 2 z +5 is negative.

 

72. |2b + 7| ≤ −6

SOLUTION:  

cannot be negative. So,  cannot be less than or equal to –6. The solution set is empty.

73. |3 − 2y | ≥ 8

SOLUTION:  

The solution set is {y |y  ≥ 5.5 or y  ≤ −2.5}.

Case 1 3 – 2y is positive.

and Case 2 3 – 2y is negative.

 

74. |9 − 4m| < −1

SOLUTION:  

cannot be negative. So, cannot be less than or equal to –1. The solution set is empty.

75. |5c − 2| ≤ 13

SOLUTION:  

The solution set is {c|−2.2 ≤ c ≤ 3}.

Case 1 5c – 2 is positive.

and Case 2 5c – 2 is negative.

 

Evaluate  for each set of values. Round to the nearest tenth if necessary.

76. a = 2, b = −5, c = 2

SOLUTION:  

77. a = 1, b = 12, c = 11

SOLUTION:  

78. a = −9, b = 10, c = −1

SOLUTION:  

79. a = 1, b = 7, c = −3

SOLUTION:  

80. a = 2, b = −4, c = −6

SOLUTION:  

81. a = 3, b = 1, c = 2

SOLUTION:  

This value is not a real number.

eSolutions Manual - Powered by Cognero Page 31

9-4 Solving Quadratic Equations by Completing the Square

Find the value of c that makes each trinomial a perfect square.

1. x2 − 18x + c

SOLUTION:  In this trinomial, b = –18.

So, c must be 81 to make the trinomial a perfect square.

2. x2 + 22x + c

SOLUTION:  In this trinomial, b = 22.

So, c must be 121 to make the trinomial a perfect square.

3. x2 + 9x + c

SOLUTION:  In this trinomial, b = 9.

 

So, c must be  to make the trinomial a perfect square.

4. x2 − 7x + c

SOLUTION:  In this trinomial, b = –7.

 

So, c must be  to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

5. x2 + 4x = 6

SOLUTION:  

The solutions are about –5.2 and 1.2.

6. x2 − 8x = −9

SOLUTION:  

The solutions are about 1.4 and 6.6.

7. 4x2 + 9x − 1 = 0

SOLUTION:  

The solutions are about –2.4 and 0.1.

8. −2x2 + 10x + 22 = 4

SOLUTION:  

The solutions are about –1.4 and 6.4.

9. CCSS MODELING Collin is building a deck on the back of his family’s house. He has enough lumber for the deck to be 144 square feet. The length should be 10 feet more than its width. What should the dimensions of the deck be?

SOLUTION:  Let x = the width of the deck and x + 10 = the length of the deck.

The dimensions of the deck can not be negative. So, x = –5 + 13 or 8. The width of the deck is 8 feet and the length of the deck is 8 + 10 or 18 feet.

Find the value of c that makes each trinomial a perfect square.

10. x2 + 26x + c

SOLUTION:  In this trinomial, b = 26.

So, c must be 169 to make the trinomial a perfect square.

11. x2 − 24x + c

SOLUTION:  In this trinomial, b = –24.

So, c must be 144 to make the trinomial a perfect square.

12. x2 − 19x + c

SOLUTION:  In this trinomial, b = –19.

So, c must be  to make the trinomial a perfect square.

13. x2 + 17x + c

SOLUTION:  In this trinomial, b = 17.

So, c must be  to make the trinomial a perfect square.

14. x2 + 5x + c

SOLUTION:  In this trinomial, b = 5.

So, c must be  to make the trinomial a perfect square.

15. x2 − 13x + c

SOLUTION:  In this trinomial, b = –13.

So, c must be  to make the trinomial a perfect square.

16. x2 − 22x + c

SOLUTION:  In this trinomial, b = –22.

So, c must be 121 to make the trinomial a perfect square.

17. x2 − 15x + c

SOLUTION:  In this trinomial, b = –15.

So, c must be  to make the trinomial a perfect square.

18. x2 + 24x + c

SOLUTION:  In this trinomial, b = 24.

So, c must be 144 to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

19. x2 + 6x − 16 = 0

SOLUTION:  

  The solutions are –8 and 2.

20. x2 − 2x − 14 = 0

SOLUTION:  

  The solutions are about –2.9 and 4.9.

21. x2 − 8x − 1 = 8

SOLUTION:  

  The solutions are –1 and 9.

22. x2 + 3x + 21 = 22

SOLUTION:  

  The solutions are about –3.3 and 0.3.

23. x2 − 11x + 3 = 5

SOLUTION:  

  The solutions are about –0.2 and 11.2.

24. 5x2 − 10x = 23

SOLUTION:  

  The solutions are about –1.4 and 3.4.

25. 2x2 − 2x + 7 = 5

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

26. 3x2 + 12x + 81 = 15

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

27. 4x2 + 6x = 12

SOLUTION:  

  The solutions are about –2.6 and 1.1.

28. 4x2 + 5 = 10x

SOLUTION:  

  The solutions are about 0.7 and 1.8.

29. −2x2 + 10x = −14

SOLUTION:  

  The solutions are about –1.1 and 6.1.

30. −3x2 − 12 = 14x

SOLUTION:  

  The solutions are about –3.5 and –1.1.

31. STOCK The price p in dollars for a particular stock can be modeled by the quadratic equation p = 3.5t − 0.05t2, wh

the number of days after the stock is purchased. When is the stock worth $60?

SOLUTION:  Let p = 60.

  The stock is worth $60 on the 30th and 40th days after purchase.

GEOMETRY  Find the value of x for each figure. Round to the nearest tenth if necessary.

32. area = 45 in2

SOLUTION:  

  The height cannot be negative. So, x ≈ 6.3.

33. area = 110 ft2

SOLUTION:  

  The dimensions of the rectangle cannot be negative. So, x ≈ 5.3.

34. NUMBER THEORY The product of two consecutive even integers is 224. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers are 14 and 16 or –16 and –14.

35. CCSS PRECISION The product of two consecutive negative odd integers is 483. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers must be negative. So, they are –23 and –21.

36. GEOMETRY Find the area of the triangle.

SOLUTION:  

  The dimensions of the triangle cannot be negative. So, x = 18. The base of the triangle is 18 meters and the height is 18 + 6 or 24 meters.

The area of the triangle is 216 square meters.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

37. 0.2x2 − 0.2x − 0.4 = 0

SOLUTION:  

  The solutions are –1 and 2.

38. 0.5x2 = 2x − 0.3

SOLUTION:  

  The solutions are about 0.2 and 3.8.

39. 2x2 −

SOLUTION:  

  The solutions are about 0.2 and 0.9.

40. 

SOLUTION:  

  The solutions are –0.5 and 2.5.

41. 

SOLUTION:  

  The solutions are about –8.2 and 0.2.

42. 

SOLUTION:  

  The solutions are about –5.1 and 0.1.

43. ASTRONOMY The height of an object t seconds after it is dropped is given by the equation ,

where h0 is the initial height and g is the acceleration due to gravity. The acceleration due to gravity near the surface

of Mars is 3.73 m/s2, while on Earth it is 9.8 m/s

2. Suppose an object is dropped from an initial height of 120 meters

above the surface of each planet. a. On which planet would the object reach the ground first? b. How long would it take the object to reach the ground on each planet? Round each answer to the nearest tenth. c. Do the times that it takes the object to reach the ground seem reasonable? Explain your reasoning.

SOLUTION:  a. The object on Earth will reach the ground first because it is falling at a faster rate.   b. Mars:

So, t ≈ 8.0 seconds. Earth:

So, t ≈ 4.9 seconds.   c. Sample answer: Yes; the acceleration due to gravity is much greater on Earth than on Mars, so the time to reach the ground should be much less.

44. Find all values of c that make x2 + cx + 100 a perfect square trinomial.

SOLUTION:  

So, c can be –20 or 20 to make the trinomial a perfect square.

45. Find all values of c that make x2 + cx + 225 a perfect square trinomial.

SOLUTION:  

So, c can be –30 or 30 to make the trinomial a perfect square.

46. PAINTING Before she begins painting a picture, Donna stretches her canvas over a wood frame. The frame has a length of 60 inches and a width of 4 inches. She has enough canvas to cover 480 square inches. Donna decides to increase the dimensions of the frame. If the increase in the length is 10 times the increase in the width, what will the dimensions of the frame be?

SOLUTION:  Let x = the increase in the width and let 10x = the increase in the length. So, x + 4 = the new width and 10x + 60 = the new length.  

  The dimensions cannot be negative, so x = 2. The width is 2 + 4 or 6 inches and the length is 10(2) + 60 or 80 inches.

47. MULTIPLE REPRESENTATIONS In this problem, you will investigate a property of quadratic equations. a. TABULAR  Copy the table shown and complete the second column.

b. ALGEBRAIC Set each trinomial equal to zero, and solve the equation by completing the square. Complete the last column of the table with the number of roots of each equation.

c. VERBAL Compare the number of roots of each equation to the result in the b2 − 4ac column. Is there a

relationship between these values? If so, describe it.

d. ANALYTICAL Predict how many solutions 2x2 − 9x + 15 = 0 will have.  Verify your prediction by solving the 

equation.

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3    

3x2 + 6x + 9    

x2 − 2x + 7    

x2 + 10x + 25    

x2 + 3x − 12    

SOLUTION:  a.

b.

The trinomial x2 − 8x + 16 has 1 root.

 

The trinomial 2x

2 − 11x + 3 has 2 roots.

 

The square root of a negative number has no real roots. So, the trinomial 3x2 + 6x + 9 has 0 roots.

 

The square root of a negative number has no real roots. So, the trinomial x2 − 2x + 7.

 

The trinomial x2 + 10x + 25 has 1 root.

 

The trinomial x2 + 3x − 12 has 2 roots.

 

c. If b2 − 4ac is negative, the equation has no real solutions. If b

2 − 4ac is zero, the equation has one solution. If b2

− 4ac is positive, the equation has 2 solutions. d.

The equation 2x2 − 9x + 15 = 0 has 0 real solutions because b

2 − 4ac is negative.

 

The equation cannot be solved because the square root of a negative number has no real roots.

Trinomial b2 − 4ac Number

of Roots

x2 − 8x + 16 (–8)

2 – 4(1)(16) = 0 1

2x2 − 11x + 3 (–11)

2 – 4(2)(3) = 97  

3x2 + 6x + 9 (6)

2 – 4(3)(9) = −72  

x2 − 2x + 7 (–2)

2 – 4(1)(7) = −24  

x2 + 10x + 25 (10)

2 – 4(1)(25) = 0  

x2 + 3x − 12 (3)

2 – 4(1)(–12) = 57  

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3 97 2

3x2 + 6x + 9 −72 0

x2 − 2x + 7 −24 0

x2 + 10x + 25 0 1

x2 + 3x − 12 57 2

48. CCSS PERSEVERANCE Given y = ax2 + bx + c with a ≠ 0, derive the equation for the axis of symmetry by 

completing the square and rewriting the equation in the form y = a(x − h)2 + k .

SOLUTION:  

Let    and   , then the equation becomes y = (x - h)2 + k .

For an equation in the form y = ax2 + bx + c,  the equation for the axis of symmetry is   .

So, for an equation in the form  y = (x - h)2 + k, the equation for the axis of symmetry becomes x = h.

49. REASONING Determine the number of solutions x2 + bx = c has if . Explain.

SOLUTION:  

None; Sample answer: If you add to each side of the equation and each side of the inequality, you get

. Since the left side of the last equation is a perfect square, it cannot

equal the negative number . So, there are no real solutions.

50. WHICH ONE DOESN’T BELONG? Identify the expression that does not belong with the other three. Explain your reasoning.

 

SOLUTION:  The first 3 trinomials are perfect squares.

The trinomial  is not a perfect square. So, it does not belong.

51. OPEN ENDED Write a quadratic equation for which the only solution is 4.

SOLUTION:  Find a quadratic that has two roots of 4.

Verify that the solution is 4.

So, the quadratic equation x2 − 8x + 16 = 0 has a solution of 4.

52. WRITING IN MATH Compare and contrast the following strategies for solving x2 − 5x − 7 = 0: completing the

square, graphing, and factoring.

SOLUTION:  Sample answer: Because the leading coefficient is 1, solving the equation by completing the square is simpler and yields exact answers for the solutions.   

 

To solve the equation by graphing, use a graphing calculator to graph the related function y = x2 - 5x - 7. Select the

zero option from the 2nd [CALC] menu to determine the roots. 

   

  The roots are given as decimal approximations. So, for this equation the solutions would have to be given as estimations.   There are no factors of –7 that have a sum of –5, so solving by factoring is not possible.

53. The length of a rectangle is 3 times its width. The area of the rectangle is 75 square feet. Find the length of the rectangle in feet. A 25 B 15 C 10 D 5

SOLUTION:  Let x = the width of the rectangle and let 3x = the length of the rectangle.

The length of the rectangle is 3(5) or 15 feet. Choice B is the correct answer.

54. PROBABILITY At a festival, winners of a game draw a token for a prize. There is one token for each prize. The prizes include 9 movie passes, 8 stuffed animals, 5 hats, 10 jump ropes, and 4 glow necklaces. What is the probabilitythat the first person to draw a token will win a movie pass?

F  

G

H

J  

SOLUTION:  There are 9 + 8 + 5 + 10 + 4 or 36 possible outcomes.

P(movie pass) =  or  . Choice J is the correct answer.

55. GRIDDED RESPONSE The population of a town can be modeled by P = 22,000 + 125t, where P represents the population and t represents the number of years from 2000. How many years after 2000 will the population be 26,000?

SOLUTION:  

In 32 years the population of the town will be 26,000.

56. Percy delivers pizzas for Pizza King. He is paid $6 an hour plus $2.50 for each pizza he delivers. Percy earned $280 last week. If he worked a total of 30 hours, how many pizzas did he deliver? A 250 pizzas B 184 pizzas C 40 pizzas D 34 pizzas

SOLUTION:  Let p = the number of pizzas Percy delivered.

Percy delivered 40 pizzas. Choice C is the correct answer.

Describe how the graph of each function is related to the graph of f (x) = x2.

57. g(x) = −12 + x2

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –12, the translation is

down. So, the graph is shifted down 12 units from the parent function. 

58. h(x) = (x + 2)2

SOLUTION:  

The graph of f (x) = (x – c)2 represents a translation left or right from the parent graph. Since c = –2, the translation

is to the left by 2 units. 

 

59. g(x) = 2x2 + 5

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 2 and c = 5. Since 2 > 0 and  > 1, the graph of y = 2x

2 + 5

is the graph of y = x2 vertically stretched and shifted up 5 units.

60. 

SOLUTION:  

The function can be written f (x) = a(x – b)2, where a = and b = 6. Since > 0 and  < 1, the graph of 

is the graph of y = x2 vertically compressed and shifted right 6 units.

61. g(x) = 6 + x2

SOLUTION:  

The function can be written f (x) = ax2 + c, where a =  and c = 6. Since  > 0 and   > 1, the graph of y = 6 +

x2 is the graph of y = x

2 vertically stretched and shifted up 6 units.

62. h(x) = − 1 −   x2

SOLUTION:  

The function can be written f (x) = –ax2 + c, where a =  and c = –1. Since  < 0 and   > 1, the graph of y

= –1 – x2 is the graph of y = x

2 vertically stretched, shifted down 1 unit and reflected across the x-axis.

63. RIDES A popular amusement park ride whisks riders to the top of a 250-foot tower and drops them. A function for

the height of a rider is h = −16t2 + 250, where h is the height and t is the time in seconds. The ride stops the descent

of the rider 40 feet above the ground. Write an equation that models the drop of the rider. How long does it take to fall from 250 feet to 40 feet?

SOLUTION:  

The ride stops the descent of the rider at 40 feet above ground, so h = 40. Then, the equation 40 = −16t2 + 250

models the drop of the rider.

Time cannot be negative. So, it takes about 3.6 seconds to complete the ride.

Simplify. Assume that no denominator is equal to zero.

64. 

SOLUTION:  

65. 

SOLUTION:  

66. 

SOLUTION:  

67. 

SOLUTION:  

68. 

SOLUTION:  

69. b3(m

−3)(b

−6)

SOLUTION:  

Solve each open sentence.

70. |y − 2| > 7

SOLUTION:  

The solution set is {y |y > 9 or y < −5}.

Case 1 y – 2 is positive.

and Case 2 y – 2 is negative.

 

71. |z + 5| < 3

SOLUTION:  

The solution set is {z |−8 < z < −2}.

Case 1 z +5 is positive.

and Case 2 z +5 is negative.

 

72. |2b + 7| ≤ −6

SOLUTION:  

cannot be negative. So,  cannot be less than or equal to –6. The solution set is empty.

73. |3 − 2y | ≥ 8

SOLUTION:  

The solution set is {y |y  ≥ 5.5 or y  ≤ −2.5}.

Case 1 3 – 2y is positive.

and Case 2 3 – 2y is negative.

 

74. |9 − 4m| < −1

SOLUTION:  

cannot be negative. So, cannot be less than or equal to –1. The solution set is empty.

75. |5c − 2| ≤ 13

SOLUTION:  

The solution set is {c|−2.2 ≤ c ≤ 3}.

Case 1 5c – 2 is positive.

and Case 2 5c – 2 is negative.

 

Evaluate  for each set of values. Round to the nearest tenth if necessary.

76. a = 2, b = −5, c = 2

SOLUTION:  

77. a = 1, b = 12, c = 11

SOLUTION:  

78. a = −9, b = 10, c = −1

SOLUTION:  

79. a = 1, b = 7, c = −3

SOLUTION:  

80. a = 2, b = −4, c = −6

SOLUTION:  

81. a = 3, b = 1, c = 2

SOLUTION:  

This value is not a real number.

eSolutions Manual - Powered by Cognero Page 32

9-4 Solving Quadratic Equations by Completing the Square

Find the value of c that makes each trinomial a perfect square.

1. x2 − 18x + c

SOLUTION:  In this trinomial, b = –18.

So, c must be 81 to make the trinomial a perfect square.

2. x2 + 22x + c

SOLUTION:  In this trinomial, b = 22.

So, c must be 121 to make the trinomial a perfect square.

3. x2 + 9x + c

SOLUTION:  In this trinomial, b = 9.

 

So, c must be  to make the trinomial a perfect square.

4. x2 − 7x + c

SOLUTION:  In this trinomial, b = –7.

 

So, c must be  to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

5. x2 + 4x = 6

SOLUTION:  

The solutions are about –5.2 and 1.2.

6. x2 − 8x = −9

SOLUTION:  

The solutions are about 1.4 and 6.6.

7. 4x2 + 9x − 1 = 0

SOLUTION:  

The solutions are about –2.4 and 0.1.

8. −2x2 + 10x + 22 = 4

SOLUTION:  

The solutions are about –1.4 and 6.4.

9. CCSS MODELING Collin is building a deck on the back of his family’s house. He has enough lumber for the deck to be 144 square feet. The length should be 10 feet more than its width. What should the dimensions of the deck be?

SOLUTION:  Let x = the width of the deck and x + 10 = the length of the deck.

The dimensions of the deck can not be negative. So, x = –5 + 13 or 8. The width of the deck is 8 feet and the length of the deck is 8 + 10 or 18 feet.

Find the value of c that makes each trinomial a perfect square.

10. x2 + 26x + c

SOLUTION:  In this trinomial, b = 26.

So, c must be 169 to make the trinomial a perfect square.

11. x2 − 24x + c

SOLUTION:  In this trinomial, b = –24.

So, c must be 144 to make the trinomial a perfect square.

12. x2 − 19x + c

SOLUTION:  In this trinomial, b = –19.

So, c must be  to make the trinomial a perfect square.

13. x2 + 17x + c

SOLUTION:  In this trinomial, b = 17.

So, c must be  to make the trinomial a perfect square.

14. x2 + 5x + c

SOLUTION:  In this trinomial, b = 5.

So, c must be  to make the trinomial a perfect square.

15. x2 − 13x + c

SOLUTION:  In this trinomial, b = –13.

So, c must be  to make the trinomial a perfect square.

16. x2 − 22x + c

SOLUTION:  In this trinomial, b = –22.

So, c must be 121 to make the trinomial a perfect square.

17. x2 − 15x + c

SOLUTION:  In this trinomial, b = –15.

So, c must be  to make the trinomial a perfect square.

18. x2 + 24x + c

SOLUTION:  In this trinomial, b = 24.

So, c must be 144 to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

19. x2 + 6x − 16 = 0

SOLUTION:  

  The solutions are –8 and 2.

20. x2 − 2x − 14 = 0

SOLUTION:  

  The solutions are about –2.9 and 4.9.

21. x2 − 8x − 1 = 8

SOLUTION:  

  The solutions are –1 and 9.

22. x2 + 3x + 21 = 22

SOLUTION:  

  The solutions are about –3.3 and 0.3.

23. x2 − 11x + 3 = 5

SOLUTION:  

  The solutions are about –0.2 and 11.2.

24. 5x2 − 10x = 23

SOLUTION:  

  The solutions are about –1.4 and 3.4.

25. 2x2 − 2x + 7 = 5

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

26. 3x2 + 12x + 81 = 15

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

27. 4x2 + 6x = 12

SOLUTION:  

  The solutions are about –2.6 and 1.1.

28. 4x2 + 5 = 10x

SOLUTION:  

  The solutions are about 0.7 and 1.8.

29. −2x2 + 10x = −14

SOLUTION:  

  The solutions are about –1.1 and 6.1.

30. −3x2 − 12 = 14x

SOLUTION:  

  The solutions are about –3.5 and –1.1.

31. STOCK The price p in dollars for a particular stock can be modeled by the quadratic equation p = 3.5t − 0.05t2, wh

the number of days after the stock is purchased. When is the stock worth $60?

SOLUTION:  Let p = 60.

  The stock is worth $60 on the 30th and 40th days after purchase.

GEOMETRY  Find the value of x for each figure. Round to the nearest tenth if necessary.

32. area = 45 in2

SOLUTION:  

  The height cannot be negative. So, x ≈ 6.3.

33. area = 110 ft2

SOLUTION:  

  The dimensions of the rectangle cannot be negative. So, x ≈ 5.3.

34. NUMBER THEORY The product of two consecutive even integers is 224. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers are 14 and 16 or –16 and –14.

35. CCSS PRECISION The product of two consecutive negative odd integers is 483. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers must be negative. So, they are –23 and –21.

36. GEOMETRY Find the area of the triangle.

SOLUTION:  

  The dimensions of the triangle cannot be negative. So, x = 18. The base of the triangle is 18 meters and the height is 18 + 6 or 24 meters.

The area of the triangle is 216 square meters.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

37. 0.2x2 − 0.2x − 0.4 = 0

SOLUTION:  

  The solutions are –1 and 2.

38. 0.5x2 = 2x − 0.3

SOLUTION:  

  The solutions are about 0.2 and 3.8.

39. 2x2 −

SOLUTION:  

  The solutions are about 0.2 and 0.9.

40. 

SOLUTION:  

  The solutions are –0.5 and 2.5.

41. 

SOLUTION:  

  The solutions are about –8.2 and 0.2.

42. 

SOLUTION:  

  The solutions are about –5.1 and 0.1.

43. ASTRONOMY The height of an object t seconds after it is dropped is given by the equation ,

where h0 is the initial height and g is the acceleration due to gravity. The acceleration due to gravity near the surface

of Mars is 3.73 m/s2, while on Earth it is 9.8 m/s

2. Suppose an object is dropped from an initial height of 120 meters

above the surface of each planet. a. On which planet would the object reach the ground first? b. How long would it take the object to reach the ground on each planet? Round each answer to the nearest tenth. c. Do the times that it takes the object to reach the ground seem reasonable? Explain your reasoning.

SOLUTION:  a. The object on Earth will reach the ground first because it is falling at a faster rate.   b. Mars:

So, t ≈ 8.0 seconds. Earth:

So, t ≈ 4.9 seconds.   c. Sample answer: Yes; the acceleration due to gravity is much greater on Earth than on Mars, so the time to reach the ground should be much less.

44. Find all values of c that make x2 + cx + 100 a perfect square trinomial.

SOLUTION:  

So, c can be –20 or 20 to make the trinomial a perfect square.

45. Find all values of c that make x2 + cx + 225 a perfect square trinomial.

SOLUTION:  

So, c can be –30 or 30 to make the trinomial a perfect square.

46. PAINTING Before she begins painting a picture, Donna stretches her canvas over a wood frame. The frame has a length of 60 inches and a width of 4 inches. She has enough canvas to cover 480 square inches. Donna decides to increase the dimensions of the frame. If the increase in the length is 10 times the increase in the width, what will the dimensions of the frame be?

SOLUTION:  Let x = the increase in the width and let 10x = the increase in the length. So, x + 4 = the new width and 10x + 60 = the new length.  

  The dimensions cannot be negative, so x = 2. The width is 2 + 4 or 6 inches and the length is 10(2) + 60 or 80 inches.

47. MULTIPLE REPRESENTATIONS In this problem, you will investigate a property of quadratic equations. a. TABULAR  Copy the table shown and complete the second column.

b. ALGEBRAIC Set each trinomial equal to zero, and solve the equation by completing the square. Complete the last column of the table with the number of roots of each equation.

c. VERBAL Compare the number of roots of each equation to the result in the b2 − 4ac column. Is there a

relationship between these values? If so, describe it.

d. ANALYTICAL Predict how many solutions 2x2 − 9x + 15 = 0 will have.  Verify your prediction by solving the 

equation.

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3    

3x2 + 6x + 9    

x2 − 2x + 7    

x2 + 10x + 25    

x2 + 3x − 12    

SOLUTION:  a.

b.

The trinomial x2 − 8x + 16 has 1 root.

 

The trinomial 2x

2 − 11x + 3 has 2 roots.

 

The square root of a negative number has no real roots. So, the trinomial 3x2 + 6x + 9 has 0 roots.

 

The square root of a negative number has no real roots. So, the trinomial x2 − 2x + 7.

 

The trinomial x2 + 10x + 25 has 1 root.

 

The trinomial x2 + 3x − 12 has 2 roots.

 

c. If b2 − 4ac is negative, the equation has no real solutions. If b

2 − 4ac is zero, the equation has one solution. If b2

− 4ac is positive, the equation has 2 solutions. d.

The equation 2x2 − 9x + 15 = 0 has 0 real solutions because b

2 − 4ac is negative.

 

The equation cannot be solved because the square root of a negative number has no real roots.

Trinomial b2 − 4ac Number

of Roots

x2 − 8x + 16 (–8)

2 – 4(1)(16) = 0 1

2x2 − 11x + 3 (–11)

2 – 4(2)(3) = 97  

3x2 + 6x + 9 (6)

2 – 4(3)(9) = −72  

x2 − 2x + 7 (–2)

2 – 4(1)(7) = −24  

x2 + 10x + 25 (10)

2 – 4(1)(25) = 0  

x2 + 3x − 12 (3)

2 – 4(1)(–12) = 57  

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3 97 2

3x2 + 6x + 9 −72 0

x2 − 2x + 7 −24 0

x2 + 10x + 25 0 1

x2 + 3x − 12 57 2

48. CCSS PERSEVERANCE Given y = ax2 + bx + c with a ≠ 0, derive the equation for the axis of symmetry by 

completing the square and rewriting the equation in the form y = a(x − h)2 + k .

SOLUTION:  

Let    and   , then the equation becomes y = (x - h)2 + k .

For an equation in the form y = ax2 + bx + c,  the equation for the axis of symmetry is   .

So, for an equation in the form  y = (x - h)2 + k, the equation for the axis of symmetry becomes x = h.

49. REASONING Determine the number of solutions x2 + bx = c has if . Explain.

SOLUTION:  

None; Sample answer: If you add to each side of the equation and each side of the inequality, you get

. Since the left side of the last equation is a perfect square, it cannot

equal the negative number . So, there are no real solutions.

50. WHICH ONE DOESN’T BELONG? Identify the expression that does not belong with the other three. Explain your reasoning.

 

SOLUTION:  The first 3 trinomials are perfect squares.

The trinomial  is not a perfect square. So, it does not belong.

51. OPEN ENDED Write a quadratic equation for which the only solution is 4.

SOLUTION:  Find a quadratic that has two roots of 4.

Verify that the solution is 4.

So, the quadratic equation x2 − 8x + 16 = 0 has a solution of 4.

52. WRITING IN MATH Compare and contrast the following strategies for solving x2 − 5x − 7 = 0: completing the

square, graphing, and factoring.

SOLUTION:  Sample answer: Because the leading coefficient is 1, solving the equation by completing the square is simpler and yields exact answers for the solutions.   

 

To solve the equation by graphing, use a graphing calculator to graph the related function y = x2 - 5x - 7. Select the

zero option from the 2nd [CALC] menu to determine the roots. 

   

  The roots are given as decimal approximations. So, for this equation the solutions would have to be given as estimations.   There are no factors of –7 that have a sum of –5, so solving by factoring is not possible.

53. The length of a rectangle is 3 times its width. The area of the rectangle is 75 square feet. Find the length of the rectangle in feet. A 25 B 15 C 10 D 5

SOLUTION:  Let x = the width of the rectangle and let 3x = the length of the rectangle.

The length of the rectangle is 3(5) or 15 feet. Choice B is the correct answer.

54. PROBABILITY At a festival, winners of a game draw a token for a prize. There is one token for each prize. The prizes include 9 movie passes, 8 stuffed animals, 5 hats, 10 jump ropes, and 4 glow necklaces. What is the probabilitythat the first person to draw a token will win a movie pass?

F  

G

H

J  

SOLUTION:  There are 9 + 8 + 5 + 10 + 4 or 36 possible outcomes.

P(movie pass) =  or  . Choice J is the correct answer.

55. GRIDDED RESPONSE The population of a town can be modeled by P = 22,000 + 125t, where P represents the population and t represents the number of years from 2000. How many years after 2000 will the population be 26,000?

SOLUTION:  

In 32 years the population of the town will be 26,000.

56. Percy delivers pizzas for Pizza King. He is paid $6 an hour plus $2.50 for each pizza he delivers. Percy earned $280 last week. If he worked a total of 30 hours, how many pizzas did he deliver? A 250 pizzas B 184 pizzas C 40 pizzas D 34 pizzas

SOLUTION:  Let p = the number of pizzas Percy delivered.

Percy delivered 40 pizzas. Choice C is the correct answer.

Describe how the graph of each function is related to the graph of f (x) = x2.

57. g(x) = −12 + x2

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –12, the translation is

down. So, the graph is shifted down 12 units from the parent function. 

58. h(x) = (x + 2)2

SOLUTION:  

The graph of f (x) = (x – c)2 represents a translation left or right from the parent graph. Since c = –2, the translation

is to the left by 2 units. 

 

59. g(x) = 2x2 + 5

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 2 and c = 5. Since 2 > 0 and  > 1, the graph of y = 2x

2 + 5

is the graph of y = x2 vertically stretched and shifted up 5 units.

60. 

SOLUTION:  

The function can be written f (x) = a(x – b)2, where a = and b = 6. Since > 0 and  < 1, the graph of 

is the graph of y = x2 vertically compressed and shifted right 6 units.

61. g(x) = 6 + x2

SOLUTION:  

The function can be written f (x) = ax2 + c, where a =  and c = 6. Since  > 0 and   > 1, the graph of y = 6 +

x2 is the graph of y = x

2 vertically stretched and shifted up 6 units.

62. h(x) = − 1 −   x2

SOLUTION:  

The function can be written f (x) = –ax2 + c, where a =  and c = –1. Since  < 0 and   > 1, the graph of y

= –1 – x2 is the graph of y = x

2 vertically stretched, shifted down 1 unit and reflected across the x-axis.

63. RIDES A popular amusement park ride whisks riders to the top of a 250-foot tower and drops them. A function for

the height of a rider is h = −16t2 + 250, where h is the height and t is the time in seconds. The ride stops the descent

of the rider 40 feet above the ground. Write an equation that models the drop of the rider. How long does it take to fall from 250 feet to 40 feet?

SOLUTION:  

The ride stops the descent of the rider at 40 feet above ground, so h = 40. Then, the equation 40 = −16t2 + 250

models the drop of the rider.

Time cannot be negative. So, it takes about 3.6 seconds to complete the ride.

Simplify. Assume that no denominator is equal to zero.

64. 

SOLUTION:  

65. 

SOLUTION:  

66. 

SOLUTION:  

67. 

SOLUTION:  

68. 

SOLUTION:  

69. b3(m

−3)(b

−6)

SOLUTION:  

Solve each open sentence.

70. |y − 2| > 7

SOLUTION:  

The solution set is {y |y > 9 or y < −5}.

Case 1 y – 2 is positive.

and Case 2 y – 2 is negative.

 

71. |z + 5| < 3

SOLUTION:  

The solution set is {z |−8 < z < −2}.

Case 1 z +5 is positive.

and Case 2 z +5 is negative.

 

72. |2b + 7| ≤ −6

SOLUTION:  

cannot be negative. So,  cannot be less than or equal to –6. The solution set is empty.

73. |3 − 2y | ≥ 8

SOLUTION:  

The solution set is {y |y  ≥ 5.5 or y  ≤ −2.5}.

Case 1 3 – 2y is positive.

and Case 2 3 – 2y is negative.

 

74. |9 − 4m| < −1

SOLUTION:  

cannot be negative. So, cannot be less than or equal to –1. The solution set is empty.

75. |5c − 2| ≤ 13

SOLUTION:  

The solution set is {c|−2.2 ≤ c ≤ 3}.

Case 1 5c – 2 is positive.

and Case 2 5c – 2 is negative.

 

Evaluate  for each set of values. Round to the nearest tenth if necessary.

76. a = 2, b = −5, c = 2

SOLUTION:  

77. a = 1, b = 12, c = 11

SOLUTION:  

78. a = −9, b = 10, c = −1

SOLUTION:  

79. a = 1, b = 7, c = −3

SOLUTION:  

80. a = 2, b = −4, c = −6

SOLUTION:  

81. a = 3, b = 1, c = 2

SOLUTION:  

This value is not a real number.

eSolutions Manual - Powered by Cognero Page 33

9-4 Solving Quadratic Equations by Completing the Square

Find the value of c that makes each trinomial a perfect square.

1. x2 − 18x + c

SOLUTION:  In this trinomial, b = –18.

So, c must be 81 to make the trinomial a perfect square.

2. x2 + 22x + c

SOLUTION:  In this trinomial, b = 22.

So, c must be 121 to make the trinomial a perfect square.

3. x2 + 9x + c

SOLUTION:  In this trinomial, b = 9.

 

So, c must be  to make the trinomial a perfect square.

4. x2 − 7x + c

SOLUTION:  In this trinomial, b = –7.

 

So, c must be  to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

5. x2 + 4x = 6

SOLUTION:  

The solutions are about –5.2 and 1.2.

6. x2 − 8x = −9

SOLUTION:  

The solutions are about 1.4 and 6.6.

7. 4x2 + 9x − 1 = 0

SOLUTION:  

The solutions are about –2.4 and 0.1.

8. −2x2 + 10x + 22 = 4

SOLUTION:  

The solutions are about –1.4 and 6.4.

9. CCSS MODELING Collin is building a deck on the back of his family’s house. He has enough lumber for the deck to be 144 square feet. The length should be 10 feet more than its width. What should the dimensions of the deck be?

SOLUTION:  Let x = the width of the deck and x + 10 = the length of the deck.

The dimensions of the deck can not be negative. So, x = –5 + 13 or 8. The width of the deck is 8 feet and the length of the deck is 8 + 10 or 18 feet.

Find the value of c that makes each trinomial a perfect square.

10. x2 + 26x + c

SOLUTION:  In this trinomial, b = 26.

So, c must be 169 to make the trinomial a perfect square.

11. x2 − 24x + c

SOLUTION:  In this trinomial, b = –24.

So, c must be 144 to make the trinomial a perfect square.

12. x2 − 19x + c

SOLUTION:  In this trinomial, b = –19.

So, c must be  to make the trinomial a perfect square.

13. x2 + 17x + c

SOLUTION:  In this trinomial, b = 17.

So, c must be  to make the trinomial a perfect square.

14. x2 + 5x + c

SOLUTION:  In this trinomial, b = 5.

So, c must be  to make the trinomial a perfect square.

15. x2 − 13x + c

SOLUTION:  In this trinomial, b = –13.

So, c must be  to make the trinomial a perfect square.

16. x2 − 22x + c

SOLUTION:  In this trinomial, b = –22.

So, c must be 121 to make the trinomial a perfect square.

17. x2 − 15x + c

SOLUTION:  In this trinomial, b = –15.

So, c must be  to make the trinomial a perfect square.

18. x2 + 24x + c

SOLUTION:  In this trinomial, b = 24.

So, c must be 144 to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

19. x2 + 6x − 16 = 0

SOLUTION:  

  The solutions are –8 and 2.

20. x2 − 2x − 14 = 0

SOLUTION:  

  The solutions are about –2.9 and 4.9.

21. x2 − 8x − 1 = 8

SOLUTION:  

  The solutions are –1 and 9.

22. x2 + 3x + 21 = 22

SOLUTION:  

  The solutions are about –3.3 and 0.3.

23. x2 − 11x + 3 = 5

SOLUTION:  

  The solutions are about –0.2 and 11.2.

24. 5x2 − 10x = 23

SOLUTION:  

  The solutions are about –1.4 and 3.4.

25. 2x2 − 2x + 7 = 5

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

26. 3x2 + 12x + 81 = 15

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

27. 4x2 + 6x = 12

SOLUTION:  

  The solutions are about –2.6 and 1.1.

28. 4x2 + 5 = 10x

SOLUTION:  

  The solutions are about 0.7 and 1.8.

29. −2x2 + 10x = −14

SOLUTION:  

  The solutions are about –1.1 and 6.1.

30. −3x2 − 12 = 14x

SOLUTION:  

  The solutions are about –3.5 and –1.1.

31. STOCK The price p in dollars for a particular stock can be modeled by the quadratic equation p = 3.5t − 0.05t2, wh

the number of days after the stock is purchased. When is the stock worth $60?

SOLUTION:  Let p = 60.

  The stock is worth $60 on the 30th and 40th days after purchase.

GEOMETRY  Find the value of x for each figure. Round to the nearest tenth if necessary.

32. area = 45 in2

SOLUTION:  

  The height cannot be negative. So, x ≈ 6.3.

33. area = 110 ft2

SOLUTION:  

  The dimensions of the rectangle cannot be negative. So, x ≈ 5.3.

34. NUMBER THEORY The product of two consecutive even integers is 224. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers are 14 and 16 or –16 and –14.

35. CCSS PRECISION The product of two consecutive negative odd integers is 483. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers must be negative. So, they are –23 and –21.

36. GEOMETRY Find the area of the triangle.

SOLUTION:  

  The dimensions of the triangle cannot be negative. So, x = 18. The base of the triangle is 18 meters and the height is 18 + 6 or 24 meters.

The area of the triangle is 216 square meters.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

37. 0.2x2 − 0.2x − 0.4 = 0

SOLUTION:  

  The solutions are –1 and 2.

38. 0.5x2 = 2x − 0.3

SOLUTION:  

  The solutions are about 0.2 and 3.8.

39. 2x2 −

SOLUTION:  

  The solutions are about 0.2 and 0.9.

40. 

SOLUTION:  

  The solutions are –0.5 and 2.5.

41. 

SOLUTION:  

  The solutions are about –8.2 and 0.2.

42. 

SOLUTION:  

  The solutions are about –5.1 and 0.1.

43. ASTRONOMY The height of an object t seconds after it is dropped is given by the equation ,

where h0 is the initial height and g is the acceleration due to gravity. The acceleration due to gravity near the surface

of Mars is 3.73 m/s2, while on Earth it is 9.8 m/s

2. Suppose an object is dropped from an initial height of 120 meters

above the surface of each planet. a. On which planet would the object reach the ground first? b. How long would it take the object to reach the ground on each planet? Round each answer to the nearest tenth. c. Do the times that it takes the object to reach the ground seem reasonable? Explain your reasoning.

SOLUTION:  a. The object on Earth will reach the ground first because it is falling at a faster rate.   b. Mars:

So, t ≈ 8.0 seconds. Earth:

So, t ≈ 4.9 seconds.   c. Sample answer: Yes; the acceleration due to gravity is much greater on Earth than on Mars, so the time to reach the ground should be much less.

44. Find all values of c that make x2 + cx + 100 a perfect square trinomial.

SOLUTION:  

So, c can be –20 or 20 to make the trinomial a perfect square.

45. Find all values of c that make x2 + cx + 225 a perfect square trinomial.

SOLUTION:  

So, c can be –30 or 30 to make the trinomial a perfect square.

46. PAINTING Before she begins painting a picture, Donna stretches her canvas over a wood frame. The frame has a length of 60 inches and a width of 4 inches. She has enough canvas to cover 480 square inches. Donna decides to increase the dimensions of the frame. If the increase in the length is 10 times the increase in the width, what will the dimensions of the frame be?

SOLUTION:  Let x = the increase in the width and let 10x = the increase in the length. So, x + 4 = the new width and 10x + 60 = the new length.  

  The dimensions cannot be negative, so x = 2. The width is 2 + 4 or 6 inches and the length is 10(2) + 60 or 80 inches.

47. MULTIPLE REPRESENTATIONS In this problem, you will investigate a property of quadratic equations. a. TABULAR  Copy the table shown and complete the second column.

b. ALGEBRAIC Set each trinomial equal to zero, and solve the equation by completing the square. Complete the last column of the table with the number of roots of each equation.

c. VERBAL Compare the number of roots of each equation to the result in the b2 − 4ac column. Is there a

relationship between these values? If so, describe it.

d. ANALYTICAL Predict how many solutions 2x2 − 9x + 15 = 0 will have.  Verify your prediction by solving the 

equation.

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3    

3x2 + 6x + 9    

x2 − 2x + 7    

x2 + 10x + 25    

x2 + 3x − 12    

SOLUTION:  a.

b.

The trinomial x2 − 8x + 16 has 1 root.

 

The trinomial 2x

2 − 11x + 3 has 2 roots.

 

The square root of a negative number has no real roots. So, the trinomial 3x2 + 6x + 9 has 0 roots.

 

The square root of a negative number has no real roots. So, the trinomial x2 − 2x + 7.

 

The trinomial x2 + 10x + 25 has 1 root.

 

The trinomial x2 + 3x − 12 has 2 roots.

 

c. If b2 − 4ac is negative, the equation has no real solutions. If b

2 − 4ac is zero, the equation has one solution. If b2

− 4ac is positive, the equation has 2 solutions. d.

The equation 2x2 − 9x + 15 = 0 has 0 real solutions because b

2 − 4ac is negative.

 

The equation cannot be solved because the square root of a negative number has no real roots.

Trinomial b2 − 4ac Number

of Roots

x2 − 8x + 16 (–8)

2 – 4(1)(16) = 0 1

2x2 − 11x + 3 (–11)

2 – 4(2)(3) = 97  

3x2 + 6x + 9 (6)

2 – 4(3)(9) = −72  

x2 − 2x + 7 (–2)

2 – 4(1)(7) = −24  

x2 + 10x + 25 (10)

2 – 4(1)(25) = 0  

x2 + 3x − 12 (3)

2 – 4(1)(–12) = 57  

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3 97 2

3x2 + 6x + 9 −72 0

x2 − 2x + 7 −24 0

x2 + 10x + 25 0 1

x2 + 3x − 12 57 2

48. CCSS PERSEVERANCE Given y = ax2 + bx + c with a ≠ 0, derive the equation for the axis of symmetry by 

completing the square and rewriting the equation in the form y = a(x − h)2 + k .

SOLUTION:  

Let    and   , then the equation becomes y = (x - h)2 + k .

For an equation in the form y = ax2 + bx + c,  the equation for the axis of symmetry is   .

So, for an equation in the form  y = (x - h)2 + k, the equation for the axis of symmetry becomes x = h.

49. REASONING Determine the number of solutions x2 + bx = c has if . Explain.

SOLUTION:  

None; Sample answer: If you add to each side of the equation and each side of the inequality, you get

. Since the left side of the last equation is a perfect square, it cannot

equal the negative number . So, there are no real solutions.

50. WHICH ONE DOESN’T BELONG? Identify the expression that does not belong with the other three. Explain your reasoning.

 

SOLUTION:  The first 3 trinomials are perfect squares.

The trinomial  is not a perfect square. So, it does not belong.

51. OPEN ENDED Write a quadratic equation for which the only solution is 4.

SOLUTION:  Find a quadratic that has two roots of 4.

Verify that the solution is 4.

So, the quadratic equation x2 − 8x + 16 = 0 has a solution of 4.

52. WRITING IN MATH Compare and contrast the following strategies for solving x2 − 5x − 7 = 0: completing the

square, graphing, and factoring.

SOLUTION:  Sample answer: Because the leading coefficient is 1, solving the equation by completing the square is simpler and yields exact answers for the solutions.   

 

To solve the equation by graphing, use a graphing calculator to graph the related function y = x2 - 5x - 7. Select the

zero option from the 2nd [CALC] menu to determine the roots. 

   

  The roots are given as decimal approximations. So, for this equation the solutions would have to be given as estimations.   There are no factors of –7 that have a sum of –5, so solving by factoring is not possible.

53. The length of a rectangle is 3 times its width. The area of the rectangle is 75 square feet. Find the length of the rectangle in feet. A 25 B 15 C 10 D 5

SOLUTION:  Let x = the width of the rectangle and let 3x = the length of the rectangle.

The length of the rectangle is 3(5) or 15 feet. Choice B is the correct answer.

54. PROBABILITY At a festival, winners of a game draw a token for a prize. There is one token for each prize. The prizes include 9 movie passes, 8 stuffed animals, 5 hats, 10 jump ropes, and 4 glow necklaces. What is the probabilitythat the first person to draw a token will win a movie pass?

F  

G

H

J  

SOLUTION:  There are 9 + 8 + 5 + 10 + 4 or 36 possible outcomes.

P(movie pass) =  or  . Choice J is the correct answer.

55. GRIDDED RESPONSE The population of a town can be modeled by P = 22,000 + 125t, where P represents the population and t represents the number of years from 2000. How many years after 2000 will the population be 26,000?

SOLUTION:  

In 32 years the population of the town will be 26,000.

56. Percy delivers pizzas for Pizza King. He is paid $6 an hour plus $2.50 for each pizza he delivers. Percy earned $280 last week. If he worked a total of 30 hours, how many pizzas did he deliver? A 250 pizzas B 184 pizzas C 40 pizzas D 34 pizzas

SOLUTION:  Let p = the number of pizzas Percy delivered.

Percy delivered 40 pizzas. Choice C is the correct answer.

Describe how the graph of each function is related to the graph of f (x) = x2.

57. g(x) = −12 + x2

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –12, the translation is

down. So, the graph is shifted down 12 units from the parent function. 

58. h(x) = (x + 2)2

SOLUTION:  

The graph of f (x) = (x – c)2 represents a translation left or right from the parent graph. Since c = –2, the translation

is to the left by 2 units. 

 

59. g(x) = 2x2 + 5

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 2 and c = 5. Since 2 > 0 and  > 1, the graph of y = 2x

2 + 5

is the graph of y = x2 vertically stretched and shifted up 5 units.

60. 

SOLUTION:  

The function can be written f (x) = a(x – b)2, where a = and b = 6. Since > 0 and  < 1, the graph of 

is the graph of y = x2 vertically compressed and shifted right 6 units.

61. g(x) = 6 + x2

SOLUTION:  

The function can be written f (x) = ax2 + c, where a =  and c = 6. Since  > 0 and   > 1, the graph of y = 6 +

x2 is the graph of y = x

2 vertically stretched and shifted up 6 units.

62. h(x) = − 1 −   x2

SOLUTION:  

The function can be written f (x) = –ax2 + c, where a =  and c = –1. Since  < 0 and   > 1, the graph of y

= –1 – x2 is the graph of y = x

2 vertically stretched, shifted down 1 unit and reflected across the x-axis.

63. RIDES A popular amusement park ride whisks riders to the top of a 250-foot tower and drops them. A function for

the height of a rider is h = −16t2 + 250, where h is the height and t is the time in seconds. The ride stops the descent

of the rider 40 feet above the ground. Write an equation that models the drop of the rider. How long does it take to fall from 250 feet to 40 feet?

SOLUTION:  

The ride stops the descent of the rider at 40 feet above ground, so h = 40. Then, the equation 40 = −16t2 + 250

models the drop of the rider.

Time cannot be negative. So, it takes about 3.6 seconds to complete the ride.

Simplify. Assume that no denominator is equal to zero.

64. 

SOLUTION:  

65. 

SOLUTION:  

66. 

SOLUTION:  

67. 

SOLUTION:  

68. 

SOLUTION:  

69. b3(m

−3)(b

−6)

SOLUTION:  

Solve each open sentence.

70. |y − 2| > 7

SOLUTION:  

The solution set is {y |y > 9 or y < −5}.

Case 1 y – 2 is positive.

and Case 2 y – 2 is negative.

 

71. |z + 5| < 3

SOLUTION:  

The solution set is {z |−8 < z < −2}.

Case 1 z +5 is positive.

and Case 2 z +5 is negative.

 

72. |2b + 7| ≤ −6

SOLUTION:  

cannot be negative. So,  cannot be less than or equal to –6. The solution set is empty.

73. |3 − 2y | ≥ 8

SOLUTION:  

The solution set is {y |y  ≥ 5.5 or y  ≤ −2.5}.

Case 1 3 – 2y is positive.

and Case 2 3 – 2y is negative.

 

74. |9 − 4m| < −1

SOLUTION:  

cannot be negative. So, cannot be less than or equal to –1. The solution set is empty.

75. |5c − 2| ≤ 13

SOLUTION:  

The solution set is {c|−2.2 ≤ c ≤ 3}.

Case 1 5c – 2 is positive.

and Case 2 5c – 2 is negative.

 

Evaluate  for each set of values. Round to the nearest tenth if necessary.

76. a = 2, b = −5, c = 2

SOLUTION:  

77. a = 1, b = 12, c = 11

SOLUTION:  

78. a = −9, b = 10, c = −1

SOLUTION:  

79. a = 1, b = 7, c = −3

SOLUTION:  

80. a = 2, b = −4, c = −6

SOLUTION:  

81. a = 3, b = 1, c = 2

SOLUTION:  

This value is not a real number.

eSolutions Manual - Powered by Cognero Page 34

9-4 Solving Quadratic Equations by Completing the Square

Find the value of c that makes each trinomial a perfect square.

1. x2 − 18x + c

SOLUTION:  In this trinomial, b = –18.

So, c must be 81 to make the trinomial a perfect square.

2. x2 + 22x + c

SOLUTION:  In this trinomial, b = 22.

So, c must be 121 to make the trinomial a perfect square.

3. x2 + 9x + c

SOLUTION:  In this trinomial, b = 9.

 

So, c must be  to make the trinomial a perfect square.

4. x2 − 7x + c

SOLUTION:  In this trinomial, b = –7.

 

So, c must be  to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

5. x2 + 4x = 6

SOLUTION:  

The solutions are about –5.2 and 1.2.

6. x2 − 8x = −9

SOLUTION:  

The solutions are about 1.4 and 6.6.

7. 4x2 + 9x − 1 = 0

SOLUTION:  

The solutions are about –2.4 and 0.1.

8. −2x2 + 10x + 22 = 4

SOLUTION:  

The solutions are about –1.4 and 6.4.

9. CCSS MODELING Collin is building a deck on the back of his family’s house. He has enough lumber for the deck to be 144 square feet. The length should be 10 feet more than its width. What should the dimensions of the deck be?

SOLUTION:  Let x = the width of the deck and x + 10 = the length of the deck.

The dimensions of the deck can not be negative. So, x = –5 + 13 or 8. The width of the deck is 8 feet and the length of the deck is 8 + 10 or 18 feet.

Find the value of c that makes each trinomial a perfect square.

10. x2 + 26x + c

SOLUTION:  In this trinomial, b = 26.

So, c must be 169 to make the trinomial a perfect square.

11. x2 − 24x + c

SOLUTION:  In this trinomial, b = –24.

So, c must be 144 to make the trinomial a perfect square.

12. x2 − 19x + c

SOLUTION:  In this trinomial, b = –19.

So, c must be  to make the trinomial a perfect square.

13. x2 + 17x + c

SOLUTION:  In this trinomial, b = 17.

So, c must be  to make the trinomial a perfect square.

14. x2 + 5x + c

SOLUTION:  In this trinomial, b = 5.

So, c must be  to make the trinomial a perfect square.

15. x2 − 13x + c

SOLUTION:  In this trinomial, b = –13.

So, c must be  to make the trinomial a perfect square.

16. x2 − 22x + c

SOLUTION:  In this trinomial, b = –22.

So, c must be 121 to make the trinomial a perfect square.

17. x2 − 15x + c

SOLUTION:  In this trinomial, b = –15.

So, c must be  to make the trinomial a perfect square.

18. x2 + 24x + c

SOLUTION:  In this trinomial, b = 24.

So, c must be 144 to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

19. x2 + 6x − 16 = 0

SOLUTION:  

  The solutions are –8 and 2.

20. x2 − 2x − 14 = 0

SOLUTION:  

  The solutions are about –2.9 and 4.9.

21. x2 − 8x − 1 = 8

SOLUTION:  

  The solutions are –1 and 9.

22. x2 + 3x + 21 = 22

SOLUTION:  

  The solutions are about –3.3 and 0.3.

23. x2 − 11x + 3 = 5

SOLUTION:  

  The solutions are about –0.2 and 11.2.

24. 5x2 − 10x = 23

SOLUTION:  

  The solutions are about –1.4 and 3.4.

25. 2x2 − 2x + 7 = 5

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

26. 3x2 + 12x + 81 = 15

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

27. 4x2 + 6x = 12

SOLUTION:  

  The solutions are about –2.6 and 1.1.

28. 4x2 + 5 = 10x

SOLUTION:  

  The solutions are about 0.7 and 1.8.

29. −2x2 + 10x = −14

SOLUTION:  

  The solutions are about –1.1 and 6.1.

30. −3x2 − 12 = 14x

SOLUTION:  

  The solutions are about –3.5 and –1.1.

31. STOCK The price p in dollars for a particular stock can be modeled by the quadratic equation p = 3.5t − 0.05t2, wh

the number of days after the stock is purchased. When is the stock worth $60?

SOLUTION:  Let p = 60.

  The stock is worth $60 on the 30th and 40th days after purchase.

GEOMETRY  Find the value of x for each figure. Round to the nearest tenth if necessary.

32. area = 45 in2

SOLUTION:  

  The height cannot be negative. So, x ≈ 6.3.

33. area = 110 ft2

SOLUTION:  

  The dimensions of the rectangle cannot be negative. So, x ≈ 5.3.

34. NUMBER THEORY The product of two consecutive even integers is 224. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers are 14 and 16 or –16 and –14.

35. CCSS PRECISION The product of two consecutive negative odd integers is 483. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers must be negative. So, they are –23 and –21.

36. GEOMETRY Find the area of the triangle.

SOLUTION:  

  The dimensions of the triangle cannot be negative. So, x = 18. The base of the triangle is 18 meters and the height is 18 + 6 or 24 meters.

The area of the triangle is 216 square meters.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

37. 0.2x2 − 0.2x − 0.4 = 0

SOLUTION:  

  The solutions are –1 and 2.

38. 0.5x2 = 2x − 0.3

SOLUTION:  

  The solutions are about 0.2 and 3.8.

39. 2x2 −

SOLUTION:  

  The solutions are about 0.2 and 0.9.

40. 

SOLUTION:  

  The solutions are –0.5 and 2.5.

41. 

SOLUTION:  

  The solutions are about –8.2 and 0.2.

42. 

SOLUTION:  

  The solutions are about –5.1 and 0.1.

43. ASTRONOMY The height of an object t seconds after it is dropped is given by the equation ,

where h0 is the initial height and g is the acceleration due to gravity. The acceleration due to gravity near the surface

of Mars is 3.73 m/s2, while on Earth it is 9.8 m/s

2. Suppose an object is dropped from an initial height of 120 meters

above the surface of each planet. a. On which planet would the object reach the ground first? b. How long would it take the object to reach the ground on each planet? Round each answer to the nearest tenth. c. Do the times that it takes the object to reach the ground seem reasonable? Explain your reasoning.

SOLUTION:  a. The object on Earth will reach the ground first because it is falling at a faster rate.   b. Mars:

So, t ≈ 8.0 seconds. Earth:

So, t ≈ 4.9 seconds.   c. Sample answer: Yes; the acceleration due to gravity is much greater on Earth than on Mars, so the time to reach the ground should be much less.

44. Find all values of c that make x2 + cx + 100 a perfect square trinomial.

SOLUTION:  

So, c can be –20 or 20 to make the trinomial a perfect square.

45. Find all values of c that make x2 + cx + 225 a perfect square trinomial.

SOLUTION:  

So, c can be –30 or 30 to make the trinomial a perfect square.

46. PAINTING Before she begins painting a picture, Donna stretches her canvas over a wood frame. The frame has a length of 60 inches and a width of 4 inches. She has enough canvas to cover 480 square inches. Donna decides to increase the dimensions of the frame. If the increase in the length is 10 times the increase in the width, what will the dimensions of the frame be?

SOLUTION:  Let x = the increase in the width and let 10x = the increase in the length. So, x + 4 = the new width and 10x + 60 = the new length.  

  The dimensions cannot be negative, so x = 2. The width is 2 + 4 or 6 inches and the length is 10(2) + 60 or 80 inches.

47. MULTIPLE REPRESENTATIONS In this problem, you will investigate a property of quadratic equations. a. TABULAR  Copy the table shown and complete the second column.

b. ALGEBRAIC Set each trinomial equal to zero, and solve the equation by completing the square. Complete the last column of the table with the number of roots of each equation.

c. VERBAL Compare the number of roots of each equation to the result in the b2 − 4ac column. Is there a

relationship between these values? If so, describe it.

d. ANALYTICAL Predict how many solutions 2x2 − 9x + 15 = 0 will have.  Verify your prediction by solving the 

equation.

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3    

3x2 + 6x + 9    

x2 − 2x + 7    

x2 + 10x + 25    

x2 + 3x − 12    

SOLUTION:  a.

b.

The trinomial x2 − 8x + 16 has 1 root.

 

The trinomial 2x

2 − 11x + 3 has 2 roots.

 

The square root of a negative number has no real roots. So, the trinomial 3x2 + 6x + 9 has 0 roots.

 

The square root of a negative number has no real roots. So, the trinomial x2 − 2x + 7.

 

The trinomial x2 + 10x + 25 has 1 root.

 

The trinomial x2 + 3x − 12 has 2 roots.

 

c. If b2 − 4ac is negative, the equation has no real solutions. If b

2 − 4ac is zero, the equation has one solution. If b2

− 4ac is positive, the equation has 2 solutions. d.

The equation 2x2 − 9x + 15 = 0 has 0 real solutions because b

2 − 4ac is negative.

 

The equation cannot be solved because the square root of a negative number has no real roots.

Trinomial b2 − 4ac Number

of Roots

x2 − 8x + 16 (–8)

2 – 4(1)(16) = 0 1

2x2 − 11x + 3 (–11)

2 – 4(2)(3) = 97  

3x2 + 6x + 9 (6)

2 – 4(3)(9) = −72  

x2 − 2x + 7 (–2)

2 – 4(1)(7) = −24  

x2 + 10x + 25 (10)

2 – 4(1)(25) = 0  

x2 + 3x − 12 (3)

2 – 4(1)(–12) = 57  

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3 97 2

3x2 + 6x + 9 −72 0

x2 − 2x + 7 −24 0

x2 + 10x + 25 0 1

x2 + 3x − 12 57 2

48. CCSS PERSEVERANCE Given y = ax2 + bx + c with a ≠ 0, derive the equation for the axis of symmetry by 

completing the square and rewriting the equation in the form y = a(x − h)2 + k .

SOLUTION:  

Let    and   , then the equation becomes y = (x - h)2 + k .

For an equation in the form y = ax2 + bx + c,  the equation for the axis of symmetry is   .

So, for an equation in the form  y = (x - h)2 + k, the equation for the axis of symmetry becomes x = h.

49. REASONING Determine the number of solutions x2 + bx = c has if . Explain.

SOLUTION:  

None; Sample answer: If you add to each side of the equation and each side of the inequality, you get

. Since the left side of the last equation is a perfect square, it cannot

equal the negative number . So, there are no real solutions.

50. WHICH ONE DOESN’T BELONG? Identify the expression that does not belong with the other three. Explain your reasoning.

 

SOLUTION:  The first 3 trinomials are perfect squares.

The trinomial  is not a perfect square. So, it does not belong.

51. OPEN ENDED Write a quadratic equation for which the only solution is 4.

SOLUTION:  Find a quadratic that has two roots of 4.

Verify that the solution is 4.

So, the quadratic equation x2 − 8x + 16 = 0 has a solution of 4.

52. WRITING IN MATH Compare and contrast the following strategies for solving x2 − 5x − 7 = 0: completing the

square, graphing, and factoring.

SOLUTION:  Sample answer: Because the leading coefficient is 1, solving the equation by completing the square is simpler and yields exact answers for the solutions.   

 

To solve the equation by graphing, use a graphing calculator to graph the related function y = x2 - 5x - 7. Select the

zero option from the 2nd [CALC] menu to determine the roots. 

   

  The roots are given as decimal approximations. So, for this equation the solutions would have to be given as estimations.   There are no factors of –7 that have a sum of –5, so solving by factoring is not possible.

53. The length of a rectangle is 3 times its width. The area of the rectangle is 75 square feet. Find the length of the rectangle in feet. A 25 B 15 C 10 D 5

SOLUTION:  Let x = the width of the rectangle and let 3x = the length of the rectangle.

The length of the rectangle is 3(5) or 15 feet. Choice B is the correct answer.

54. PROBABILITY At a festival, winners of a game draw a token for a prize. There is one token for each prize. The prizes include 9 movie passes, 8 stuffed animals, 5 hats, 10 jump ropes, and 4 glow necklaces. What is the probabilitythat the first person to draw a token will win a movie pass?

F  

G

H

J  

SOLUTION:  There are 9 + 8 + 5 + 10 + 4 or 36 possible outcomes.

P(movie pass) =  or  . Choice J is the correct answer.

55. GRIDDED RESPONSE The population of a town can be modeled by P = 22,000 + 125t, where P represents the population and t represents the number of years from 2000. How many years after 2000 will the population be 26,000?

SOLUTION:  

In 32 years the population of the town will be 26,000.

56. Percy delivers pizzas for Pizza King. He is paid $6 an hour plus $2.50 for each pizza he delivers. Percy earned $280 last week. If he worked a total of 30 hours, how many pizzas did he deliver? A 250 pizzas B 184 pizzas C 40 pizzas D 34 pizzas

SOLUTION:  Let p = the number of pizzas Percy delivered.

Percy delivered 40 pizzas. Choice C is the correct answer.

Describe how the graph of each function is related to the graph of f (x) = x2.

57. g(x) = −12 + x2

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –12, the translation is

down. So, the graph is shifted down 12 units from the parent function. 

58. h(x) = (x + 2)2

SOLUTION:  

The graph of f (x) = (x – c)2 represents a translation left or right from the parent graph. Since c = –2, the translation

is to the left by 2 units. 

 

59. g(x) = 2x2 + 5

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 2 and c = 5. Since 2 > 0 and  > 1, the graph of y = 2x

2 + 5

is the graph of y = x2 vertically stretched and shifted up 5 units.

60. 

SOLUTION:  

The function can be written f (x) = a(x – b)2, where a = and b = 6. Since > 0 and  < 1, the graph of 

is the graph of y = x2 vertically compressed and shifted right 6 units.

61. g(x) = 6 + x2

SOLUTION:  

The function can be written f (x) = ax2 + c, where a =  and c = 6. Since  > 0 and   > 1, the graph of y = 6 +

x2 is the graph of y = x

2 vertically stretched and shifted up 6 units.

62. h(x) = − 1 −   x2

SOLUTION:  

The function can be written f (x) = –ax2 + c, where a =  and c = –1. Since  < 0 and   > 1, the graph of y

= –1 – x2 is the graph of y = x

2 vertically stretched, shifted down 1 unit and reflected across the x-axis.

63. RIDES A popular amusement park ride whisks riders to the top of a 250-foot tower and drops them. A function for

the height of a rider is h = −16t2 + 250, where h is the height and t is the time in seconds. The ride stops the descent

of the rider 40 feet above the ground. Write an equation that models the drop of the rider. How long does it take to fall from 250 feet to 40 feet?

SOLUTION:  

The ride stops the descent of the rider at 40 feet above ground, so h = 40. Then, the equation 40 = −16t2 + 250

models the drop of the rider.

Time cannot be negative. So, it takes about 3.6 seconds to complete the ride.

Simplify. Assume that no denominator is equal to zero.

64. 

SOLUTION:  

65. 

SOLUTION:  

66. 

SOLUTION:  

67. 

SOLUTION:  

68. 

SOLUTION:  

69. b3(m

−3)(b

−6)

SOLUTION:  

Solve each open sentence.

70. |y − 2| > 7

SOLUTION:  

The solution set is {y |y > 9 or y < −5}.

Case 1 y – 2 is positive.

and Case 2 y – 2 is negative.

 

71. |z + 5| < 3

SOLUTION:  

The solution set is {z |−8 < z < −2}.

Case 1 z +5 is positive.

and Case 2 z +5 is negative.

 

72. |2b + 7| ≤ −6

SOLUTION:  

cannot be negative. So,  cannot be less than or equal to –6. The solution set is empty.

73. |3 − 2y | ≥ 8

SOLUTION:  

The solution set is {y |y  ≥ 5.5 or y  ≤ −2.5}.

Case 1 3 – 2y is positive.

and Case 2 3 – 2y is negative.

 

74. |9 − 4m| < −1

SOLUTION:  

cannot be negative. So, cannot be less than or equal to –1. The solution set is empty.

75. |5c − 2| ≤ 13

SOLUTION:  

The solution set is {c|−2.2 ≤ c ≤ 3}.

Case 1 5c – 2 is positive.

and Case 2 5c – 2 is negative.

 

Evaluate  for each set of values. Round to the nearest tenth if necessary.

76. a = 2, b = −5, c = 2

SOLUTION:  

77. a = 1, b = 12, c = 11

SOLUTION:  

78. a = −9, b = 10, c = −1

SOLUTION:  

79. a = 1, b = 7, c = −3

SOLUTION:  

80. a = 2, b = −4, c = −6

SOLUTION:  

81. a = 3, b = 1, c = 2

SOLUTION:  

This value is not a real number.

eSolutions Manual - Powered by Cognero Page 35

9-4 Solving Quadratic Equations by Completing the Square

Find the value of c that makes each trinomial a perfect square.

1. x2 − 18x + c

SOLUTION:  In this trinomial, b = –18.

So, c must be 81 to make the trinomial a perfect square.

2. x2 + 22x + c

SOLUTION:  In this trinomial, b = 22.

So, c must be 121 to make the trinomial a perfect square.

3. x2 + 9x + c

SOLUTION:  In this trinomial, b = 9.

 

So, c must be  to make the trinomial a perfect square.

4. x2 − 7x + c

SOLUTION:  In this trinomial, b = –7.

 

So, c must be  to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

5. x2 + 4x = 6

SOLUTION:  

The solutions are about –5.2 and 1.2.

6. x2 − 8x = −9

SOLUTION:  

The solutions are about 1.4 and 6.6.

7. 4x2 + 9x − 1 = 0

SOLUTION:  

The solutions are about –2.4 and 0.1.

8. −2x2 + 10x + 22 = 4

SOLUTION:  

The solutions are about –1.4 and 6.4.

9. CCSS MODELING Collin is building a deck on the back of his family’s house. He has enough lumber for the deck to be 144 square feet. The length should be 10 feet more than its width. What should the dimensions of the deck be?

SOLUTION:  Let x = the width of the deck and x + 10 = the length of the deck.

The dimensions of the deck can not be negative. So, x = –5 + 13 or 8. The width of the deck is 8 feet and the length of the deck is 8 + 10 or 18 feet.

Find the value of c that makes each trinomial a perfect square.

10. x2 + 26x + c

SOLUTION:  In this trinomial, b = 26.

So, c must be 169 to make the trinomial a perfect square.

11. x2 − 24x + c

SOLUTION:  In this trinomial, b = –24.

So, c must be 144 to make the trinomial a perfect square.

12. x2 − 19x + c

SOLUTION:  In this trinomial, b = –19.

So, c must be  to make the trinomial a perfect square.

13. x2 + 17x + c

SOLUTION:  In this trinomial, b = 17.

So, c must be  to make the trinomial a perfect square.

14. x2 + 5x + c

SOLUTION:  In this trinomial, b = 5.

So, c must be  to make the trinomial a perfect square.

15. x2 − 13x + c

SOLUTION:  In this trinomial, b = –13.

So, c must be  to make the trinomial a perfect square.

16. x2 − 22x + c

SOLUTION:  In this trinomial, b = –22.

So, c must be 121 to make the trinomial a perfect square.

17. x2 − 15x + c

SOLUTION:  In this trinomial, b = –15.

So, c must be  to make the trinomial a perfect square.

18. x2 + 24x + c

SOLUTION:  In this trinomial, b = 24.

So, c must be 144 to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

19. x2 + 6x − 16 = 0

SOLUTION:  

  The solutions are –8 and 2.

20. x2 − 2x − 14 = 0

SOLUTION:  

  The solutions are about –2.9 and 4.9.

21. x2 − 8x − 1 = 8

SOLUTION:  

  The solutions are –1 and 9.

22. x2 + 3x + 21 = 22

SOLUTION:  

  The solutions are about –3.3 and 0.3.

23. x2 − 11x + 3 = 5

SOLUTION:  

  The solutions are about –0.2 and 11.2.

24. 5x2 − 10x = 23

SOLUTION:  

  The solutions are about –1.4 and 3.4.

25. 2x2 − 2x + 7 = 5

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

26. 3x2 + 12x + 81 = 15

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

27. 4x2 + 6x = 12

SOLUTION:  

  The solutions are about –2.6 and 1.1.

28. 4x2 + 5 = 10x

SOLUTION:  

  The solutions are about 0.7 and 1.8.

29. −2x2 + 10x = −14

SOLUTION:  

  The solutions are about –1.1 and 6.1.

30. −3x2 − 12 = 14x

SOLUTION:  

  The solutions are about –3.5 and –1.1.

31. STOCK The price p in dollars for a particular stock can be modeled by the quadratic equation p = 3.5t − 0.05t2, wh

the number of days after the stock is purchased. When is the stock worth $60?

SOLUTION:  Let p = 60.

  The stock is worth $60 on the 30th and 40th days after purchase.

GEOMETRY  Find the value of x for each figure. Round to the nearest tenth if necessary.

32. area = 45 in2

SOLUTION:  

  The height cannot be negative. So, x ≈ 6.3.

33. area = 110 ft2

SOLUTION:  

  The dimensions of the rectangle cannot be negative. So, x ≈ 5.3.

34. NUMBER THEORY The product of two consecutive even integers is 224. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers are 14 and 16 or –16 and –14.

35. CCSS PRECISION The product of two consecutive negative odd integers is 483. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers must be negative. So, they are –23 and –21.

36. GEOMETRY Find the area of the triangle.

SOLUTION:  

  The dimensions of the triangle cannot be negative. So, x = 18. The base of the triangle is 18 meters and the height is 18 + 6 or 24 meters.

The area of the triangle is 216 square meters.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

37. 0.2x2 − 0.2x − 0.4 = 0

SOLUTION:  

  The solutions are –1 and 2.

38. 0.5x2 = 2x − 0.3

SOLUTION:  

  The solutions are about 0.2 and 3.8.

39. 2x2 −

SOLUTION:  

  The solutions are about 0.2 and 0.9.

40. 

SOLUTION:  

  The solutions are –0.5 and 2.5.

41. 

SOLUTION:  

  The solutions are about –8.2 and 0.2.

42. 

SOLUTION:  

  The solutions are about –5.1 and 0.1.

43. ASTRONOMY The height of an object t seconds after it is dropped is given by the equation ,

where h0 is the initial height and g is the acceleration due to gravity. The acceleration due to gravity near the surface

of Mars is 3.73 m/s2, while on Earth it is 9.8 m/s

2. Suppose an object is dropped from an initial height of 120 meters

above the surface of each planet. a. On which planet would the object reach the ground first? b. How long would it take the object to reach the ground on each planet? Round each answer to the nearest tenth. c. Do the times that it takes the object to reach the ground seem reasonable? Explain your reasoning.

SOLUTION:  a. The object on Earth will reach the ground first because it is falling at a faster rate.   b. Mars:

So, t ≈ 8.0 seconds. Earth:

So, t ≈ 4.9 seconds.   c. Sample answer: Yes; the acceleration due to gravity is much greater on Earth than on Mars, so the time to reach the ground should be much less.

44. Find all values of c that make x2 + cx + 100 a perfect square trinomial.

SOLUTION:  

So, c can be –20 or 20 to make the trinomial a perfect square.

45. Find all values of c that make x2 + cx + 225 a perfect square trinomial.

SOLUTION:  

So, c can be –30 or 30 to make the trinomial a perfect square.

46. PAINTING Before she begins painting a picture, Donna stretches her canvas over a wood frame. The frame has a length of 60 inches and a width of 4 inches. She has enough canvas to cover 480 square inches. Donna decides to increase the dimensions of the frame. If the increase in the length is 10 times the increase in the width, what will the dimensions of the frame be?

SOLUTION:  Let x = the increase in the width and let 10x = the increase in the length. So, x + 4 = the new width and 10x + 60 = the new length.  

  The dimensions cannot be negative, so x = 2. The width is 2 + 4 or 6 inches and the length is 10(2) + 60 or 80 inches.

47. MULTIPLE REPRESENTATIONS In this problem, you will investigate a property of quadratic equations. a. TABULAR  Copy the table shown and complete the second column.

b. ALGEBRAIC Set each trinomial equal to zero, and solve the equation by completing the square. Complete the last column of the table with the number of roots of each equation.

c. VERBAL Compare the number of roots of each equation to the result in the b2 − 4ac column. Is there a

relationship between these values? If so, describe it.

d. ANALYTICAL Predict how many solutions 2x2 − 9x + 15 = 0 will have.  Verify your prediction by solving the 

equation.

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3    

3x2 + 6x + 9    

x2 − 2x + 7    

x2 + 10x + 25    

x2 + 3x − 12    

SOLUTION:  a.

b.

The trinomial x2 − 8x + 16 has 1 root.

 

The trinomial 2x

2 − 11x + 3 has 2 roots.

 

The square root of a negative number has no real roots. So, the trinomial 3x2 + 6x + 9 has 0 roots.

 

The square root of a negative number has no real roots. So, the trinomial x2 − 2x + 7.

 

The trinomial x2 + 10x + 25 has 1 root.

 

The trinomial x2 + 3x − 12 has 2 roots.

 

c. If b2 − 4ac is negative, the equation has no real solutions. If b

2 − 4ac is zero, the equation has one solution. If b2

− 4ac is positive, the equation has 2 solutions. d.

The equation 2x2 − 9x + 15 = 0 has 0 real solutions because b

2 − 4ac is negative.

 

The equation cannot be solved because the square root of a negative number has no real roots.

Trinomial b2 − 4ac Number

of Roots

x2 − 8x + 16 (–8)

2 – 4(1)(16) = 0 1

2x2 − 11x + 3 (–11)

2 – 4(2)(3) = 97  

3x2 + 6x + 9 (6)

2 – 4(3)(9) = −72  

x2 − 2x + 7 (–2)

2 – 4(1)(7) = −24  

x2 + 10x + 25 (10)

2 – 4(1)(25) = 0  

x2 + 3x − 12 (3)

2 – 4(1)(–12) = 57  

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3 97 2

3x2 + 6x + 9 −72 0

x2 − 2x + 7 −24 0

x2 + 10x + 25 0 1

x2 + 3x − 12 57 2

48. CCSS PERSEVERANCE Given y = ax2 + bx + c with a ≠ 0, derive the equation for the axis of symmetry by 

completing the square and rewriting the equation in the form y = a(x − h)2 + k .

SOLUTION:  

Let    and   , then the equation becomes y = (x - h)2 + k .

For an equation in the form y = ax2 + bx + c,  the equation for the axis of symmetry is   .

So, for an equation in the form  y = (x - h)2 + k, the equation for the axis of symmetry becomes x = h.

49. REASONING Determine the number of solutions x2 + bx = c has if . Explain.

SOLUTION:  

None; Sample answer: If you add to each side of the equation and each side of the inequality, you get

. Since the left side of the last equation is a perfect square, it cannot

equal the negative number . So, there are no real solutions.

50. WHICH ONE DOESN’T BELONG? Identify the expression that does not belong with the other three. Explain your reasoning.

 

SOLUTION:  The first 3 trinomials are perfect squares.

The trinomial  is not a perfect square. So, it does not belong.

51. OPEN ENDED Write a quadratic equation for which the only solution is 4.

SOLUTION:  Find a quadratic that has two roots of 4.

Verify that the solution is 4.

So, the quadratic equation x2 − 8x + 16 = 0 has a solution of 4.

52. WRITING IN MATH Compare and contrast the following strategies for solving x2 − 5x − 7 = 0: completing the

square, graphing, and factoring.

SOLUTION:  Sample answer: Because the leading coefficient is 1, solving the equation by completing the square is simpler and yields exact answers for the solutions.   

 

To solve the equation by graphing, use a graphing calculator to graph the related function y = x2 - 5x - 7. Select the

zero option from the 2nd [CALC] menu to determine the roots. 

   

  The roots are given as decimal approximations. So, for this equation the solutions would have to be given as estimations.   There are no factors of –7 that have a sum of –5, so solving by factoring is not possible.

53. The length of a rectangle is 3 times its width. The area of the rectangle is 75 square feet. Find the length of the rectangle in feet. A 25 B 15 C 10 D 5

SOLUTION:  Let x = the width of the rectangle and let 3x = the length of the rectangle.

The length of the rectangle is 3(5) or 15 feet. Choice B is the correct answer.

54. PROBABILITY At a festival, winners of a game draw a token for a prize. There is one token for each prize. The prizes include 9 movie passes, 8 stuffed animals, 5 hats, 10 jump ropes, and 4 glow necklaces. What is the probabilitythat the first person to draw a token will win a movie pass?

F  

G

H

J  

SOLUTION:  There are 9 + 8 + 5 + 10 + 4 or 36 possible outcomes.

P(movie pass) =  or  . Choice J is the correct answer.

55. GRIDDED RESPONSE The population of a town can be modeled by P = 22,000 + 125t, where P represents the population and t represents the number of years from 2000. How many years after 2000 will the population be 26,000?

SOLUTION:  

In 32 years the population of the town will be 26,000.

56. Percy delivers pizzas for Pizza King. He is paid $6 an hour plus $2.50 for each pizza he delivers. Percy earned $280 last week. If he worked a total of 30 hours, how many pizzas did he deliver? A 250 pizzas B 184 pizzas C 40 pizzas D 34 pizzas

SOLUTION:  Let p = the number of pizzas Percy delivered.

Percy delivered 40 pizzas. Choice C is the correct answer.

Describe how the graph of each function is related to the graph of f (x) = x2.

57. g(x) = −12 + x2

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –12, the translation is

down. So, the graph is shifted down 12 units from the parent function. 

58. h(x) = (x + 2)2

SOLUTION:  

The graph of f (x) = (x – c)2 represents a translation left or right from the parent graph. Since c = –2, the translation

is to the left by 2 units. 

 

59. g(x) = 2x2 + 5

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 2 and c = 5. Since 2 > 0 and  > 1, the graph of y = 2x

2 + 5

is the graph of y = x2 vertically stretched and shifted up 5 units.

60. 

SOLUTION:  

The function can be written f (x) = a(x – b)2, where a = and b = 6. Since > 0 and  < 1, the graph of 

is the graph of y = x2 vertically compressed and shifted right 6 units.

61. g(x) = 6 + x2

SOLUTION:  

The function can be written f (x) = ax2 + c, where a =  and c = 6. Since  > 0 and   > 1, the graph of y = 6 +

x2 is the graph of y = x

2 vertically stretched and shifted up 6 units.

62. h(x) = − 1 −   x2

SOLUTION:  

The function can be written f (x) = –ax2 + c, where a =  and c = –1. Since  < 0 and   > 1, the graph of y

= –1 – x2 is the graph of y = x

2 vertically stretched, shifted down 1 unit and reflected across the x-axis.

63. RIDES A popular amusement park ride whisks riders to the top of a 250-foot tower and drops them. A function for

the height of a rider is h = −16t2 + 250, where h is the height and t is the time in seconds. The ride stops the descent

of the rider 40 feet above the ground. Write an equation that models the drop of the rider. How long does it take to fall from 250 feet to 40 feet?

SOLUTION:  

The ride stops the descent of the rider at 40 feet above ground, so h = 40. Then, the equation 40 = −16t2 + 250

models the drop of the rider.

Time cannot be negative. So, it takes about 3.6 seconds to complete the ride.

Simplify. Assume that no denominator is equal to zero.

64. 

SOLUTION:  

65. 

SOLUTION:  

66. 

SOLUTION:  

67. 

SOLUTION:  

68. 

SOLUTION:  

69. b3(m

−3)(b

−6)

SOLUTION:  

Solve each open sentence.

70. |y − 2| > 7

SOLUTION:  

The solution set is {y |y > 9 or y < −5}.

Case 1 y – 2 is positive.

and Case 2 y – 2 is negative.

 

71. |z + 5| < 3

SOLUTION:  

The solution set is {z |−8 < z < −2}.

Case 1 z +5 is positive.

and Case 2 z +5 is negative.

 

72. |2b + 7| ≤ −6

SOLUTION:  

cannot be negative. So,  cannot be less than or equal to –6. The solution set is empty.

73. |3 − 2y | ≥ 8

SOLUTION:  

The solution set is {y |y  ≥ 5.5 or y  ≤ −2.5}.

Case 1 3 – 2y is positive.

and Case 2 3 – 2y is negative.

 

74. |9 − 4m| < −1

SOLUTION:  

cannot be negative. So, cannot be less than or equal to –1. The solution set is empty.

75. |5c − 2| ≤ 13

SOLUTION:  

The solution set is {c|−2.2 ≤ c ≤ 3}.

Case 1 5c – 2 is positive.

and Case 2 5c – 2 is negative.

 

Evaluate  for each set of values. Round to the nearest tenth if necessary.

76. a = 2, b = −5, c = 2

SOLUTION:  

77. a = 1, b = 12, c = 11

SOLUTION:  

78. a = −9, b = 10, c = −1

SOLUTION:  

79. a = 1, b = 7, c = −3

SOLUTION:  

80. a = 2, b = −4, c = −6

SOLUTION:  

81. a = 3, b = 1, c = 2

SOLUTION:  

This value is not a real number.

eSolutions Manual - Powered by Cognero Page 36

9-4 Solving Quadratic Equations by Completing the Square

Find the value of c that makes each trinomial a perfect square.

1. x2 − 18x + c

SOLUTION:  In this trinomial, b = –18.

So, c must be 81 to make the trinomial a perfect square.

2. x2 + 22x + c

SOLUTION:  In this trinomial, b = 22.

So, c must be 121 to make the trinomial a perfect square.

3. x2 + 9x + c

SOLUTION:  In this trinomial, b = 9.

 

So, c must be  to make the trinomial a perfect square.

4. x2 − 7x + c

SOLUTION:  In this trinomial, b = –7.

 

So, c must be  to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

5. x2 + 4x = 6

SOLUTION:  

The solutions are about –5.2 and 1.2.

6. x2 − 8x = −9

SOLUTION:  

The solutions are about 1.4 and 6.6.

7. 4x2 + 9x − 1 = 0

SOLUTION:  

The solutions are about –2.4 and 0.1.

8. −2x2 + 10x + 22 = 4

SOLUTION:  

The solutions are about –1.4 and 6.4.

9. CCSS MODELING Collin is building a deck on the back of his family’s house. He has enough lumber for the deck to be 144 square feet. The length should be 10 feet more than its width. What should the dimensions of the deck be?

SOLUTION:  Let x = the width of the deck and x + 10 = the length of the deck.

The dimensions of the deck can not be negative. So, x = –5 + 13 or 8. The width of the deck is 8 feet and the length of the deck is 8 + 10 or 18 feet.

Find the value of c that makes each trinomial a perfect square.

10. x2 + 26x + c

SOLUTION:  In this trinomial, b = 26.

So, c must be 169 to make the trinomial a perfect square.

11. x2 − 24x + c

SOLUTION:  In this trinomial, b = –24.

So, c must be 144 to make the trinomial a perfect square.

12. x2 − 19x + c

SOLUTION:  In this trinomial, b = –19.

So, c must be  to make the trinomial a perfect square.

13. x2 + 17x + c

SOLUTION:  In this trinomial, b = 17.

So, c must be  to make the trinomial a perfect square.

14. x2 + 5x + c

SOLUTION:  In this trinomial, b = 5.

So, c must be  to make the trinomial a perfect square.

15. x2 − 13x + c

SOLUTION:  In this trinomial, b = –13.

So, c must be  to make the trinomial a perfect square.

16. x2 − 22x + c

SOLUTION:  In this trinomial, b = –22.

So, c must be 121 to make the trinomial a perfect square.

17. x2 − 15x + c

SOLUTION:  In this trinomial, b = –15.

So, c must be  to make the trinomial a perfect square.

18. x2 + 24x + c

SOLUTION:  In this trinomial, b = 24.

So, c must be 144 to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

19. x2 + 6x − 16 = 0

SOLUTION:  

  The solutions are –8 and 2.

20. x2 − 2x − 14 = 0

SOLUTION:  

  The solutions are about –2.9 and 4.9.

21. x2 − 8x − 1 = 8

SOLUTION:  

  The solutions are –1 and 9.

22. x2 + 3x + 21 = 22

SOLUTION:  

  The solutions are about –3.3 and 0.3.

23. x2 − 11x + 3 = 5

SOLUTION:  

  The solutions are about –0.2 and 11.2.

24. 5x2 − 10x = 23

SOLUTION:  

  The solutions are about –1.4 and 3.4.

25. 2x2 − 2x + 7 = 5

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

26. 3x2 + 12x + 81 = 15

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

27. 4x2 + 6x = 12

SOLUTION:  

  The solutions are about –2.6 and 1.1.

28. 4x2 + 5 = 10x

SOLUTION:  

  The solutions are about 0.7 and 1.8.

29. −2x2 + 10x = −14

SOLUTION:  

  The solutions are about –1.1 and 6.1.

30. −3x2 − 12 = 14x

SOLUTION:  

  The solutions are about –3.5 and –1.1.

31. STOCK The price p in dollars for a particular stock can be modeled by the quadratic equation p = 3.5t − 0.05t2, wh

the number of days after the stock is purchased. When is the stock worth $60?

SOLUTION:  Let p = 60.

  The stock is worth $60 on the 30th and 40th days after purchase.

GEOMETRY  Find the value of x for each figure. Round to the nearest tenth if necessary.

32. area = 45 in2

SOLUTION:  

  The height cannot be negative. So, x ≈ 6.3.

33. area = 110 ft2

SOLUTION:  

  The dimensions of the rectangle cannot be negative. So, x ≈ 5.3.

34. NUMBER THEORY The product of two consecutive even integers is 224. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers are 14 and 16 or –16 and –14.

35. CCSS PRECISION The product of two consecutive negative odd integers is 483. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers must be negative. So, they are –23 and –21.

36. GEOMETRY Find the area of the triangle.

SOLUTION:  

  The dimensions of the triangle cannot be negative. So, x = 18. The base of the triangle is 18 meters and the height is 18 + 6 or 24 meters.

The area of the triangle is 216 square meters.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

37. 0.2x2 − 0.2x − 0.4 = 0

SOLUTION:  

  The solutions are –1 and 2.

38. 0.5x2 = 2x − 0.3

SOLUTION:  

  The solutions are about 0.2 and 3.8.

39. 2x2 −

SOLUTION:  

  The solutions are about 0.2 and 0.9.

40. 

SOLUTION:  

  The solutions are –0.5 and 2.5.

41. 

SOLUTION:  

  The solutions are about –8.2 and 0.2.

42. 

SOLUTION:  

  The solutions are about –5.1 and 0.1.

43. ASTRONOMY The height of an object t seconds after it is dropped is given by the equation ,

where h0 is the initial height and g is the acceleration due to gravity. The acceleration due to gravity near the surface

of Mars is 3.73 m/s2, while on Earth it is 9.8 m/s

2. Suppose an object is dropped from an initial height of 120 meters

above the surface of each planet. a. On which planet would the object reach the ground first? b. How long would it take the object to reach the ground on each planet? Round each answer to the nearest tenth. c. Do the times that it takes the object to reach the ground seem reasonable? Explain your reasoning.

SOLUTION:  a. The object on Earth will reach the ground first because it is falling at a faster rate.   b. Mars:

So, t ≈ 8.0 seconds. Earth:

So, t ≈ 4.9 seconds.   c. Sample answer: Yes; the acceleration due to gravity is much greater on Earth than on Mars, so the time to reach the ground should be much less.

44. Find all values of c that make x2 + cx + 100 a perfect square trinomial.

SOLUTION:  

So, c can be –20 or 20 to make the trinomial a perfect square.

45. Find all values of c that make x2 + cx + 225 a perfect square trinomial.

SOLUTION:  

So, c can be –30 or 30 to make the trinomial a perfect square.

46. PAINTING Before she begins painting a picture, Donna stretches her canvas over a wood frame. The frame has a length of 60 inches and a width of 4 inches. She has enough canvas to cover 480 square inches. Donna decides to increase the dimensions of the frame. If the increase in the length is 10 times the increase in the width, what will the dimensions of the frame be?

SOLUTION:  Let x = the increase in the width and let 10x = the increase in the length. So, x + 4 = the new width and 10x + 60 = the new length.  

  The dimensions cannot be negative, so x = 2. The width is 2 + 4 or 6 inches and the length is 10(2) + 60 or 80 inches.

47. MULTIPLE REPRESENTATIONS In this problem, you will investigate a property of quadratic equations. a. TABULAR  Copy the table shown and complete the second column.

b. ALGEBRAIC Set each trinomial equal to zero, and solve the equation by completing the square. Complete the last column of the table with the number of roots of each equation.

c. VERBAL Compare the number of roots of each equation to the result in the b2 − 4ac column. Is there a

relationship between these values? If so, describe it.

d. ANALYTICAL Predict how many solutions 2x2 − 9x + 15 = 0 will have.  Verify your prediction by solving the 

equation.

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3    

3x2 + 6x + 9    

x2 − 2x + 7    

x2 + 10x + 25    

x2 + 3x − 12    

SOLUTION:  a.

b.

The trinomial x2 − 8x + 16 has 1 root.

 

The trinomial 2x

2 − 11x + 3 has 2 roots.

 

The square root of a negative number has no real roots. So, the trinomial 3x2 + 6x + 9 has 0 roots.

 

The square root of a negative number has no real roots. So, the trinomial x2 − 2x + 7.

 

The trinomial x2 + 10x + 25 has 1 root.

 

The trinomial x2 + 3x − 12 has 2 roots.

 

c. If b2 − 4ac is negative, the equation has no real solutions. If b

2 − 4ac is zero, the equation has one solution. If b2

− 4ac is positive, the equation has 2 solutions. d.

The equation 2x2 − 9x + 15 = 0 has 0 real solutions because b

2 − 4ac is negative.

 

The equation cannot be solved because the square root of a negative number has no real roots.

Trinomial b2 − 4ac Number

of Roots

x2 − 8x + 16 (–8)

2 – 4(1)(16) = 0 1

2x2 − 11x + 3 (–11)

2 – 4(2)(3) = 97  

3x2 + 6x + 9 (6)

2 – 4(3)(9) = −72  

x2 − 2x + 7 (–2)

2 – 4(1)(7) = −24  

x2 + 10x + 25 (10)

2 – 4(1)(25) = 0  

x2 + 3x − 12 (3)

2 – 4(1)(–12) = 57  

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3 97 2

3x2 + 6x + 9 −72 0

x2 − 2x + 7 −24 0

x2 + 10x + 25 0 1

x2 + 3x − 12 57 2

48. CCSS PERSEVERANCE Given y = ax2 + bx + c with a ≠ 0, derive the equation for the axis of symmetry by 

completing the square and rewriting the equation in the form y = a(x − h)2 + k .

SOLUTION:  

Let    and   , then the equation becomes y = (x - h)2 + k .

For an equation in the form y = ax2 + bx + c,  the equation for the axis of symmetry is   .

So, for an equation in the form  y = (x - h)2 + k, the equation for the axis of symmetry becomes x = h.

49. REASONING Determine the number of solutions x2 + bx = c has if . Explain.

SOLUTION:  

None; Sample answer: If you add to each side of the equation and each side of the inequality, you get

. Since the left side of the last equation is a perfect square, it cannot

equal the negative number . So, there are no real solutions.

50. WHICH ONE DOESN’T BELONG? Identify the expression that does not belong with the other three. Explain your reasoning.

 

SOLUTION:  The first 3 trinomials are perfect squares.

The trinomial  is not a perfect square. So, it does not belong.

51. OPEN ENDED Write a quadratic equation for which the only solution is 4.

SOLUTION:  Find a quadratic that has two roots of 4.

Verify that the solution is 4.

So, the quadratic equation x2 − 8x + 16 = 0 has a solution of 4.

52. WRITING IN MATH Compare and contrast the following strategies for solving x2 − 5x − 7 = 0: completing the

square, graphing, and factoring.

SOLUTION:  Sample answer: Because the leading coefficient is 1, solving the equation by completing the square is simpler and yields exact answers for the solutions.   

 

To solve the equation by graphing, use a graphing calculator to graph the related function y = x2 - 5x - 7. Select the

zero option from the 2nd [CALC] menu to determine the roots. 

   

  The roots are given as decimal approximations. So, for this equation the solutions would have to be given as estimations.   There are no factors of –7 that have a sum of –5, so solving by factoring is not possible.

53. The length of a rectangle is 3 times its width. The area of the rectangle is 75 square feet. Find the length of the rectangle in feet. A 25 B 15 C 10 D 5

SOLUTION:  Let x = the width of the rectangle and let 3x = the length of the rectangle.

The length of the rectangle is 3(5) or 15 feet. Choice B is the correct answer.

54. PROBABILITY At a festival, winners of a game draw a token for a prize. There is one token for each prize. The prizes include 9 movie passes, 8 stuffed animals, 5 hats, 10 jump ropes, and 4 glow necklaces. What is the probabilitythat the first person to draw a token will win a movie pass?

F  

G

H

J  

SOLUTION:  There are 9 + 8 + 5 + 10 + 4 or 36 possible outcomes.

P(movie pass) =  or  . Choice J is the correct answer.

55. GRIDDED RESPONSE The population of a town can be modeled by P = 22,000 + 125t, where P represents the population and t represents the number of years from 2000. How many years after 2000 will the population be 26,000?

SOLUTION:  

In 32 years the population of the town will be 26,000.

56. Percy delivers pizzas for Pizza King. He is paid $6 an hour plus $2.50 for each pizza he delivers. Percy earned $280 last week. If he worked a total of 30 hours, how many pizzas did he deliver? A 250 pizzas B 184 pizzas C 40 pizzas D 34 pizzas

SOLUTION:  Let p = the number of pizzas Percy delivered.

Percy delivered 40 pizzas. Choice C is the correct answer.

Describe how the graph of each function is related to the graph of f (x) = x2.

57. g(x) = −12 + x2

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –12, the translation is

down. So, the graph is shifted down 12 units from the parent function. 

58. h(x) = (x + 2)2

SOLUTION:  

The graph of f (x) = (x – c)2 represents a translation left or right from the parent graph. Since c = –2, the translation

is to the left by 2 units. 

 

59. g(x) = 2x2 + 5

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 2 and c = 5. Since 2 > 0 and  > 1, the graph of y = 2x

2 + 5

is the graph of y = x2 vertically stretched and shifted up 5 units.

60. 

SOLUTION:  

The function can be written f (x) = a(x – b)2, where a = and b = 6. Since > 0 and  < 1, the graph of 

is the graph of y = x2 vertically compressed and shifted right 6 units.

61. g(x) = 6 + x2

SOLUTION:  

The function can be written f (x) = ax2 + c, where a =  and c = 6. Since  > 0 and   > 1, the graph of y = 6 +

x2 is the graph of y = x

2 vertically stretched and shifted up 6 units.

62. h(x) = − 1 −   x2

SOLUTION:  

The function can be written f (x) = –ax2 + c, where a =  and c = –1. Since  < 0 and   > 1, the graph of y

= –1 – x2 is the graph of y = x

2 vertically stretched, shifted down 1 unit and reflected across the x-axis.

63. RIDES A popular amusement park ride whisks riders to the top of a 250-foot tower and drops them. A function for

the height of a rider is h = −16t2 + 250, where h is the height and t is the time in seconds. The ride stops the descent

of the rider 40 feet above the ground. Write an equation that models the drop of the rider. How long does it take to fall from 250 feet to 40 feet?

SOLUTION:  

The ride stops the descent of the rider at 40 feet above ground, so h = 40. Then, the equation 40 = −16t2 + 250

models the drop of the rider.

Time cannot be negative. So, it takes about 3.6 seconds to complete the ride.

Simplify. Assume that no denominator is equal to zero.

64. 

SOLUTION:  

65. 

SOLUTION:  

66. 

SOLUTION:  

67. 

SOLUTION:  

68. 

SOLUTION:  

69. b3(m

−3)(b

−6)

SOLUTION:  

Solve each open sentence.

70. |y − 2| > 7

SOLUTION:  

The solution set is {y |y > 9 or y < −5}.

Case 1 y – 2 is positive.

and Case 2 y – 2 is negative.

 

71. |z + 5| < 3

SOLUTION:  

The solution set is {z |−8 < z < −2}.

Case 1 z +5 is positive.

and Case 2 z +5 is negative.

 

72. |2b + 7| ≤ −6

SOLUTION:  

cannot be negative. So,  cannot be less than or equal to –6. The solution set is empty.

73. |3 − 2y | ≥ 8

SOLUTION:  

The solution set is {y |y  ≥ 5.5 or y  ≤ −2.5}.

Case 1 3 – 2y is positive.

and Case 2 3 – 2y is negative.

 

74. |9 − 4m| < −1

SOLUTION:  

cannot be negative. So, cannot be less than or equal to –1. The solution set is empty.

75. |5c − 2| ≤ 13

SOLUTION:  

The solution set is {c|−2.2 ≤ c ≤ 3}.

Case 1 5c – 2 is positive.

and Case 2 5c – 2 is negative.

 

Evaluate  for each set of values. Round to the nearest tenth if necessary.

76. a = 2, b = −5, c = 2

SOLUTION:  

77. a = 1, b = 12, c = 11

SOLUTION:  

78. a = −9, b = 10, c = −1

SOLUTION:  

79. a = 1, b = 7, c = −3

SOLUTION:  

80. a = 2, b = −4, c = −6

SOLUTION:  

81. a = 3, b = 1, c = 2

SOLUTION:  

This value is not a real number.

eSolutions Manual - Powered by Cognero Page 37

9-4 Solving Quadratic Equations by Completing the Square

Find the value of c that makes each trinomial a perfect square.

1. x2 − 18x + c

SOLUTION:  In this trinomial, b = –18.

So, c must be 81 to make the trinomial a perfect square.

2. x2 + 22x + c

SOLUTION:  In this trinomial, b = 22.

So, c must be 121 to make the trinomial a perfect square.

3. x2 + 9x + c

SOLUTION:  In this trinomial, b = 9.

 

So, c must be  to make the trinomial a perfect square.

4. x2 − 7x + c

SOLUTION:  In this trinomial, b = –7.

 

So, c must be  to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

5. x2 + 4x = 6

SOLUTION:  

The solutions are about –5.2 and 1.2.

6. x2 − 8x = −9

SOLUTION:  

The solutions are about 1.4 and 6.6.

7. 4x2 + 9x − 1 = 0

SOLUTION:  

The solutions are about –2.4 and 0.1.

8. −2x2 + 10x + 22 = 4

SOLUTION:  

The solutions are about –1.4 and 6.4.

9. CCSS MODELING Collin is building a deck on the back of his family’s house. He has enough lumber for the deck to be 144 square feet. The length should be 10 feet more than its width. What should the dimensions of the deck be?

SOLUTION:  Let x = the width of the deck and x + 10 = the length of the deck.

The dimensions of the deck can not be negative. So, x = –5 + 13 or 8. The width of the deck is 8 feet and the length of the deck is 8 + 10 or 18 feet.

Find the value of c that makes each trinomial a perfect square.

10. x2 + 26x + c

SOLUTION:  In this trinomial, b = 26.

So, c must be 169 to make the trinomial a perfect square.

11. x2 − 24x + c

SOLUTION:  In this trinomial, b = –24.

So, c must be 144 to make the trinomial a perfect square.

12. x2 − 19x + c

SOLUTION:  In this trinomial, b = –19.

So, c must be  to make the trinomial a perfect square.

13. x2 + 17x + c

SOLUTION:  In this trinomial, b = 17.

So, c must be  to make the trinomial a perfect square.

14. x2 + 5x + c

SOLUTION:  In this trinomial, b = 5.

So, c must be  to make the trinomial a perfect square.

15. x2 − 13x + c

SOLUTION:  In this trinomial, b = –13.

So, c must be  to make the trinomial a perfect square.

16. x2 − 22x + c

SOLUTION:  In this trinomial, b = –22.

So, c must be 121 to make the trinomial a perfect square.

17. x2 − 15x + c

SOLUTION:  In this trinomial, b = –15.

So, c must be  to make the trinomial a perfect square.

18. x2 + 24x + c

SOLUTION:  In this trinomial, b = 24.

So, c must be 144 to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

19. x2 + 6x − 16 = 0

SOLUTION:  

  The solutions are –8 and 2.

20. x2 − 2x − 14 = 0

SOLUTION:  

  The solutions are about –2.9 and 4.9.

21. x2 − 8x − 1 = 8

SOLUTION:  

  The solutions are –1 and 9.

22. x2 + 3x + 21 = 22

SOLUTION:  

  The solutions are about –3.3 and 0.3.

23. x2 − 11x + 3 = 5

SOLUTION:  

  The solutions are about –0.2 and 11.2.

24. 5x2 − 10x = 23

SOLUTION:  

  The solutions are about –1.4 and 3.4.

25. 2x2 − 2x + 7 = 5

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

26. 3x2 + 12x + 81 = 15

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

27. 4x2 + 6x = 12

SOLUTION:  

  The solutions are about –2.6 and 1.1.

28. 4x2 + 5 = 10x

SOLUTION:  

  The solutions are about 0.7 and 1.8.

29. −2x2 + 10x = −14

SOLUTION:  

  The solutions are about –1.1 and 6.1.

30. −3x2 − 12 = 14x

SOLUTION:  

  The solutions are about –3.5 and –1.1.

31. STOCK The price p in dollars for a particular stock can be modeled by the quadratic equation p = 3.5t − 0.05t2, wh

the number of days after the stock is purchased. When is the stock worth $60?

SOLUTION:  Let p = 60.

  The stock is worth $60 on the 30th and 40th days after purchase.

GEOMETRY  Find the value of x for each figure. Round to the nearest tenth if necessary.

32. area = 45 in2

SOLUTION:  

  The height cannot be negative. So, x ≈ 6.3.

33. area = 110 ft2

SOLUTION:  

  The dimensions of the rectangle cannot be negative. So, x ≈ 5.3.

34. NUMBER THEORY The product of two consecutive even integers is 224. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers are 14 and 16 or –16 and –14.

35. CCSS PRECISION The product of two consecutive negative odd integers is 483. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers must be negative. So, they are –23 and –21.

36. GEOMETRY Find the area of the triangle.

SOLUTION:  

  The dimensions of the triangle cannot be negative. So, x = 18. The base of the triangle is 18 meters and the height is 18 + 6 or 24 meters.

The area of the triangle is 216 square meters.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

37. 0.2x2 − 0.2x − 0.4 = 0

SOLUTION:  

  The solutions are –1 and 2.

38. 0.5x2 = 2x − 0.3

SOLUTION:  

  The solutions are about 0.2 and 3.8.

39. 2x2 −

SOLUTION:  

  The solutions are about 0.2 and 0.9.

40. 

SOLUTION:  

  The solutions are –0.5 and 2.5.

41. 

SOLUTION:  

  The solutions are about –8.2 and 0.2.

42. 

SOLUTION:  

  The solutions are about –5.1 and 0.1.

43. ASTRONOMY The height of an object t seconds after it is dropped is given by the equation ,

where h0 is the initial height and g is the acceleration due to gravity. The acceleration due to gravity near the surface

of Mars is 3.73 m/s2, while on Earth it is 9.8 m/s

2. Suppose an object is dropped from an initial height of 120 meters

above the surface of each planet. a. On which planet would the object reach the ground first? b. How long would it take the object to reach the ground on each planet? Round each answer to the nearest tenth. c. Do the times that it takes the object to reach the ground seem reasonable? Explain your reasoning.

SOLUTION:  a. The object on Earth will reach the ground first because it is falling at a faster rate.   b. Mars:

So, t ≈ 8.0 seconds. Earth:

So, t ≈ 4.9 seconds.   c. Sample answer: Yes; the acceleration due to gravity is much greater on Earth than on Mars, so the time to reach the ground should be much less.

44. Find all values of c that make x2 + cx + 100 a perfect square trinomial.

SOLUTION:  

So, c can be –20 or 20 to make the trinomial a perfect square.

45. Find all values of c that make x2 + cx + 225 a perfect square trinomial.

SOLUTION:  

So, c can be –30 or 30 to make the trinomial a perfect square.

46. PAINTING Before she begins painting a picture, Donna stretches her canvas over a wood frame. The frame has a length of 60 inches and a width of 4 inches. She has enough canvas to cover 480 square inches. Donna decides to increase the dimensions of the frame. If the increase in the length is 10 times the increase in the width, what will the dimensions of the frame be?

SOLUTION:  Let x = the increase in the width and let 10x = the increase in the length. So, x + 4 = the new width and 10x + 60 = the new length.  

  The dimensions cannot be negative, so x = 2. The width is 2 + 4 or 6 inches and the length is 10(2) + 60 or 80 inches.

47. MULTIPLE REPRESENTATIONS In this problem, you will investigate a property of quadratic equations. a. TABULAR  Copy the table shown and complete the second column.

b. ALGEBRAIC Set each trinomial equal to zero, and solve the equation by completing the square. Complete the last column of the table with the number of roots of each equation.

c. VERBAL Compare the number of roots of each equation to the result in the b2 − 4ac column. Is there a

relationship between these values? If so, describe it.

d. ANALYTICAL Predict how many solutions 2x2 − 9x + 15 = 0 will have.  Verify your prediction by solving the 

equation.

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3    

3x2 + 6x + 9    

x2 − 2x + 7    

x2 + 10x + 25    

x2 + 3x − 12    

SOLUTION:  a.

b.

The trinomial x2 − 8x + 16 has 1 root.

 

The trinomial 2x

2 − 11x + 3 has 2 roots.

 

The square root of a negative number has no real roots. So, the trinomial 3x2 + 6x + 9 has 0 roots.

 

The square root of a negative number has no real roots. So, the trinomial x2 − 2x + 7.

 

The trinomial x2 + 10x + 25 has 1 root.

 

The trinomial x2 + 3x − 12 has 2 roots.

 

c. If b2 − 4ac is negative, the equation has no real solutions. If b

2 − 4ac is zero, the equation has one solution. If b2

− 4ac is positive, the equation has 2 solutions. d.

The equation 2x2 − 9x + 15 = 0 has 0 real solutions because b

2 − 4ac is negative.

 

The equation cannot be solved because the square root of a negative number has no real roots.

Trinomial b2 − 4ac Number

of Roots

x2 − 8x + 16 (–8)

2 – 4(1)(16) = 0 1

2x2 − 11x + 3 (–11)

2 – 4(2)(3) = 97  

3x2 + 6x + 9 (6)

2 – 4(3)(9) = −72  

x2 − 2x + 7 (–2)

2 – 4(1)(7) = −24  

x2 + 10x + 25 (10)

2 – 4(1)(25) = 0  

x2 + 3x − 12 (3)

2 – 4(1)(–12) = 57  

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3 97 2

3x2 + 6x + 9 −72 0

x2 − 2x + 7 −24 0

x2 + 10x + 25 0 1

x2 + 3x − 12 57 2

48. CCSS PERSEVERANCE Given y = ax2 + bx + c with a ≠ 0, derive the equation for the axis of symmetry by 

completing the square and rewriting the equation in the form y = a(x − h)2 + k .

SOLUTION:  

Let    and   , then the equation becomes y = (x - h)2 + k .

For an equation in the form y = ax2 + bx + c,  the equation for the axis of symmetry is   .

So, for an equation in the form  y = (x - h)2 + k, the equation for the axis of symmetry becomes x = h.

49. REASONING Determine the number of solutions x2 + bx = c has if . Explain.

SOLUTION:  

None; Sample answer: If you add to each side of the equation and each side of the inequality, you get

. Since the left side of the last equation is a perfect square, it cannot

equal the negative number . So, there are no real solutions.

50. WHICH ONE DOESN’T BELONG? Identify the expression that does not belong with the other three. Explain your reasoning.

 

SOLUTION:  The first 3 trinomials are perfect squares.

The trinomial  is not a perfect square. So, it does not belong.

51. OPEN ENDED Write a quadratic equation for which the only solution is 4.

SOLUTION:  Find a quadratic that has two roots of 4.

Verify that the solution is 4.

So, the quadratic equation x2 − 8x + 16 = 0 has a solution of 4.

52. WRITING IN MATH Compare and contrast the following strategies for solving x2 − 5x − 7 = 0: completing the

square, graphing, and factoring.

SOLUTION:  Sample answer: Because the leading coefficient is 1, solving the equation by completing the square is simpler and yields exact answers for the solutions.   

 

To solve the equation by graphing, use a graphing calculator to graph the related function y = x2 - 5x - 7. Select the

zero option from the 2nd [CALC] menu to determine the roots. 

   

  The roots are given as decimal approximations. So, for this equation the solutions would have to be given as estimations.   There are no factors of –7 that have a sum of –5, so solving by factoring is not possible.

53. The length of a rectangle is 3 times its width. The area of the rectangle is 75 square feet. Find the length of the rectangle in feet. A 25 B 15 C 10 D 5

SOLUTION:  Let x = the width of the rectangle and let 3x = the length of the rectangle.

The length of the rectangle is 3(5) or 15 feet. Choice B is the correct answer.

54. PROBABILITY At a festival, winners of a game draw a token for a prize. There is one token for each prize. The prizes include 9 movie passes, 8 stuffed animals, 5 hats, 10 jump ropes, and 4 glow necklaces. What is the probabilitythat the first person to draw a token will win a movie pass?

F  

G

H

J  

SOLUTION:  There are 9 + 8 + 5 + 10 + 4 or 36 possible outcomes.

P(movie pass) =  or  . Choice J is the correct answer.

55. GRIDDED RESPONSE The population of a town can be modeled by P = 22,000 + 125t, where P represents the population and t represents the number of years from 2000. How many years after 2000 will the population be 26,000?

SOLUTION:  

In 32 years the population of the town will be 26,000.

56. Percy delivers pizzas for Pizza King. He is paid $6 an hour plus $2.50 for each pizza he delivers. Percy earned $280 last week. If he worked a total of 30 hours, how many pizzas did he deliver? A 250 pizzas B 184 pizzas C 40 pizzas D 34 pizzas

SOLUTION:  Let p = the number of pizzas Percy delivered.

Percy delivered 40 pizzas. Choice C is the correct answer.

Describe how the graph of each function is related to the graph of f (x) = x2.

57. g(x) = −12 + x2

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –12, the translation is

down. So, the graph is shifted down 12 units from the parent function. 

58. h(x) = (x + 2)2

SOLUTION:  

The graph of f (x) = (x – c)2 represents a translation left or right from the parent graph. Since c = –2, the translation

is to the left by 2 units. 

 

59. g(x) = 2x2 + 5

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 2 and c = 5. Since 2 > 0 and  > 1, the graph of y = 2x

2 + 5

is the graph of y = x2 vertically stretched and shifted up 5 units.

60. 

SOLUTION:  

The function can be written f (x) = a(x – b)2, where a = and b = 6. Since > 0 and  < 1, the graph of 

is the graph of y = x2 vertically compressed and shifted right 6 units.

61. g(x) = 6 + x2

SOLUTION:  

The function can be written f (x) = ax2 + c, where a =  and c = 6. Since  > 0 and   > 1, the graph of y = 6 +

x2 is the graph of y = x

2 vertically stretched and shifted up 6 units.

62. h(x) = − 1 −   x2

SOLUTION:  

The function can be written f (x) = –ax2 + c, where a =  and c = –1. Since  < 0 and   > 1, the graph of y

= –1 – x2 is the graph of y = x

2 vertically stretched, shifted down 1 unit and reflected across the x-axis.

63. RIDES A popular amusement park ride whisks riders to the top of a 250-foot tower and drops them. A function for

the height of a rider is h = −16t2 + 250, where h is the height and t is the time in seconds. The ride stops the descent

of the rider 40 feet above the ground. Write an equation that models the drop of the rider. How long does it take to fall from 250 feet to 40 feet?

SOLUTION:  

The ride stops the descent of the rider at 40 feet above ground, so h = 40. Then, the equation 40 = −16t2 + 250

models the drop of the rider.

Time cannot be negative. So, it takes about 3.6 seconds to complete the ride.

Simplify. Assume that no denominator is equal to zero.

64. 

SOLUTION:  

65. 

SOLUTION:  

66. 

SOLUTION:  

67. 

SOLUTION:  

68. 

SOLUTION:  

69. b3(m

−3)(b

−6)

SOLUTION:  

Solve each open sentence.

70. |y − 2| > 7

SOLUTION:  

The solution set is {y |y > 9 or y < −5}.

Case 1 y – 2 is positive.

and Case 2 y – 2 is negative.

 

71. |z + 5| < 3

SOLUTION:  

The solution set is {z |−8 < z < −2}.

Case 1 z +5 is positive.

and Case 2 z +5 is negative.

 

72. |2b + 7| ≤ −6

SOLUTION:  

cannot be negative. So,  cannot be less than or equal to –6. The solution set is empty.

73. |3 − 2y | ≥ 8

SOLUTION:  

The solution set is {y |y  ≥ 5.5 or y  ≤ −2.5}.

Case 1 3 – 2y is positive.

and Case 2 3 – 2y is negative.

 

74. |9 − 4m| < −1

SOLUTION:  

cannot be negative. So, cannot be less than or equal to –1. The solution set is empty.

75. |5c − 2| ≤ 13

SOLUTION:  

The solution set is {c|−2.2 ≤ c ≤ 3}.

Case 1 5c – 2 is positive.

and Case 2 5c – 2 is negative.

 

Evaluate  for each set of values. Round to the nearest tenth if necessary.

76. a = 2, b = −5, c = 2

SOLUTION:  

77. a = 1, b = 12, c = 11

SOLUTION:  

78. a = −9, b = 10, c = −1

SOLUTION:  

79. a = 1, b = 7, c = −3

SOLUTION:  

80. a = 2, b = −4, c = −6

SOLUTION:  

81. a = 3, b = 1, c = 2

SOLUTION:  

This value is not a real number.

eSolutions Manual - Powered by Cognero Page 38

9-4 Solving Quadratic Equations by Completing the Square

Find the value of c that makes each trinomial a perfect square.

1. x2 − 18x + c

SOLUTION:  In this trinomial, b = –18.

So, c must be 81 to make the trinomial a perfect square.

2. x2 + 22x + c

SOLUTION:  In this trinomial, b = 22.

So, c must be 121 to make the trinomial a perfect square.

3. x2 + 9x + c

SOLUTION:  In this trinomial, b = 9.

 

So, c must be  to make the trinomial a perfect square.

4. x2 − 7x + c

SOLUTION:  In this trinomial, b = –7.

 

So, c must be  to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

5. x2 + 4x = 6

SOLUTION:  

The solutions are about –5.2 and 1.2.

6. x2 − 8x = −9

SOLUTION:  

The solutions are about 1.4 and 6.6.

7. 4x2 + 9x − 1 = 0

SOLUTION:  

The solutions are about –2.4 and 0.1.

8. −2x2 + 10x + 22 = 4

SOLUTION:  

The solutions are about –1.4 and 6.4.

9. CCSS MODELING Collin is building a deck on the back of his family’s house. He has enough lumber for the deck to be 144 square feet. The length should be 10 feet more than its width. What should the dimensions of the deck be?

SOLUTION:  Let x = the width of the deck and x + 10 = the length of the deck.

The dimensions of the deck can not be negative. So, x = –5 + 13 or 8. The width of the deck is 8 feet and the length of the deck is 8 + 10 or 18 feet.

Find the value of c that makes each trinomial a perfect square.

10. x2 + 26x + c

SOLUTION:  In this trinomial, b = 26.

So, c must be 169 to make the trinomial a perfect square.

11. x2 − 24x + c

SOLUTION:  In this trinomial, b = –24.

So, c must be 144 to make the trinomial a perfect square.

12. x2 − 19x + c

SOLUTION:  In this trinomial, b = –19.

So, c must be  to make the trinomial a perfect square.

13. x2 + 17x + c

SOLUTION:  In this trinomial, b = 17.

So, c must be  to make the trinomial a perfect square.

14. x2 + 5x + c

SOLUTION:  In this trinomial, b = 5.

So, c must be  to make the trinomial a perfect square.

15. x2 − 13x + c

SOLUTION:  In this trinomial, b = –13.

So, c must be  to make the trinomial a perfect square.

16. x2 − 22x + c

SOLUTION:  In this trinomial, b = –22.

So, c must be 121 to make the trinomial a perfect square.

17. x2 − 15x + c

SOLUTION:  In this trinomial, b = –15.

So, c must be  to make the trinomial a perfect square.

18. x2 + 24x + c

SOLUTION:  In this trinomial, b = 24.

So, c must be 144 to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

19. x2 + 6x − 16 = 0

SOLUTION:  

  The solutions are –8 and 2.

20. x2 − 2x − 14 = 0

SOLUTION:  

  The solutions are about –2.9 and 4.9.

21. x2 − 8x − 1 = 8

SOLUTION:  

  The solutions are –1 and 9.

22. x2 + 3x + 21 = 22

SOLUTION:  

  The solutions are about –3.3 and 0.3.

23. x2 − 11x + 3 = 5

SOLUTION:  

  The solutions are about –0.2 and 11.2.

24. 5x2 − 10x = 23

SOLUTION:  

  The solutions are about –1.4 and 3.4.

25. 2x2 − 2x + 7 = 5

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

26. 3x2 + 12x + 81 = 15

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

27. 4x2 + 6x = 12

SOLUTION:  

  The solutions are about –2.6 and 1.1.

28. 4x2 + 5 = 10x

SOLUTION:  

  The solutions are about 0.7 and 1.8.

29. −2x2 + 10x = −14

SOLUTION:  

  The solutions are about –1.1 and 6.1.

30. −3x2 − 12 = 14x

SOLUTION:  

  The solutions are about –3.5 and –1.1.

31. STOCK The price p in dollars for a particular stock can be modeled by the quadratic equation p = 3.5t − 0.05t2, wh

the number of days after the stock is purchased. When is the stock worth $60?

SOLUTION:  Let p = 60.

  The stock is worth $60 on the 30th and 40th days after purchase.

GEOMETRY  Find the value of x for each figure. Round to the nearest tenth if necessary.

32. area = 45 in2

SOLUTION:  

  The height cannot be negative. So, x ≈ 6.3.

33. area = 110 ft2

SOLUTION:  

  The dimensions of the rectangle cannot be negative. So, x ≈ 5.3.

34. NUMBER THEORY The product of two consecutive even integers is 224. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers are 14 and 16 or –16 and –14.

35. CCSS PRECISION The product of two consecutive negative odd integers is 483. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers must be negative. So, they are –23 and –21.

36. GEOMETRY Find the area of the triangle.

SOLUTION:  

  The dimensions of the triangle cannot be negative. So, x = 18. The base of the triangle is 18 meters and the height is 18 + 6 or 24 meters.

The area of the triangle is 216 square meters.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

37. 0.2x2 − 0.2x − 0.4 = 0

SOLUTION:  

  The solutions are –1 and 2.

38. 0.5x2 = 2x − 0.3

SOLUTION:  

  The solutions are about 0.2 and 3.8.

39. 2x2 −

SOLUTION:  

  The solutions are about 0.2 and 0.9.

40. 

SOLUTION:  

  The solutions are –0.5 and 2.5.

41. 

SOLUTION:  

  The solutions are about –8.2 and 0.2.

42. 

SOLUTION:  

  The solutions are about –5.1 and 0.1.

43. ASTRONOMY The height of an object t seconds after it is dropped is given by the equation ,

where h0 is the initial height and g is the acceleration due to gravity. The acceleration due to gravity near the surface

of Mars is 3.73 m/s2, while on Earth it is 9.8 m/s

2. Suppose an object is dropped from an initial height of 120 meters

above the surface of each planet. a. On which planet would the object reach the ground first? b. How long would it take the object to reach the ground on each planet? Round each answer to the nearest tenth. c. Do the times that it takes the object to reach the ground seem reasonable? Explain your reasoning.

SOLUTION:  a. The object on Earth will reach the ground first because it is falling at a faster rate.   b. Mars:

So, t ≈ 8.0 seconds. Earth:

So, t ≈ 4.9 seconds.   c. Sample answer: Yes; the acceleration due to gravity is much greater on Earth than on Mars, so the time to reach the ground should be much less.

44. Find all values of c that make x2 + cx + 100 a perfect square trinomial.

SOLUTION:  

So, c can be –20 or 20 to make the trinomial a perfect square.

45. Find all values of c that make x2 + cx + 225 a perfect square trinomial.

SOLUTION:  

So, c can be –30 or 30 to make the trinomial a perfect square.

46. PAINTING Before she begins painting a picture, Donna stretches her canvas over a wood frame. The frame has a length of 60 inches and a width of 4 inches. She has enough canvas to cover 480 square inches. Donna decides to increase the dimensions of the frame. If the increase in the length is 10 times the increase in the width, what will the dimensions of the frame be?

SOLUTION:  Let x = the increase in the width and let 10x = the increase in the length. So, x + 4 = the new width and 10x + 60 = the new length.  

  The dimensions cannot be negative, so x = 2. The width is 2 + 4 or 6 inches and the length is 10(2) + 60 or 80 inches.

47. MULTIPLE REPRESENTATIONS In this problem, you will investigate a property of quadratic equations. a. TABULAR  Copy the table shown and complete the second column.

b. ALGEBRAIC Set each trinomial equal to zero, and solve the equation by completing the square. Complete the last column of the table with the number of roots of each equation.

c. VERBAL Compare the number of roots of each equation to the result in the b2 − 4ac column. Is there a

relationship between these values? If so, describe it.

d. ANALYTICAL Predict how many solutions 2x2 − 9x + 15 = 0 will have.  Verify your prediction by solving the 

equation.

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3    

3x2 + 6x + 9    

x2 − 2x + 7    

x2 + 10x + 25    

x2 + 3x − 12    

SOLUTION:  a.

b.

The trinomial x2 − 8x + 16 has 1 root.

 

The trinomial 2x

2 − 11x + 3 has 2 roots.

 

The square root of a negative number has no real roots. So, the trinomial 3x2 + 6x + 9 has 0 roots.

 

The square root of a negative number has no real roots. So, the trinomial x2 − 2x + 7.

 

The trinomial x2 + 10x + 25 has 1 root.

 

The trinomial x2 + 3x − 12 has 2 roots.

 

c. If b2 − 4ac is negative, the equation has no real solutions. If b

2 − 4ac is zero, the equation has one solution. If b2

− 4ac is positive, the equation has 2 solutions. d.

The equation 2x2 − 9x + 15 = 0 has 0 real solutions because b

2 − 4ac is negative.

 

The equation cannot be solved because the square root of a negative number has no real roots.

Trinomial b2 − 4ac Number

of Roots

x2 − 8x + 16 (–8)

2 – 4(1)(16) = 0 1

2x2 − 11x + 3 (–11)

2 – 4(2)(3) = 97  

3x2 + 6x + 9 (6)

2 – 4(3)(9) = −72  

x2 − 2x + 7 (–2)

2 – 4(1)(7) = −24  

x2 + 10x + 25 (10)

2 – 4(1)(25) = 0  

x2 + 3x − 12 (3)

2 – 4(1)(–12) = 57  

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3 97 2

3x2 + 6x + 9 −72 0

x2 − 2x + 7 −24 0

x2 + 10x + 25 0 1

x2 + 3x − 12 57 2

48. CCSS PERSEVERANCE Given y = ax2 + bx + c with a ≠ 0, derive the equation for the axis of symmetry by 

completing the square and rewriting the equation in the form y = a(x − h)2 + k .

SOLUTION:  

Let    and   , then the equation becomes y = (x - h)2 + k .

For an equation in the form y = ax2 + bx + c,  the equation for the axis of symmetry is   .

So, for an equation in the form  y = (x - h)2 + k, the equation for the axis of symmetry becomes x = h.

49. REASONING Determine the number of solutions x2 + bx = c has if . Explain.

SOLUTION:  

None; Sample answer: If you add to each side of the equation and each side of the inequality, you get

. Since the left side of the last equation is a perfect square, it cannot

equal the negative number . So, there are no real solutions.

50. WHICH ONE DOESN’T BELONG? Identify the expression that does not belong with the other three. Explain your reasoning.

 

SOLUTION:  The first 3 trinomials are perfect squares.

The trinomial  is not a perfect square. So, it does not belong.

51. OPEN ENDED Write a quadratic equation for which the only solution is 4.

SOLUTION:  Find a quadratic that has two roots of 4.

Verify that the solution is 4.

So, the quadratic equation x2 − 8x + 16 = 0 has a solution of 4.

52. WRITING IN MATH Compare and contrast the following strategies for solving x2 − 5x − 7 = 0: completing the

square, graphing, and factoring.

SOLUTION:  Sample answer: Because the leading coefficient is 1, solving the equation by completing the square is simpler and yields exact answers for the solutions.   

 

To solve the equation by graphing, use a graphing calculator to graph the related function y = x2 - 5x - 7. Select the

zero option from the 2nd [CALC] menu to determine the roots. 

   

  The roots are given as decimal approximations. So, for this equation the solutions would have to be given as estimations.   There are no factors of –7 that have a sum of –5, so solving by factoring is not possible.

53. The length of a rectangle is 3 times its width. The area of the rectangle is 75 square feet. Find the length of the rectangle in feet. A 25 B 15 C 10 D 5

SOLUTION:  Let x = the width of the rectangle and let 3x = the length of the rectangle.

The length of the rectangle is 3(5) or 15 feet. Choice B is the correct answer.

54. PROBABILITY At a festival, winners of a game draw a token for a prize. There is one token for each prize. The prizes include 9 movie passes, 8 stuffed animals, 5 hats, 10 jump ropes, and 4 glow necklaces. What is the probabilitythat the first person to draw a token will win a movie pass?

F  

G

H

J  

SOLUTION:  There are 9 + 8 + 5 + 10 + 4 or 36 possible outcomes.

P(movie pass) =  or  . Choice J is the correct answer.

55. GRIDDED RESPONSE The population of a town can be modeled by P = 22,000 + 125t, where P represents the population and t represents the number of years from 2000. How many years after 2000 will the population be 26,000?

SOLUTION:  

In 32 years the population of the town will be 26,000.

56. Percy delivers pizzas for Pizza King. He is paid $6 an hour plus $2.50 for each pizza he delivers. Percy earned $280 last week. If he worked a total of 30 hours, how many pizzas did he deliver? A 250 pizzas B 184 pizzas C 40 pizzas D 34 pizzas

SOLUTION:  Let p = the number of pizzas Percy delivered.

Percy delivered 40 pizzas. Choice C is the correct answer.

Describe how the graph of each function is related to the graph of f (x) = x2.

57. g(x) = −12 + x2

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –12, the translation is

down. So, the graph is shifted down 12 units from the parent function. 

58. h(x) = (x + 2)2

SOLUTION:  

The graph of f (x) = (x – c)2 represents a translation left or right from the parent graph. Since c = –2, the translation

is to the left by 2 units. 

 

59. g(x) = 2x2 + 5

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 2 and c = 5. Since 2 > 0 and  > 1, the graph of y = 2x

2 + 5

is the graph of y = x2 vertically stretched and shifted up 5 units.

60. 

SOLUTION:  

The function can be written f (x) = a(x – b)2, where a = and b = 6. Since > 0 and  < 1, the graph of 

is the graph of y = x2 vertically compressed and shifted right 6 units.

61. g(x) = 6 + x2

SOLUTION:  

The function can be written f (x) = ax2 + c, where a =  and c = 6. Since  > 0 and   > 1, the graph of y = 6 +

x2 is the graph of y = x

2 vertically stretched and shifted up 6 units.

62. h(x) = − 1 −   x2

SOLUTION:  

The function can be written f (x) = –ax2 + c, where a =  and c = –1. Since  < 0 and   > 1, the graph of y

= –1 – x2 is the graph of y = x

2 vertically stretched, shifted down 1 unit and reflected across the x-axis.

63. RIDES A popular amusement park ride whisks riders to the top of a 250-foot tower and drops them. A function for

the height of a rider is h = −16t2 + 250, where h is the height and t is the time in seconds. The ride stops the descent

of the rider 40 feet above the ground. Write an equation that models the drop of the rider. How long does it take to fall from 250 feet to 40 feet?

SOLUTION:  

The ride stops the descent of the rider at 40 feet above ground, so h = 40. Then, the equation 40 = −16t2 + 250

models the drop of the rider.

Time cannot be negative. So, it takes about 3.6 seconds to complete the ride.

Simplify. Assume that no denominator is equal to zero.

64. 

SOLUTION:  

65. 

SOLUTION:  

66. 

SOLUTION:  

67. 

SOLUTION:  

68. 

SOLUTION:  

69. b3(m

−3)(b

−6)

SOLUTION:  

Solve each open sentence.

70. |y − 2| > 7

SOLUTION:  

The solution set is {y |y > 9 or y < −5}.

Case 1 y – 2 is positive.

and Case 2 y – 2 is negative.

 

71. |z + 5| < 3

SOLUTION:  

The solution set is {z |−8 < z < −2}.

Case 1 z +5 is positive.

and Case 2 z +5 is negative.

 

72. |2b + 7| ≤ −6

SOLUTION:  

cannot be negative. So,  cannot be less than or equal to –6. The solution set is empty.

73. |3 − 2y | ≥ 8

SOLUTION:  

The solution set is {y |y  ≥ 5.5 or y  ≤ −2.5}.

Case 1 3 – 2y is positive.

and Case 2 3 – 2y is negative.

 

74. |9 − 4m| < −1

SOLUTION:  

cannot be negative. So, cannot be less than or equal to –1. The solution set is empty.

75. |5c − 2| ≤ 13

SOLUTION:  

The solution set is {c|−2.2 ≤ c ≤ 3}.

Case 1 5c – 2 is positive.

and Case 2 5c – 2 is negative.

 

Evaluate  for each set of values. Round to the nearest tenth if necessary.

76. a = 2, b = −5, c = 2

SOLUTION:  

77. a = 1, b = 12, c = 11

SOLUTION:  

78. a = −9, b = 10, c = −1

SOLUTION:  

79. a = 1, b = 7, c = −3

SOLUTION:  

80. a = 2, b = −4, c = −6

SOLUTION:  

81. a = 3, b = 1, c = 2

SOLUTION:  

This value is not a real number.

eSolutions Manual - Powered by Cognero Page 39

9-4 Solving Quadratic Equations by Completing the Square

Find the value of c that makes each trinomial a perfect square.

1. x2 − 18x + c

SOLUTION:  In this trinomial, b = –18.

So, c must be 81 to make the trinomial a perfect square.

2. x2 + 22x + c

SOLUTION:  In this trinomial, b = 22.

So, c must be 121 to make the trinomial a perfect square.

3. x2 + 9x + c

SOLUTION:  In this trinomial, b = 9.

 

So, c must be  to make the trinomial a perfect square.

4. x2 − 7x + c

SOLUTION:  In this trinomial, b = –7.

 

So, c must be  to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

5. x2 + 4x = 6

SOLUTION:  

The solutions are about –5.2 and 1.2.

6. x2 − 8x = −9

SOLUTION:  

The solutions are about 1.4 and 6.6.

7. 4x2 + 9x − 1 = 0

SOLUTION:  

The solutions are about –2.4 and 0.1.

8. −2x2 + 10x + 22 = 4

SOLUTION:  

The solutions are about –1.4 and 6.4.

9. CCSS MODELING Collin is building a deck on the back of his family’s house. He has enough lumber for the deck to be 144 square feet. The length should be 10 feet more than its width. What should the dimensions of the deck be?

SOLUTION:  Let x = the width of the deck and x + 10 = the length of the deck.

The dimensions of the deck can not be negative. So, x = –5 + 13 or 8. The width of the deck is 8 feet and the length of the deck is 8 + 10 or 18 feet.

Find the value of c that makes each trinomial a perfect square.

10. x2 + 26x + c

SOLUTION:  In this trinomial, b = 26.

So, c must be 169 to make the trinomial a perfect square.

11. x2 − 24x + c

SOLUTION:  In this trinomial, b = –24.

So, c must be 144 to make the trinomial a perfect square.

12. x2 − 19x + c

SOLUTION:  In this trinomial, b = –19.

So, c must be  to make the trinomial a perfect square.

13. x2 + 17x + c

SOLUTION:  In this trinomial, b = 17.

So, c must be  to make the trinomial a perfect square.

14. x2 + 5x + c

SOLUTION:  In this trinomial, b = 5.

So, c must be  to make the trinomial a perfect square.

15. x2 − 13x + c

SOLUTION:  In this trinomial, b = –13.

So, c must be  to make the trinomial a perfect square.

16. x2 − 22x + c

SOLUTION:  In this trinomial, b = –22.

So, c must be 121 to make the trinomial a perfect square.

17. x2 − 15x + c

SOLUTION:  In this trinomial, b = –15.

So, c must be  to make the trinomial a perfect square.

18. x2 + 24x + c

SOLUTION:  In this trinomial, b = 24.

So, c must be 144 to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

19. x2 + 6x − 16 = 0

SOLUTION:  

  The solutions are –8 and 2.

20. x2 − 2x − 14 = 0

SOLUTION:  

  The solutions are about –2.9 and 4.9.

21. x2 − 8x − 1 = 8

SOLUTION:  

  The solutions are –1 and 9.

22. x2 + 3x + 21 = 22

SOLUTION:  

  The solutions are about –3.3 and 0.3.

23. x2 − 11x + 3 = 5

SOLUTION:  

  The solutions are about –0.2 and 11.2.

24. 5x2 − 10x = 23

SOLUTION:  

  The solutions are about –1.4 and 3.4.

25. 2x2 − 2x + 7 = 5

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

26. 3x2 + 12x + 81 = 15

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

27. 4x2 + 6x = 12

SOLUTION:  

  The solutions are about –2.6 and 1.1.

28. 4x2 + 5 = 10x

SOLUTION:  

  The solutions are about 0.7 and 1.8.

29. −2x2 + 10x = −14

SOLUTION:  

  The solutions are about –1.1 and 6.1.

30. −3x2 − 12 = 14x

SOLUTION:  

  The solutions are about –3.5 and –1.1.

31. STOCK The price p in dollars for a particular stock can be modeled by the quadratic equation p = 3.5t − 0.05t2, wh

the number of days after the stock is purchased. When is the stock worth $60?

SOLUTION:  Let p = 60.

  The stock is worth $60 on the 30th and 40th days after purchase.

GEOMETRY  Find the value of x for each figure. Round to the nearest tenth if necessary.

32. area = 45 in2

SOLUTION:  

  The height cannot be negative. So, x ≈ 6.3.

33. area = 110 ft2

SOLUTION:  

  The dimensions of the rectangle cannot be negative. So, x ≈ 5.3.

34. NUMBER THEORY The product of two consecutive even integers is 224. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers are 14 and 16 or –16 and –14.

35. CCSS PRECISION The product of two consecutive negative odd integers is 483. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers must be negative. So, they are –23 and –21.

36. GEOMETRY Find the area of the triangle.

SOLUTION:  

  The dimensions of the triangle cannot be negative. So, x = 18. The base of the triangle is 18 meters and the height is 18 + 6 or 24 meters.

The area of the triangle is 216 square meters.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

37. 0.2x2 − 0.2x − 0.4 = 0

SOLUTION:  

  The solutions are –1 and 2.

38. 0.5x2 = 2x − 0.3

SOLUTION:  

  The solutions are about 0.2 and 3.8.

39. 2x2 −

SOLUTION:  

  The solutions are about 0.2 and 0.9.

40. 

SOLUTION:  

  The solutions are –0.5 and 2.5.

41. 

SOLUTION:  

  The solutions are about –8.2 and 0.2.

42. 

SOLUTION:  

  The solutions are about –5.1 and 0.1.

43. ASTRONOMY The height of an object t seconds after it is dropped is given by the equation ,

where h0 is the initial height and g is the acceleration due to gravity. The acceleration due to gravity near the surface

of Mars is 3.73 m/s2, while on Earth it is 9.8 m/s

2. Suppose an object is dropped from an initial height of 120 meters

above the surface of each planet. a. On which planet would the object reach the ground first? b. How long would it take the object to reach the ground on each planet? Round each answer to the nearest tenth. c. Do the times that it takes the object to reach the ground seem reasonable? Explain your reasoning.

SOLUTION:  a. The object on Earth will reach the ground first because it is falling at a faster rate.   b. Mars:

So, t ≈ 8.0 seconds. Earth:

So, t ≈ 4.9 seconds.   c. Sample answer: Yes; the acceleration due to gravity is much greater on Earth than on Mars, so the time to reach the ground should be much less.

44. Find all values of c that make x2 + cx + 100 a perfect square trinomial.

SOLUTION:  

So, c can be –20 or 20 to make the trinomial a perfect square.

45. Find all values of c that make x2 + cx + 225 a perfect square trinomial.

SOLUTION:  

So, c can be –30 or 30 to make the trinomial a perfect square.

46. PAINTING Before she begins painting a picture, Donna stretches her canvas over a wood frame. The frame has a length of 60 inches and a width of 4 inches. She has enough canvas to cover 480 square inches. Donna decides to increase the dimensions of the frame. If the increase in the length is 10 times the increase in the width, what will the dimensions of the frame be?

SOLUTION:  Let x = the increase in the width and let 10x = the increase in the length. So, x + 4 = the new width and 10x + 60 = the new length.  

  The dimensions cannot be negative, so x = 2. The width is 2 + 4 or 6 inches and the length is 10(2) + 60 or 80 inches.

47. MULTIPLE REPRESENTATIONS In this problem, you will investigate a property of quadratic equations. a. TABULAR  Copy the table shown and complete the second column.

b. ALGEBRAIC Set each trinomial equal to zero, and solve the equation by completing the square. Complete the last column of the table with the number of roots of each equation.

c. VERBAL Compare the number of roots of each equation to the result in the b2 − 4ac column. Is there a

relationship between these values? If so, describe it.

d. ANALYTICAL Predict how many solutions 2x2 − 9x + 15 = 0 will have.  Verify your prediction by solving the 

equation.

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3    

3x2 + 6x + 9    

x2 − 2x + 7    

x2 + 10x + 25    

x2 + 3x − 12    

SOLUTION:  a.

b.

The trinomial x2 − 8x + 16 has 1 root.

 

The trinomial 2x

2 − 11x + 3 has 2 roots.

 

The square root of a negative number has no real roots. So, the trinomial 3x2 + 6x + 9 has 0 roots.

 

The square root of a negative number has no real roots. So, the trinomial x2 − 2x + 7.

 

The trinomial x2 + 10x + 25 has 1 root.

 

The trinomial x2 + 3x − 12 has 2 roots.

 

c. If b2 − 4ac is negative, the equation has no real solutions. If b

2 − 4ac is zero, the equation has one solution. If b2

− 4ac is positive, the equation has 2 solutions. d.

The equation 2x2 − 9x + 15 = 0 has 0 real solutions because b

2 − 4ac is negative.

 

The equation cannot be solved because the square root of a negative number has no real roots.

Trinomial b2 − 4ac Number

of Roots

x2 − 8x + 16 (–8)

2 – 4(1)(16) = 0 1

2x2 − 11x + 3 (–11)

2 – 4(2)(3) = 97  

3x2 + 6x + 9 (6)

2 – 4(3)(9) = −72  

x2 − 2x + 7 (–2)

2 – 4(1)(7) = −24  

x2 + 10x + 25 (10)

2 – 4(1)(25) = 0  

x2 + 3x − 12 (3)

2 – 4(1)(–12) = 57  

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3 97 2

3x2 + 6x + 9 −72 0

x2 − 2x + 7 −24 0

x2 + 10x + 25 0 1

x2 + 3x − 12 57 2

48. CCSS PERSEVERANCE Given y = ax2 + bx + c with a ≠ 0, derive the equation for the axis of symmetry by 

completing the square and rewriting the equation in the form y = a(x − h)2 + k .

SOLUTION:  

Let    and   , then the equation becomes y = (x - h)2 + k .

For an equation in the form y = ax2 + bx + c,  the equation for the axis of symmetry is   .

So, for an equation in the form  y = (x - h)2 + k, the equation for the axis of symmetry becomes x = h.

49. REASONING Determine the number of solutions x2 + bx = c has if . Explain.

SOLUTION:  

None; Sample answer: If you add to each side of the equation and each side of the inequality, you get

. Since the left side of the last equation is a perfect square, it cannot

equal the negative number . So, there are no real solutions.

50. WHICH ONE DOESN’T BELONG? Identify the expression that does not belong with the other three. Explain your reasoning.

 

SOLUTION:  The first 3 trinomials are perfect squares.

The trinomial  is not a perfect square. So, it does not belong.

51. OPEN ENDED Write a quadratic equation for which the only solution is 4.

SOLUTION:  Find a quadratic that has two roots of 4.

Verify that the solution is 4.

So, the quadratic equation x2 − 8x + 16 = 0 has a solution of 4.

52. WRITING IN MATH Compare and contrast the following strategies for solving x2 − 5x − 7 = 0: completing the

square, graphing, and factoring.

SOLUTION:  Sample answer: Because the leading coefficient is 1, solving the equation by completing the square is simpler and yields exact answers for the solutions.   

 

To solve the equation by graphing, use a graphing calculator to graph the related function y = x2 - 5x - 7. Select the

zero option from the 2nd [CALC] menu to determine the roots. 

   

  The roots are given as decimal approximations. So, for this equation the solutions would have to be given as estimations.   There are no factors of –7 that have a sum of –5, so solving by factoring is not possible.

53. The length of a rectangle is 3 times its width. The area of the rectangle is 75 square feet. Find the length of the rectangle in feet. A 25 B 15 C 10 D 5

SOLUTION:  Let x = the width of the rectangle and let 3x = the length of the rectangle.

The length of the rectangle is 3(5) or 15 feet. Choice B is the correct answer.

54. PROBABILITY At a festival, winners of a game draw a token for a prize. There is one token for each prize. The prizes include 9 movie passes, 8 stuffed animals, 5 hats, 10 jump ropes, and 4 glow necklaces. What is the probabilitythat the first person to draw a token will win a movie pass?

F  

G

H

J  

SOLUTION:  There are 9 + 8 + 5 + 10 + 4 or 36 possible outcomes.

P(movie pass) =  or  . Choice J is the correct answer.

55. GRIDDED RESPONSE The population of a town can be modeled by P = 22,000 + 125t, where P represents the population and t represents the number of years from 2000. How many years after 2000 will the population be 26,000?

SOLUTION:  

In 32 years the population of the town will be 26,000.

56. Percy delivers pizzas for Pizza King. He is paid $6 an hour plus $2.50 for each pizza he delivers. Percy earned $280 last week. If he worked a total of 30 hours, how many pizzas did he deliver? A 250 pizzas B 184 pizzas C 40 pizzas D 34 pizzas

SOLUTION:  Let p = the number of pizzas Percy delivered.

Percy delivered 40 pizzas. Choice C is the correct answer.

Describe how the graph of each function is related to the graph of f (x) = x2.

57. g(x) = −12 + x2

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –12, the translation is

down. So, the graph is shifted down 12 units from the parent function. 

58. h(x) = (x + 2)2

SOLUTION:  

The graph of f (x) = (x – c)2 represents a translation left or right from the parent graph. Since c = –2, the translation

is to the left by 2 units. 

 

59. g(x) = 2x2 + 5

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 2 and c = 5. Since 2 > 0 and  > 1, the graph of y = 2x

2 + 5

is the graph of y = x2 vertically stretched and shifted up 5 units.

60. 

SOLUTION:  

The function can be written f (x) = a(x – b)2, where a = and b = 6. Since > 0 and  < 1, the graph of 

is the graph of y = x2 vertically compressed and shifted right 6 units.

61. g(x) = 6 + x2

SOLUTION:  

The function can be written f (x) = ax2 + c, where a =  and c = 6. Since  > 0 and   > 1, the graph of y = 6 +

x2 is the graph of y = x

2 vertically stretched and shifted up 6 units.

62. h(x) = − 1 −   x2

SOLUTION:  

The function can be written f (x) = –ax2 + c, where a =  and c = –1. Since  < 0 and   > 1, the graph of y

= –1 – x2 is the graph of y = x

2 vertically stretched, shifted down 1 unit and reflected across the x-axis.

63. RIDES A popular amusement park ride whisks riders to the top of a 250-foot tower and drops them. A function for

the height of a rider is h = −16t2 + 250, where h is the height and t is the time in seconds. The ride stops the descent

of the rider 40 feet above the ground. Write an equation that models the drop of the rider. How long does it take to fall from 250 feet to 40 feet?

SOLUTION:  

The ride stops the descent of the rider at 40 feet above ground, so h = 40. Then, the equation 40 = −16t2 + 250

models the drop of the rider.

Time cannot be negative. So, it takes about 3.6 seconds to complete the ride.

Simplify. Assume that no denominator is equal to zero.

64. 

SOLUTION:  

65. 

SOLUTION:  

66. 

SOLUTION:  

67. 

SOLUTION:  

68. 

SOLUTION:  

69. b3(m

−3)(b

−6)

SOLUTION:  

Solve each open sentence.

70. |y − 2| > 7

SOLUTION:  

The solution set is {y |y > 9 or y < −5}.

Case 1 y – 2 is positive.

and Case 2 y – 2 is negative.

 

71. |z + 5| < 3

SOLUTION:  

The solution set is {z |−8 < z < −2}.

Case 1 z +5 is positive.

and Case 2 z +5 is negative.

 

72. |2b + 7| ≤ −6

SOLUTION:  

cannot be negative. So,  cannot be less than or equal to –6. The solution set is empty.

73. |3 − 2y | ≥ 8

SOLUTION:  

The solution set is {y |y  ≥ 5.5 or y  ≤ −2.5}.

Case 1 3 – 2y is positive.

and Case 2 3 – 2y is negative.

 

74. |9 − 4m| < −1

SOLUTION:  

cannot be negative. So, cannot be less than or equal to –1. The solution set is empty.

75. |5c − 2| ≤ 13

SOLUTION:  

The solution set is {c|−2.2 ≤ c ≤ 3}.

Case 1 5c – 2 is positive.

and Case 2 5c – 2 is negative.

 

Evaluate  for each set of values. Round to the nearest tenth if necessary.

76. a = 2, b = −5, c = 2

SOLUTION:  

77. a = 1, b = 12, c = 11

SOLUTION:  

78. a = −9, b = 10, c = −1

SOLUTION:  

79. a = 1, b = 7, c = −3

SOLUTION:  

80. a = 2, b = −4, c = −6

SOLUTION:  

81. a = 3, b = 1, c = 2

SOLUTION:  

This value is not a real number.

eSolutions Manual - Powered by Cognero Page 40

9-4 Solving Quadratic Equations by Completing the Square

Find the value of c that makes each trinomial a perfect square.

1. x2 − 18x + c

SOLUTION:  In this trinomial, b = –18.

So, c must be 81 to make the trinomial a perfect square.

2. x2 + 22x + c

SOLUTION:  In this trinomial, b = 22.

So, c must be 121 to make the trinomial a perfect square.

3. x2 + 9x + c

SOLUTION:  In this trinomial, b = 9.

 

So, c must be  to make the trinomial a perfect square.

4. x2 − 7x + c

SOLUTION:  In this trinomial, b = –7.

 

So, c must be  to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

5. x2 + 4x = 6

SOLUTION:  

The solutions are about –5.2 and 1.2.

6. x2 − 8x = −9

SOLUTION:  

The solutions are about 1.4 and 6.6.

7. 4x2 + 9x − 1 = 0

SOLUTION:  

The solutions are about –2.4 and 0.1.

8. −2x2 + 10x + 22 = 4

SOLUTION:  

The solutions are about –1.4 and 6.4.

9. CCSS MODELING Collin is building a deck on the back of his family’s house. He has enough lumber for the deck to be 144 square feet. The length should be 10 feet more than its width. What should the dimensions of the deck be?

SOLUTION:  Let x = the width of the deck and x + 10 = the length of the deck.

The dimensions of the deck can not be negative. So, x = –5 + 13 or 8. The width of the deck is 8 feet and the length of the deck is 8 + 10 or 18 feet.

Find the value of c that makes each trinomial a perfect square.

10. x2 + 26x + c

SOLUTION:  In this trinomial, b = 26.

So, c must be 169 to make the trinomial a perfect square.

11. x2 − 24x + c

SOLUTION:  In this trinomial, b = –24.

So, c must be 144 to make the trinomial a perfect square.

12. x2 − 19x + c

SOLUTION:  In this trinomial, b = –19.

So, c must be  to make the trinomial a perfect square.

13. x2 + 17x + c

SOLUTION:  In this trinomial, b = 17.

So, c must be  to make the trinomial a perfect square.

14. x2 + 5x + c

SOLUTION:  In this trinomial, b = 5.

So, c must be  to make the trinomial a perfect square.

15. x2 − 13x + c

SOLUTION:  In this trinomial, b = –13.

So, c must be  to make the trinomial a perfect square.

16. x2 − 22x + c

SOLUTION:  In this trinomial, b = –22.

So, c must be 121 to make the trinomial a perfect square.

17. x2 − 15x + c

SOLUTION:  In this trinomial, b = –15.

So, c must be  to make the trinomial a perfect square.

18. x2 + 24x + c

SOLUTION:  In this trinomial, b = 24.

So, c must be 144 to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

19. x2 + 6x − 16 = 0

SOLUTION:  

  The solutions are –8 and 2.

20. x2 − 2x − 14 = 0

SOLUTION:  

  The solutions are about –2.9 and 4.9.

21. x2 − 8x − 1 = 8

SOLUTION:  

  The solutions are –1 and 9.

22. x2 + 3x + 21 = 22

SOLUTION:  

  The solutions are about –3.3 and 0.3.

23. x2 − 11x + 3 = 5

SOLUTION:  

  The solutions are about –0.2 and 11.2.

24. 5x2 − 10x = 23

SOLUTION:  

  The solutions are about –1.4 and 3.4.

25. 2x2 − 2x + 7 = 5

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

26. 3x2 + 12x + 81 = 15

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

27. 4x2 + 6x = 12

SOLUTION:  

  The solutions are about –2.6 and 1.1.

28. 4x2 + 5 = 10x

SOLUTION:  

  The solutions are about 0.7 and 1.8.

29. −2x2 + 10x = −14

SOLUTION:  

  The solutions are about –1.1 and 6.1.

30. −3x2 − 12 = 14x

SOLUTION:  

  The solutions are about –3.5 and –1.1.

31. STOCK The price p in dollars for a particular stock can be modeled by the quadratic equation p = 3.5t − 0.05t2, wh

the number of days after the stock is purchased. When is the stock worth $60?

SOLUTION:  Let p = 60.

  The stock is worth $60 on the 30th and 40th days after purchase.

GEOMETRY  Find the value of x for each figure. Round to the nearest tenth if necessary.

32. area = 45 in2

SOLUTION:  

  The height cannot be negative. So, x ≈ 6.3.

33. area = 110 ft2

SOLUTION:  

  The dimensions of the rectangle cannot be negative. So, x ≈ 5.3.

34. NUMBER THEORY The product of two consecutive even integers is 224. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers are 14 and 16 or –16 and –14.

35. CCSS PRECISION The product of two consecutive negative odd integers is 483. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers must be negative. So, they are –23 and –21.

36. GEOMETRY Find the area of the triangle.

SOLUTION:  

  The dimensions of the triangle cannot be negative. So, x = 18. The base of the triangle is 18 meters and the height is 18 + 6 or 24 meters.

The area of the triangle is 216 square meters.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

37. 0.2x2 − 0.2x − 0.4 = 0

SOLUTION:  

  The solutions are –1 and 2.

38. 0.5x2 = 2x − 0.3

SOLUTION:  

  The solutions are about 0.2 and 3.8.

39. 2x2 −

SOLUTION:  

  The solutions are about 0.2 and 0.9.

40. 

SOLUTION:  

  The solutions are –0.5 and 2.5.

41. 

SOLUTION:  

  The solutions are about –8.2 and 0.2.

42. 

SOLUTION:  

  The solutions are about –5.1 and 0.1.

43. ASTRONOMY The height of an object t seconds after it is dropped is given by the equation ,

where h0 is the initial height and g is the acceleration due to gravity. The acceleration due to gravity near the surface

of Mars is 3.73 m/s2, while on Earth it is 9.8 m/s

2. Suppose an object is dropped from an initial height of 120 meters

above the surface of each planet. a. On which planet would the object reach the ground first? b. How long would it take the object to reach the ground on each planet? Round each answer to the nearest tenth. c. Do the times that it takes the object to reach the ground seem reasonable? Explain your reasoning.

SOLUTION:  a. The object on Earth will reach the ground first because it is falling at a faster rate.   b. Mars:

So, t ≈ 8.0 seconds. Earth:

So, t ≈ 4.9 seconds.   c. Sample answer: Yes; the acceleration due to gravity is much greater on Earth than on Mars, so the time to reach the ground should be much less.

44. Find all values of c that make x2 + cx + 100 a perfect square trinomial.

SOLUTION:  

So, c can be –20 or 20 to make the trinomial a perfect square.

45. Find all values of c that make x2 + cx + 225 a perfect square trinomial.

SOLUTION:  

So, c can be –30 or 30 to make the trinomial a perfect square.

46. PAINTING Before she begins painting a picture, Donna stretches her canvas over a wood frame. The frame has a length of 60 inches and a width of 4 inches. She has enough canvas to cover 480 square inches. Donna decides to increase the dimensions of the frame. If the increase in the length is 10 times the increase in the width, what will the dimensions of the frame be?

SOLUTION:  Let x = the increase in the width and let 10x = the increase in the length. So, x + 4 = the new width and 10x + 60 = the new length.  

  The dimensions cannot be negative, so x = 2. The width is 2 + 4 or 6 inches and the length is 10(2) + 60 or 80 inches.

47. MULTIPLE REPRESENTATIONS In this problem, you will investigate a property of quadratic equations. a. TABULAR  Copy the table shown and complete the second column.

b. ALGEBRAIC Set each trinomial equal to zero, and solve the equation by completing the square. Complete the last column of the table with the number of roots of each equation.

c. VERBAL Compare the number of roots of each equation to the result in the b2 − 4ac column. Is there a

relationship between these values? If so, describe it.

d. ANALYTICAL Predict how many solutions 2x2 − 9x + 15 = 0 will have.  Verify your prediction by solving the 

equation.

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3    

3x2 + 6x + 9    

x2 − 2x + 7    

x2 + 10x + 25    

x2 + 3x − 12    

SOLUTION:  a.

b.

The trinomial x2 − 8x + 16 has 1 root.

 

The trinomial 2x

2 − 11x + 3 has 2 roots.

 

The square root of a negative number has no real roots. So, the trinomial 3x2 + 6x + 9 has 0 roots.

 

The square root of a negative number has no real roots. So, the trinomial x2 − 2x + 7.

 

The trinomial x2 + 10x + 25 has 1 root.

 

The trinomial x2 + 3x − 12 has 2 roots.

 

c. If b2 − 4ac is negative, the equation has no real solutions. If b

2 − 4ac is zero, the equation has one solution. If b2

− 4ac is positive, the equation has 2 solutions. d.

The equation 2x2 − 9x + 15 = 0 has 0 real solutions because b

2 − 4ac is negative.

 

The equation cannot be solved because the square root of a negative number has no real roots.

Trinomial b2 − 4ac Number

of Roots

x2 − 8x + 16 (–8)

2 – 4(1)(16) = 0 1

2x2 − 11x + 3 (–11)

2 – 4(2)(3) = 97  

3x2 + 6x + 9 (6)

2 – 4(3)(9) = −72  

x2 − 2x + 7 (–2)

2 – 4(1)(7) = −24  

x2 + 10x + 25 (10)

2 – 4(1)(25) = 0  

x2 + 3x − 12 (3)

2 – 4(1)(–12) = 57  

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3 97 2

3x2 + 6x + 9 −72 0

x2 − 2x + 7 −24 0

x2 + 10x + 25 0 1

x2 + 3x − 12 57 2

48. CCSS PERSEVERANCE Given y = ax2 + bx + c with a ≠ 0, derive the equation for the axis of symmetry by 

completing the square and rewriting the equation in the form y = a(x − h)2 + k .

SOLUTION:  

Let    and   , then the equation becomes y = (x - h)2 + k .

For an equation in the form y = ax2 + bx + c,  the equation for the axis of symmetry is   .

So, for an equation in the form  y = (x - h)2 + k, the equation for the axis of symmetry becomes x = h.

49. REASONING Determine the number of solutions x2 + bx = c has if . Explain.

SOLUTION:  

None; Sample answer: If you add to each side of the equation and each side of the inequality, you get

. Since the left side of the last equation is a perfect square, it cannot

equal the negative number . So, there are no real solutions.

50. WHICH ONE DOESN’T BELONG? Identify the expression that does not belong with the other three. Explain your reasoning.

 

SOLUTION:  The first 3 trinomials are perfect squares.

The trinomial  is not a perfect square. So, it does not belong.

51. OPEN ENDED Write a quadratic equation for which the only solution is 4.

SOLUTION:  Find a quadratic that has two roots of 4.

Verify that the solution is 4.

So, the quadratic equation x2 − 8x + 16 = 0 has a solution of 4.

52. WRITING IN MATH Compare and contrast the following strategies for solving x2 − 5x − 7 = 0: completing the

square, graphing, and factoring.

SOLUTION:  Sample answer: Because the leading coefficient is 1, solving the equation by completing the square is simpler and yields exact answers for the solutions.   

 

To solve the equation by graphing, use a graphing calculator to graph the related function y = x2 - 5x - 7. Select the

zero option from the 2nd [CALC] menu to determine the roots. 

   

  The roots are given as decimal approximations. So, for this equation the solutions would have to be given as estimations.   There are no factors of –7 that have a sum of –5, so solving by factoring is not possible.

53. The length of a rectangle is 3 times its width. The area of the rectangle is 75 square feet. Find the length of the rectangle in feet. A 25 B 15 C 10 D 5

SOLUTION:  Let x = the width of the rectangle and let 3x = the length of the rectangle.

The length of the rectangle is 3(5) or 15 feet. Choice B is the correct answer.

54. PROBABILITY At a festival, winners of a game draw a token for a prize. There is one token for each prize. The prizes include 9 movie passes, 8 stuffed animals, 5 hats, 10 jump ropes, and 4 glow necklaces. What is the probabilitythat the first person to draw a token will win a movie pass?

F  

G

H

J  

SOLUTION:  There are 9 + 8 + 5 + 10 + 4 or 36 possible outcomes.

P(movie pass) =  or  . Choice J is the correct answer.

55. GRIDDED RESPONSE The population of a town can be modeled by P = 22,000 + 125t, where P represents the population and t represents the number of years from 2000. How many years after 2000 will the population be 26,000?

SOLUTION:  

In 32 years the population of the town will be 26,000.

56. Percy delivers pizzas for Pizza King. He is paid $6 an hour plus $2.50 for each pizza he delivers. Percy earned $280 last week. If he worked a total of 30 hours, how many pizzas did he deliver? A 250 pizzas B 184 pizzas C 40 pizzas D 34 pizzas

SOLUTION:  Let p = the number of pizzas Percy delivered.

Percy delivered 40 pizzas. Choice C is the correct answer.

Describe how the graph of each function is related to the graph of f (x) = x2.

57. g(x) = −12 + x2

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –12, the translation is

down. So, the graph is shifted down 12 units from the parent function. 

58. h(x) = (x + 2)2

SOLUTION:  

The graph of f (x) = (x – c)2 represents a translation left or right from the parent graph. Since c = –2, the translation

is to the left by 2 units. 

 

59. g(x) = 2x2 + 5

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 2 and c = 5. Since 2 > 0 and  > 1, the graph of y = 2x

2 + 5

is the graph of y = x2 vertically stretched and shifted up 5 units.

60. 

SOLUTION:  

The function can be written f (x) = a(x – b)2, where a = and b = 6. Since > 0 and  < 1, the graph of 

is the graph of y = x2 vertically compressed and shifted right 6 units.

61. g(x) = 6 + x2

SOLUTION:  

The function can be written f (x) = ax2 + c, where a =  and c = 6. Since  > 0 and   > 1, the graph of y = 6 +

x2 is the graph of y = x

2 vertically stretched and shifted up 6 units.

62. h(x) = − 1 −   x2

SOLUTION:  

The function can be written f (x) = –ax2 + c, where a =  and c = –1. Since  < 0 and   > 1, the graph of y

= –1 – x2 is the graph of y = x

2 vertically stretched, shifted down 1 unit and reflected across the x-axis.

63. RIDES A popular amusement park ride whisks riders to the top of a 250-foot tower and drops them. A function for

the height of a rider is h = −16t2 + 250, where h is the height and t is the time in seconds. The ride stops the descent

of the rider 40 feet above the ground. Write an equation that models the drop of the rider. How long does it take to fall from 250 feet to 40 feet?

SOLUTION:  

The ride stops the descent of the rider at 40 feet above ground, so h = 40. Then, the equation 40 = −16t2 + 250

models the drop of the rider.

Time cannot be negative. So, it takes about 3.6 seconds to complete the ride.

Simplify. Assume that no denominator is equal to zero.

64. 

SOLUTION:  

65. 

SOLUTION:  

66. 

SOLUTION:  

67. 

SOLUTION:  

68. 

SOLUTION:  

69. b3(m

−3)(b

−6)

SOLUTION:  

Solve each open sentence.

70. |y − 2| > 7

SOLUTION:  

The solution set is {y |y > 9 or y < −5}.

Case 1 y – 2 is positive.

and Case 2 y – 2 is negative.

 

71. |z + 5| < 3

SOLUTION:  

The solution set is {z |−8 < z < −2}.

Case 1 z +5 is positive.

and Case 2 z +5 is negative.

 

72. |2b + 7| ≤ −6

SOLUTION:  

cannot be negative. So,  cannot be less than or equal to –6. The solution set is empty.

73. |3 − 2y | ≥ 8

SOLUTION:  

The solution set is {y |y  ≥ 5.5 or y  ≤ −2.5}.

Case 1 3 – 2y is positive.

and Case 2 3 – 2y is negative.

 

74. |9 − 4m| < −1

SOLUTION:  

cannot be negative. So, cannot be less than or equal to –1. The solution set is empty.

75. |5c − 2| ≤ 13

SOLUTION:  

The solution set is {c|−2.2 ≤ c ≤ 3}.

Case 1 5c – 2 is positive.

and Case 2 5c – 2 is negative.

 

Evaluate  for each set of values. Round to the nearest tenth if necessary.

76. a = 2, b = −5, c = 2

SOLUTION:  

77. a = 1, b = 12, c = 11

SOLUTION:  

78. a = −9, b = 10, c = −1

SOLUTION:  

79. a = 1, b = 7, c = −3

SOLUTION:  

80. a = 2, b = −4, c = −6

SOLUTION:  

81. a = 3, b = 1, c = 2

SOLUTION:  

This value is not a real number.

eSolutions Manual - Powered by Cognero Page 41

9-4 Solving Quadratic Equations by Completing the Square

Find the value of c that makes each trinomial a perfect square.

1. x2 − 18x + c

SOLUTION:  In this trinomial, b = –18.

So, c must be 81 to make the trinomial a perfect square.

2. x2 + 22x + c

SOLUTION:  In this trinomial, b = 22.

So, c must be 121 to make the trinomial a perfect square.

3. x2 + 9x + c

SOLUTION:  In this trinomial, b = 9.

 

So, c must be  to make the trinomial a perfect square.

4. x2 − 7x + c

SOLUTION:  In this trinomial, b = –7.

 

So, c must be  to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

5. x2 + 4x = 6

SOLUTION:  

The solutions are about –5.2 and 1.2.

6. x2 − 8x = −9

SOLUTION:  

The solutions are about 1.4 and 6.6.

7. 4x2 + 9x − 1 = 0

SOLUTION:  

The solutions are about –2.4 and 0.1.

8. −2x2 + 10x + 22 = 4

SOLUTION:  

The solutions are about –1.4 and 6.4.

9. CCSS MODELING Collin is building a deck on the back of his family’s house. He has enough lumber for the deck to be 144 square feet. The length should be 10 feet more than its width. What should the dimensions of the deck be?

SOLUTION:  Let x = the width of the deck and x + 10 = the length of the deck.

The dimensions of the deck can not be negative. So, x = –5 + 13 or 8. The width of the deck is 8 feet and the length of the deck is 8 + 10 or 18 feet.

Find the value of c that makes each trinomial a perfect square.

10. x2 + 26x + c

SOLUTION:  In this trinomial, b = 26.

So, c must be 169 to make the trinomial a perfect square.

11. x2 − 24x + c

SOLUTION:  In this trinomial, b = –24.

So, c must be 144 to make the trinomial a perfect square.

12. x2 − 19x + c

SOLUTION:  In this trinomial, b = –19.

So, c must be  to make the trinomial a perfect square.

13. x2 + 17x + c

SOLUTION:  In this trinomial, b = 17.

So, c must be  to make the trinomial a perfect square.

14. x2 + 5x + c

SOLUTION:  In this trinomial, b = 5.

So, c must be  to make the trinomial a perfect square.

15. x2 − 13x + c

SOLUTION:  In this trinomial, b = –13.

So, c must be  to make the trinomial a perfect square.

16. x2 − 22x + c

SOLUTION:  In this trinomial, b = –22.

So, c must be 121 to make the trinomial a perfect square.

17. x2 − 15x + c

SOLUTION:  In this trinomial, b = –15.

So, c must be  to make the trinomial a perfect square.

18. x2 + 24x + c

SOLUTION:  In this trinomial, b = 24.

So, c must be 144 to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

19. x2 + 6x − 16 = 0

SOLUTION:  

  The solutions are –8 and 2.

20. x2 − 2x − 14 = 0

SOLUTION:  

  The solutions are about –2.9 and 4.9.

21. x2 − 8x − 1 = 8

SOLUTION:  

  The solutions are –1 and 9.

22. x2 + 3x + 21 = 22

SOLUTION:  

  The solutions are about –3.3 and 0.3.

23. x2 − 11x + 3 = 5

SOLUTION:  

  The solutions are about –0.2 and 11.2.

24. 5x2 − 10x = 23

SOLUTION:  

  The solutions are about –1.4 and 3.4.

25. 2x2 − 2x + 7 = 5

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

26. 3x2 + 12x + 81 = 15

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

27. 4x2 + 6x = 12

SOLUTION:  

  The solutions are about –2.6 and 1.1.

28. 4x2 + 5 = 10x

SOLUTION:  

  The solutions are about 0.7 and 1.8.

29. −2x2 + 10x = −14

SOLUTION:  

  The solutions are about –1.1 and 6.1.

30. −3x2 − 12 = 14x

SOLUTION:  

  The solutions are about –3.5 and –1.1.

31. STOCK The price p in dollars for a particular stock can be modeled by the quadratic equation p = 3.5t − 0.05t2, wh

the number of days after the stock is purchased. When is the stock worth $60?

SOLUTION:  Let p = 60.

  The stock is worth $60 on the 30th and 40th days after purchase.

GEOMETRY  Find the value of x for each figure. Round to the nearest tenth if necessary.

32. area = 45 in2

SOLUTION:  

  The height cannot be negative. So, x ≈ 6.3.

33. area = 110 ft2

SOLUTION:  

  The dimensions of the rectangle cannot be negative. So, x ≈ 5.3.

34. NUMBER THEORY The product of two consecutive even integers is 224. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers are 14 and 16 or –16 and –14.

35. CCSS PRECISION The product of two consecutive negative odd integers is 483. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers must be negative. So, they are –23 and –21.

36. GEOMETRY Find the area of the triangle.

SOLUTION:  

  The dimensions of the triangle cannot be negative. So, x = 18. The base of the triangle is 18 meters and the height is 18 + 6 or 24 meters.

The area of the triangle is 216 square meters.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

37. 0.2x2 − 0.2x − 0.4 = 0

SOLUTION:  

  The solutions are –1 and 2.

38. 0.5x2 = 2x − 0.3

SOLUTION:  

  The solutions are about 0.2 and 3.8.

39. 2x2 −

SOLUTION:  

  The solutions are about 0.2 and 0.9.

40. 

SOLUTION:  

  The solutions are –0.5 and 2.5.

41. 

SOLUTION:  

  The solutions are about –8.2 and 0.2.

42. 

SOLUTION:  

  The solutions are about –5.1 and 0.1.

43. ASTRONOMY The height of an object t seconds after it is dropped is given by the equation ,

where h0 is the initial height and g is the acceleration due to gravity. The acceleration due to gravity near the surface

of Mars is 3.73 m/s2, while on Earth it is 9.8 m/s

2. Suppose an object is dropped from an initial height of 120 meters

above the surface of each planet. a. On which planet would the object reach the ground first? b. How long would it take the object to reach the ground on each planet? Round each answer to the nearest tenth. c. Do the times that it takes the object to reach the ground seem reasonable? Explain your reasoning.

SOLUTION:  a. The object on Earth will reach the ground first because it is falling at a faster rate.   b. Mars:

So, t ≈ 8.0 seconds. Earth:

So, t ≈ 4.9 seconds.   c. Sample answer: Yes; the acceleration due to gravity is much greater on Earth than on Mars, so the time to reach the ground should be much less.

44. Find all values of c that make x2 + cx + 100 a perfect square trinomial.

SOLUTION:  

So, c can be –20 or 20 to make the trinomial a perfect square.

45. Find all values of c that make x2 + cx + 225 a perfect square trinomial.

SOLUTION:  

So, c can be –30 or 30 to make the trinomial a perfect square.

46. PAINTING Before she begins painting a picture, Donna stretches her canvas over a wood frame. The frame has a length of 60 inches and a width of 4 inches. She has enough canvas to cover 480 square inches. Donna decides to increase the dimensions of the frame. If the increase in the length is 10 times the increase in the width, what will the dimensions of the frame be?

SOLUTION:  Let x = the increase in the width and let 10x = the increase in the length. So, x + 4 = the new width and 10x + 60 = the new length.  

  The dimensions cannot be negative, so x = 2. The width is 2 + 4 or 6 inches and the length is 10(2) + 60 or 80 inches.

47. MULTIPLE REPRESENTATIONS In this problem, you will investigate a property of quadratic equations. a. TABULAR  Copy the table shown and complete the second column.

b. ALGEBRAIC Set each trinomial equal to zero, and solve the equation by completing the square. Complete the last column of the table with the number of roots of each equation.

c. VERBAL Compare the number of roots of each equation to the result in the b2 − 4ac column. Is there a

relationship between these values? If so, describe it.

d. ANALYTICAL Predict how many solutions 2x2 − 9x + 15 = 0 will have.  Verify your prediction by solving the 

equation.

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3    

3x2 + 6x + 9    

x2 − 2x + 7    

x2 + 10x + 25    

x2 + 3x − 12    

SOLUTION:  a.

b.

The trinomial x2 − 8x + 16 has 1 root.

 

The trinomial 2x

2 − 11x + 3 has 2 roots.

 

The square root of a negative number has no real roots. So, the trinomial 3x2 + 6x + 9 has 0 roots.

 

The square root of a negative number has no real roots. So, the trinomial x2 − 2x + 7.

 

The trinomial x2 + 10x + 25 has 1 root.

 

The trinomial x2 + 3x − 12 has 2 roots.

 

c. If b2 − 4ac is negative, the equation has no real solutions. If b

2 − 4ac is zero, the equation has one solution. If b2

− 4ac is positive, the equation has 2 solutions. d.

The equation 2x2 − 9x + 15 = 0 has 0 real solutions because b

2 − 4ac is negative.

 

The equation cannot be solved because the square root of a negative number has no real roots.

Trinomial b2 − 4ac Number

of Roots

x2 − 8x + 16 (–8)

2 – 4(1)(16) = 0 1

2x2 − 11x + 3 (–11)

2 – 4(2)(3) = 97  

3x2 + 6x + 9 (6)

2 – 4(3)(9) = −72  

x2 − 2x + 7 (–2)

2 – 4(1)(7) = −24  

x2 + 10x + 25 (10)

2 – 4(1)(25) = 0  

x2 + 3x − 12 (3)

2 – 4(1)(–12) = 57  

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3 97 2

3x2 + 6x + 9 −72 0

x2 − 2x + 7 −24 0

x2 + 10x + 25 0 1

x2 + 3x − 12 57 2

48. CCSS PERSEVERANCE Given y = ax2 + bx + c with a ≠ 0, derive the equation for the axis of symmetry by 

completing the square and rewriting the equation in the form y = a(x − h)2 + k .

SOLUTION:  

Let    and   , then the equation becomes y = (x - h)2 + k .

For an equation in the form y = ax2 + bx + c,  the equation for the axis of symmetry is   .

So, for an equation in the form  y = (x - h)2 + k, the equation for the axis of symmetry becomes x = h.

49. REASONING Determine the number of solutions x2 + bx = c has if . Explain.

SOLUTION:  

None; Sample answer: If you add to each side of the equation and each side of the inequality, you get

. Since the left side of the last equation is a perfect square, it cannot

equal the negative number . So, there are no real solutions.

50. WHICH ONE DOESN’T BELONG? Identify the expression that does not belong with the other three. Explain your reasoning.

 

SOLUTION:  The first 3 trinomials are perfect squares.

The trinomial  is not a perfect square. So, it does not belong.

51. OPEN ENDED Write a quadratic equation for which the only solution is 4.

SOLUTION:  Find a quadratic that has two roots of 4.

Verify that the solution is 4.

So, the quadratic equation x2 − 8x + 16 = 0 has a solution of 4.

52. WRITING IN MATH Compare and contrast the following strategies for solving x2 − 5x − 7 = 0: completing the

square, graphing, and factoring.

SOLUTION:  Sample answer: Because the leading coefficient is 1, solving the equation by completing the square is simpler and yields exact answers for the solutions.   

 

To solve the equation by graphing, use a graphing calculator to graph the related function y = x2 - 5x - 7. Select the

zero option from the 2nd [CALC] menu to determine the roots. 

   

  The roots are given as decimal approximations. So, for this equation the solutions would have to be given as estimations.   There are no factors of –7 that have a sum of –5, so solving by factoring is not possible.

53. The length of a rectangle is 3 times its width. The area of the rectangle is 75 square feet. Find the length of the rectangle in feet. A 25 B 15 C 10 D 5

SOLUTION:  Let x = the width of the rectangle and let 3x = the length of the rectangle.

The length of the rectangle is 3(5) or 15 feet. Choice B is the correct answer.

54. PROBABILITY At a festival, winners of a game draw a token for a prize. There is one token for each prize. The prizes include 9 movie passes, 8 stuffed animals, 5 hats, 10 jump ropes, and 4 glow necklaces. What is the probabilitythat the first person to draw a token will win a movie pass?

F  

G

H

J  

SOLUTION:  There are 9 + 8 + 5 + 10 + 4 or 36 possible outcomes.

P(movie pass) =  or  . Choice J is the correct answer.

55. GRIDDED RESPONSE The population of a town can be modeled by P = 22,000 + 125t, where P represents the population and t represents the number of years from 2000. How many years after 2000 will the population be 26,000?

SOLUTION:  

In 32 years the population of the town will be 26,000.

56. Percy delivers pizzas for Pizza King. He is paid $6 an hour plus $2.50 for each pizza he delivers. Percy earned $280 last week. If he worked a total of 30 hours, how many pizzas did he deliver? A 250 pizzas B 184 pizzas C 40 pizzas D 34 pizzas

SOLUTION:  Let p = the number of pizzas Percy delivered.

Percy delivered 40 pizzas. Choice C is the correct answer.

Describe how the graph of each function is related to the graph of f (x) = x2.

57. g(x) = −12 + x2

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –12, the translation is

down. So, the graph is shifted down 12 units from the parent function. 

58. h(x) = (x + 2)2

SOLUTION:  

The graph of f (x) = (x – c)2 represents a translation left or right from the parent graph. Since c = –2, the translation

is to the left by 2 units. 

 

59. g(x) = 2x2 + 5

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 2 and c = 5. Since 2 > 0 and  > 1, the graph of y = 2x

2 + 5

is the graph of y = x2 vertically stretched and shifted up 5 units.

60. 

SOLUTION:  

The function can be written f (x) = a(x – b)2, where a = and b = 6. Since > 0 and  < 1, the graph of 

is the graph of y = x2 vertically compressed and shifted right 6 units.

61. g(x) = 6 + x2

SOLUTION:  

The function can be written f (x) = ax2 + c, where a =  and c = 6. Since  > 0 and   > 1, the graph of y = 6 +

x2 is the graph of y = x

2 vertically stretched and shifted up 6 units.

62. h(x) = − 1 −   x2

SOLUTION:  

The function can be written f (x) = –ax2 + c, where a =  and c = –1. Since  < 0 and   > 1, the graph of y

= –1 – x2 is the graph of y = x

2 vertically stretched, shifted down 1 unit and reflected across the x-axis.

63. RIDES A popular amusement park ride whisks riders to the top of a 250-foot tower and drops them. A function for

the height of a rider is h = −16t2 + 250, where h is the height and t is the time in seconds. The ride stops the descent

of the rider 40 feet above the ground. Write an equation that models the drop of the rider. How long does it take to fall from 250 feet to 40 feet?

SOLUTION:  

The ride stops the descent of the rider at 40 feet above ground, so h = 40. Then, the equation 40 = −16t2 + 250

models the drop of the rider.

Time cannot be negative. So, it takes about 3.6 seconds to complete the ride.

Simplify. Assume that no denominator is equal to zero.

64. 

SOLUTION:  

65. 

SOLUTION:  

66. 

SOLUTION:  

67. 

SOLUTION:  

68. 

SOLUTION:  

69. b3(m

−3)(b

−6)

SOLUTION:  

Solve each open sentence.

70. |y − 2| > 7

SOLUTION:  

The solution set is {y |y > 9 or y < −5}.

Case 1 y – 2 is positive.

and Case 2 y – 2 is negative.

 

71. |z + 5| < 3

SOLUTION:  

The solution set is {z |−8 < z < −2}.

Case 1 z +5 is positive.

and Case 2 z +5 is negative.

 

72. |2b + 7| ≤ −6

SOLUTION:  

cannot be negative. So,  cannot be less than or equal to –6. The solution set is empty.

73. |3 − 2y | ≥ 8

SOLUTION:  

The solution set is {y |y  ≥ 5.5 or y  ≤ −2.5}.

Case 1 3 – 2y is positive.

and Case 2 3 – 2y is negative.

 

74. |9 − 4m| < −1

SOLUTION:  

cannot be negative. So, cannot be less than or equal to –1. The solution set is empty.

75. |5c − 2| ≤ 13

SOLUTION:  

The solution set is {c|−2.2 ≤ c ≤ 3}.

Case 1 5c – 2 is positive.

and Case 2 5c – 2 is negative.

 

Evaluate  for each set of values. Round to the nearest tenth if necessary.

76. a = 2, b = −5, c = 2

SOLUTION:  

77. a = 1, b = 12, c = 11

SOLUTION:  

78. a = −9, b = 10, c = −1

SOLUTION:  

79. a = 1, b = 7, c = −3

SOLUTION:  

80. a = 2, b = −4, c = −6

SOLUTION:  

81. a = 3, b = 1, c = 2

SOLUTION:  

This value is not a real number.

eSolutions Manual - Powered by Cognero Page 42

9-4 Solving Quadratic Equations by Completing the Square

Find the value of c that makes each trinomial a perfect square.

1. x2 − 18x + c

SOLUTION:  In this trinomial, b = –18.

So, c must be 81 to make the trinomial a perfect square.

2. x2 + 22x + c

SOLUTION:  In this trinomial, b = 22.

So, c must be 121 to make the trinomial a perfect square.

3. x2 + 9x + c

SOLUTION:  In this trinomial, b = 9.

 

So, c must be  to make the trinomial a perfect square.

4. x2 − 7x + c

SOLUTION:  In this trinomial, b = –7.

 

So, c must be  to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

5. x2 + 4x = 6

SOLUTION:  

The solutions are about –5.2 and 1.2.

6. x2 − 8x = −9

SOLUTION:  

The solutions are about 1.4 and 6.6.

7. 4x2 + 9x − 1 = 0

SOLUTION:  

The solutions are about –2.4 and 0.1.

8. −2x2 + 10x + 22 = 4

SOLUTION:  

The solutions are about –1.4 and 6.4.

9. CCSS MODELING Collin is building a deck on the back of his family’s house. He has enough lumber for the deck to be 144 square feet. The length should be 10 feet more than its width. What should the dimensions of the deck be?

SOLUTION:  Let x = the width of the deck and x + 10 = the length of the deck.

The dimensions of the deck can not be negative. So, x = –5 + 13 or 8. The width of the deck is 8 feet and the length of the deck is 8 + 10 or 18 feet.

Find the value of c that makes each trinomial a perfect square.

10. x2 + 26x + c

SOLUTION:  In this trinomial, b = 26.

So, c must be 169 to make the trinomial a perfect square.

11. x2 − 24x + c

SOLUTION:  In this trinomial, b = –24.

So, c must be 144 to make the trinomial a perfect square.

12. x2 − 19x + c

SOLUTION:  In this trinomial, b = –19.

So, c must be  to make the trinomial a perfect square.

13. x2 + 17x + c

SOLUTION:  In this trinomial, b = 17.

So, c must be  to make the trinomial a perfect square.

14. x2 + 5x + c

SOLUTION:  In this trinomial, b = 5.

So, c must be  to make the trinomial a perfect square.

15. x2 − 13x + c

SOLUTION:  In this trinomial, b = –13.

So, c must be  to make the trinomial a perfect square.

16. x2 − 22x + c

SOLUTION:  In this trinomial, b = –22.

So, c must be 121 to make the trinomial a perfect square.

17. x2 − 15x + c

SOLUTION:  In this trinomial, b = –15.

So, c must be  to make the trinomial a perfect square.

18. x2 + 24x + c

SOLUTION:  In this trinomial, b = 24.

So, c must be 144 to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

19. x2 + 6x − 16 = 0

SOLUTION:  

  The solutions are –8 and 2.

20. x2 − 2x − 14 = 0

SOLUTION:  

  The solutions are about –2.9 and 4.9.

21. x2 − 8x − 1 = 8

SOLUTION:  

  The solutions are –1 and 9.

22. x2 + 3x + 21 = 22

SOLUTION:  

  The solutions are about –3.3 and 0.3.

23. x2 − 11x + 3 = 5

SOLUTION:  

  The solutions are about –0.2 and 11.2.

24. 5x2 − 10x = 23

SOLUTION:  

  The solutions are about –1.4 and 3.4.

25. 2x2 − 2x + 7 = 5

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

26. 3x2 + 12x + 81 = 15

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

27. 4x2 + 6x = 12

SOLUTION:  

  The solutions are about –2.6 and 1.1.

28. 4x2 + 5 = 10x

SOLUTION:  

  The solutions are about 0.7 and 1.8.

29. −2x2 + 10x = −14

SOLUTION:  

  The solutions are about –1.1 and 6.1.

30. −3x2 − 12 = 14x

SOLUTION:  

  The solutions are about –3.5 and –1.1.

31. STOCK The price p in dollars for a particular stock can be modeled by the quadratic equation p = 3.5t − 0.05t2, wh

the number of days after the stock is purchased. When is the stock worth $60?

SOLUTION:  Let p = 60.

  The stock is worth $60 on the 30th and 40th days after purchase.

GEOMETRY  Find the value of x for each figure. Round to the nearest tenth if necessary.

32. area = 45 in2

SOLUTION:  

  The height cannot be negative. So, x ≈ 6.3.

33. area = 110 ft2

SOLUTION:  

  The dimensions of the rectangle cannot be negative. So, x ≈ 5.3.

34. NUMBER THEORY The product of two consecutive even integers is 224. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers are 14 and 16 or –16 and –14.

35. CCSS PRECISION The product of two consecutive negative odd integers is 483. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers must be negative. So, they are –23 and –21.

36. GEOMETRY Find the area of the triangle.

SOLUTION:  

  The dimensions of the triangle cannot be negative. So, x = 18. The base of the triangle is 18 meters and the height is 18 + 6 or 24 meters.

The area of the triangle is 216 square meters.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

37. 0.2x2 − 0.2x − 0.4 = 0

SOLUTION:  

  The solutions are –1 and 2.

38. 0.5x2 = 2x − 0.3

SOLUTION:  

  The solutions are about 0.2 and 3.8.

39. 2x2 −

SOLUTION:  

  The solutions are about 0.2 and 0.9.

40. 

SOLUTION:  

  The solutions are –0.5 and 2.5.

41. 

SOLUTION:  

  The solutions are about –8.2 and 0.2.

42. 

SOLUTION:  

  The solutions are about –5.1 and 0.1.

43. ASTRONOMY The height of an object t seconds after it is dropped is given by the equation ,

where h0 is the initial height and g is the acceleration due to gravity. The acceleration due to gravity near the surface

of Mars is 3.73 m/s2, while on Earth it is 9.8 m/s

2. Suppose an object is dropped from an initial height of 120 meters

above the surface of each planet. a. On which planet would the object reach the ground first? b. How long would it take the object to reach the ground on each planet? Round each answer to the nearest tenth. c. Do the times that it takes the object to reach the ground seem reasonable? Explain your reasoning.

SOLUTION:  a. The object on Earth will reach the ground first because it is falling at a faster rate.   b. Mars:

So, t ≈ 8.0 seconds. Earth:

So, t ≈ 4.9 seconds.   c. Sample answer: Yes; the acceleration due to gravity is much greater on Earth than on Mars, so the time to reach the ground should be much less.

44. Find all values of c that make x2 + cx + 100 a perfect square trinomial.

SOLUTION:  

So, c can be –20 or 20 to make the trinomial a perfect square.

45. Find all values of c that make x2 + cx + 225 a perfect square trinomial.

SOLUTION:  

So, c can be –30 or 30 to make the trinomial a perfect square.

46. PAINTING Before she begins painting a picture, Donna stretches her canvas over a wood frame. The frame has a length of 60 inches and a width of 4 inches. She has enough canvas to cover 480 square inches. Donna decides to increase the dimensions of the frame. If the increase in the length is 10 times the increase in the width, what will the dimensions of the frame be?

SOLUTION:  Let x = the increase in the width and let 10x = the increase in the length. So, x + 4 = the new width and 10x + 60 = the new length.  

  The dimensions cannot be negative, so x = 2. The width is 2 + 4 or 6 inches and the length is 10(2) + 60 or 80 inches.

47. MULTIPLE REPRESENTATIONS In this problem, you will investigate a property of quadratic equations. a. TABULAR  Copy the table shown and complete the second column.

b. ALGEBRAIC Set each trinomial equal to zero, and solve the equation by completing the square. Complete the last column of the table with the number of roots of each equation.

c. VERBAL Compare the number of roots of each equation to the result in the b2 − 4ac column. Is there a

relationship between these values? If so, describe it.

d. ANALYTICAL Predict how many solutions 2x2 − 9x + 15 = 0 will have.  Verify your prediction by solving the 

equation.

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3    

3x2 + 6x + 9    

x2 − 2x + 7    

x2 + 10x + 25    

x2 + 3x − 12    

SOLUTION:  a.

b.

The trinomial x2 − 8x + 16 has 1 root.

 

The trinomial 2x

2 − 11x + 3 has 2 roots.

 

The square root of a negative number has no real roots. So, the trinomial 3x2 + 6x + 9 has 0 roots.

 

The square root of a negative number has no real roots. So, the trinomial x2 − 2x + 7.

 

The trinomial x2 + 10x + 25 has 1 root.

 

The trinomial x2 + 3x − 12 has 2 roots.

 

c. If b2 − 4ac is negative, the equation has no real solutions. If b

2 − 4ac is zero, the equation has one solution. If b2

− 4ac is positive, the equation has 2 solutions. d.

The equation 2x2 − 9x + 15 = 0 has 0 real solutions because b

2 − 4ac is negative.

 

The equation cannot be solved because the square root of a negative number has no real roots.

Trinomial b2 − 4ac Number

of Roots

x2 − 8x + 16 (–8)

2 – 4(1)(16) = 0 1

2x2 − 11x + 3 (–11)

2 – 4(2)(3) = 97  

3x2 + 6x + 9 (6)

2 – 4(3)(9) = −72  

x2 − 2x + 7 (–2)

2 – 4(1)(7) = −24  

x2 + 10x + 25 (10)

2 – 4(1)(25) = 0  

x2 + 3x − 12 (3)

2 – 4(1)(–12) = 57  

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3 97 2

3x2 + 6x + 9 −72 0

x2 − 2x + 7 −24 0

x2 + 10x + 25 0 1

x2 + 3x − 12 57 2

48. CCSS PERSEVERANCE Given y = ax2 + bx + c with a ≠ 0, derive the equation for the axis of symmetry by 

completing the square and rewriting the equation in the form y = a(x − h)2 + k .

SOLUTION:  

Let    and   , then the equation becomes y = (x - h)2 + k .

For an equation in the form y = ax2 + bx + c,  the equation for the axis of symmetry is   .

So, for an equation in the form  y = (x - h)2 + k, the equation for the axis of symmetry becomes x = h.

49. REASONING Determine the number of solutions x2 + bx = c has if . Explain.

SOLUTION:  

None; Sample answer: If you add to each side of the equation and each side of the inequality, you get

. Since the left side of the last equation is a perfect square, it cannot

equal the negative number . So, there are no real solutions.

50. WHICH ONE DOESN’T BELONG? Identify the expression that does not belong with the other three. Explain your reasoning.

 

SOLUTION:  The first 3 trinomials are perfect squares.

The trinomial  is not a perfect square. So, it does not belong.

51. OPEN ENDED Write a quadratic equation for which the only solution is 4.

SOLUTION:  Find a quadratic that has two roots of 4.

Verify that the solution is 4.

So, the quadratic equation x2 − 8x + 16 = 0 has a solution of 4.

52. WRITING IN MATH Compare and contrast the following strategies for solving x2 − 5x − 7 = 0: completing the

square, graphing, and factoring.

SOLUTION:  Sample answer: Because the leading coefficient is 1, solving the equation by completing the square is simpler and yields exact answers for the solutions.   

 

To solve the equation by graphing, use a graphing calculator to graph the related function y = x2 - 5x - 7. Select the

zero option from the 2nd [CALC] menu to determine the roots. 

   

  The roots are given as decimal approximations. So, for this equation the solutions would have to be given as estimations.   There are no factors of –7 that have a sum of –5, so solving by factoring is not possible.

53. The length of a rectangle is 3 times its width. The area of the rectangle is 75 square feet. Find the length of the rectangle in feet. A 25 B 15 C 10 D 5

SOLUTION:  Let x = the width of the rectangle and let 3x = the length of the rectangle.

The length of the rectangle is 3(5) or 15 feet. Choice B is the correct answer.

54. PROBABILITY At a festival, winners of a game draw a token for a prize. There is one token for each prize. The prizes include 9 movie passes, 8 stuffed animals, 5 hats, 10 jump ropes, and 4 glow necklaces. What is the probabilitythat the first person to draw a token will win a movie pass?

F  

G

H

J  

SOLUTION:  There are 9 + 8 + 5 + 10 + 4 or 36 possible outcomes.

P(movie pass) =  or  . Choice J is the correct answer.

55. GRIDDED RESPONSE The population of a town can be modeled by P = 22,000 + 125t, where P represents the population and t represents the number of years from 2000. How many years after 2000 will the population be 26,000?

SOLUTION:  

In 32 years the population of the town will be 26,000.

56. Percy delivers pizzas for Pizza King. He is paid $6 an hour plus $2.50 for each pizza he delivers. Percy earned $280 last week. If he worked a total of 30 hours, how many pizzas did he deliver? A 250 pizzas B 184 pizzas C 40 pizzas D 34 pizzas

SOLUTION:  Let p = the number of pizzas Percy delivered.

Percy delivered 40 pizzas. Choice C is the correct answer.

Describe how the graph of each function is related to the graph of f (x) = x2.

57. g(x) = −12 + x2

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –12, the translation is

down. So, the graph is shifted down 12 units from the parent function. 

58. h(x) = (x + 2)2

SOLUTION:  

The graph of f (x) = (x – c)2 represents a translation left or right from the parent graph. Since c = –2, the translation

is to the left by 2 units. 

 

59. g(x) = 2x2 + 5

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 2 and c = 5. Since 2 > 0 and  > 1, the graph of y = 2x

2 + 5

is the graph of y = x2 vertically stretched and shifted up 5 units.

60. 

SOLUTION:  

The function can be written f (x) = a(x – b)2, where a = and b = 6. Since > 0 and  < 1, the graph of 

is the graph of y = x2 vertically compressed and shifted right 6 units.

61. g(x) = 6 + x2

SOLUTION:  

The function can be written f (x) = ax2 + c, where a =  and c = 6. Since  > 0 and   > 1, the graph of y = 6 +

x2 is the graph of y = x

2 vertically stretched and shifted up 6 units.

62. h(x) = − 1 −   x2

SOLUTION:  

The function can be written f (x) = –ax2 + c, where a =  and c = –1. Since  < 0 and   > 1, the graph of y

= –1 – x2 is the graph of y = x

2 vertically stretched, shifted down 1 unit and reflected across the x-axis.

63. RIDES A popular amusement park ride whisks riders to the top of a 250-foot tower and drops them. A function for

the height of a rider is h = −16t2 + 250, where h is the height and t is the time in seconds. The ride stops the descent

of the rider 40 feet above the ground. Write an equation that models the drop of the rider. How long does it take to fall from 250 feet to 40 feet?

SOLUTION:  

The ride stops the descent of the rider at 40 feet above ground, so h = 40. Then, the equation 40 = −16t2 + 250

models the drop of the rider.

Time cannot be negative. So, it takes about 3.6 seconds to complete the ride.

Simplify. Assume that no denominator is equal to zero.

64. 

SOLUTION:  

65. 

SOLUTION:  

66. 

SOLUTION:  

67. 

SOLUTION:  

68. 

SOLUTION:  

69. b3(m

−3)(b

−6)

SOLUTION:  

Solve each open sentence.

70. |y − 2| > 7

SOLUTION:  

The solution set is {y |y > 9 or y < −5}.

Case 1 y – 2 is positive.

and Case 2 y – 2 is negative.

 

71. |z + 5| < 3

SOLUTION:  

The solution set is {z |−8 < z < −2}.

Case 1 z +5 is positive.

and Case 2 z +5 is negative.

 

72. |2b + 7| ≤ −6

SOLUTION:  

cannot be negative. So,  cannot be less than or equal to –6. The solution set is empty.

73. |3 − 2y | ≥ 8

SOLUTION:  

The solution set is {y |y  ≥ 5.5 or y  ≤ −2.5}.

Case 1 3 – 2y is positive.

and Case 2 3 – 2y is negative.

 

74. |9 − 4m| < −1

SOLUTION:  

cannot be negative. So, cannot be less than or equal to –1. The solution set is empty.

75. |5c − 2| ≤ 13

SOLUTION:  

The solution set is {c|−2.2 ≤ c ≤ 3}.

Case 1 5c – 2 is positive.

and Case 2 5c – 2 is negative.

 

Evaluate  for each set of values. Round to the nearest tenth if necessary.

76. a = 2, b = −5, c = 2

SOLUTION:  

77. a = 1, b = 12, c = 11

SOLUTION:  

78. a = −9, b = 10, c = −1

SOLUTION:  

79. a = 1, b = 7, c = −3

SOLUTION:  

80. a = 2, b = −4, c = −6

SOLUTION:  

81. a = 3, b = 1, c = 2

SOLUTION:  

This value is not a real number.

eSolutions Manual - Powered by Cognero Page 43

9-4 Solving Quadratic Equations by Completing the Square

Find the value of c that makes each trinomial a perfect square.

1. x2 − 18x + c

SOLUTION:  In this trinomial, b = –18.

So, c must be 81 to make the trinomial a perfect square.

2. x2 + 22x + c

SOLUTION:  In this trinomial, b = 22.

So, c must be 121 to make the trinomial a perfect square.

3. x2 + 9x + c

SOLUTION:  In this trinomial, b = 9.

 

So, c must be  to make the trinomial a perfect square.

4. x2 − 7x + c

SOLUTION:  In this trinomial, b = –7.

 

So, c must be  to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

5. x2 + 4x = 6

SOLUTION:  

The solutions are about –5.2 and 1.2.

6. x2 − 8x = −9

SOLUTION:  

The solutions are about 1.4 and 6.6.

7. 4x2 + 9x − 1 = 0

SOLUTION:  

The solutions are about –2.4 and 0.1.

8. −2x2 + 10x + 22 = 4

SOLUTION:  

The solutions are about –1.4 and 6.4.

9. CCSS MODELING Collin is building a deck on the back of his family’s house. He has enough lumber for the deck to be 144 square feet. The length should be 10 feet more than its width. What should the dimensions of the deck be?

SOLUTION:  Let x = the width of the deck and x + 10 = the length of the deck.

The dimensions of the deck can not be negative. So, x = –5 + 13 or 8. The width of the deck is 8 feet and the length of the deck is 8 + 10 or 18 feet.

Find the value of c that makes each trinomial a perfect square.

10. x2 + 26x + c

SOLUTION:  In this trinomial, b = 26.

So, c must be 169 to make the trinomial a perfect square.

11. x2 − 24x + c

SOLUTION:  In this trinomial, b = –24.

So, c must be 144 to make the trinomial a perfect square.

12. x2 − 19x + c

SOLUTION:  In this trinomial, b = –19.

So, c must be  to make the trinomial a perfect square.

13. x2 + 17x + c

SOLUTION:  In this trinomial, b = 17.

So, c must be  to make the trinomial a perfect square.

14. x2 + 5x + c

SOLUTION:  In this trinomial, b = 5.

So, c must be  to make the trinomial a perfect square.

15. x2 − 13x + c

SOLUTION:  In this trinomial, b = –13.

So, c must be  to make the trinomial a perfect square.

16. x2 − 22x + c

SOLUTION:  In this trinomial, b = –22.

So, c must be 121 to make the trinomial a perfect square.

17. x2 − 15x + c

SOLUTION:  In this trinomial, b = –15.

So, c must be  to make the trinomial a perfect square.

18. x2 + 24x + c

SOLUTION:  In this trinomial, b = 24.

So, c must be 144 to make the trinomial a perfect square.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

19. x2 + 6x − 16 = 0

SOLUTION:  

  The solutions are –8 and 2.

20. x2 − 2x − 14 = 0

SOLUTION:  

  The solutions are about –2.9 and 4.9.

21. x2 − 8x − 1 = 8

SOLUTION:  

  The solutions are –1 and 9.

22. x2 + 3x + 21 = 22

SOLUTION:  

  The solutions are about –3.3 and 0.3.

23. x2 − 11x + 3 = 5

SOLUTION:  

  The solutions are about –0.2 and 11.2.

24. 5x2 − 10x = 23

SOLUTION:  

  The solutions are about –1.4 and 3.4.

25. 2x2 − 2x + 7 = 5

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

26. 3x2 + 12x + 81 = 15

SOLUTION:  

  The square root of a negative number has no real roots. So, there is no solution.

27. 4x2 + 6x = 12

SOLUTION:  

  The solutions are about –2.6 and 1.1.

28. 4x2 + 5 = 10x

SOLUTION:  

  The solutions are about 0.7 and 1.8.

29. −2x2 + 10x = −14

SOLUTION:  

  The solutions are about –1.1 and 6.1.

30. −3x2 − 12 = 14x

SOLUTION:  

  The solutions are about –3.5 and –1.1.

31. STOCK The price p in dollars for a particular stock can be modeled by the quadratic equation p = 3.5t − 0.05t2, wh

the number of days after the stock is purchased. When is the stock worth $60?

SOLUTION:  Let p = 60.

  The stock is worth $60 on the 30th and 40th days after purchase.

GEOMETRY  Find the value of x for each figure. Round to the nearest tenth if necessary.

32. area = 45 in2

SOLUTION:  

  The height cannot be negative. So, x ≈ 6.3.

33. area = 110 ft2

SOLUTION:  

  The dimensions of the rectangle cannot be negative. So, x ≈ 5.3.

34. NUMBER THEORY The product of two consecutive even integers is 224. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers are 14 and 16 or –16 and –14.

35. CCSS PRECISION The product of two consecutive negative odd integers is 483. Find the integers.

SOLUTION:  Let x = the first integer and x + 2 = the second integer.

  The integers must be negative. So, they are –23 and –21.

36. GEOMETRY Find the area of the triangle.

SOLUTION:  

  The dimensions of the triangle cannot be negative. So, x = 18. The base of the triangle is 18 meters and the height is 18 + 6 or 24 meters.

The area of the triangle is 216 square meters.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

37. 0.2x2 − 0.2x − 0.4 = 0

SOLUTION:  

  The solutions are –1 and 2.

38. 0.5x2 = 2x − 0.3

SOLUTION:  

  The solutions are about 0.2 and 3.8.

39. 2x2 −

SOLUTION:  

  The solutions are about 0.2 and 0.9.

40. 

SOLUTION:  

  The solutions are –0.5 and 2.5.

41. 

SOLUTION:  

  The solutions are about –8.2 and 0.2.

42. 

SOLUTION:  

  The solutions are about –5.1 and 0.1.

43. ASTRONOMY The height of an object t seconds after it is dropped is given by the equation ,

where h0 is the initial height and g is the acceleration due to gravity. The acceleration due to gravity near the surface

of Mars is 3.73 m/s2, while on Earth it is 9.8 m/s

2. Suppose an object is dropped from an initial height of 120 meters

above the surface of each planet. a. On which planet would the object reach the ground first? b. How long would it take the object to reach the ground on each planet? Round each answer to the nearest tenth. c. Do the times that it takes the object to reach the ground seem reasonable? Explain your reasoning.

SOLUTION:  a. The object on Earth will reach the ground first because it is falling at a faster rate.   b. Mars:

So, t ≈ 8.0 seconds. Earth:

So, t ≈ 4.9 seconds.   c. Sample answer: Yes; the acceleration due to gravity is much greater on Earth than on Mars, so the time to reach the ground should be much less.

44. Find all values of c that make x2 + cx + 100 a perfect square trinomial.

SOLUTION:  

So, c can be –20 or 20 to make the trinomial a perfect square.

45. Find all values of c that make x2 + cx + 225 a perfect square trinomial.

SOLUTION:  

So, c can be –30 or 30 to make the trinomial a perfect square.

46. PAINTING Before she begins painting a picture, Donna stretches her canvas over a wood frame. The frame has a length of 60 inches and a width of 4 inches. She has enough canvas to cover 480 square inches. Donna decides to increase the dimensions of the frame. If the increase in the length is 10 times the increase in the width, what will the dimensions of the frame be?

SOLUTION:  Let x = the increase in the width and let 10x = the increase in the length. So, x + 4 = the new width and 10x + 60 = the new length.  

  The dimensions cannot be negative, so x = 2. The width is 2 + 4 or 6 inches and the length is 10(2) + 60 or 80 inches.

47. MULTIPLE REPRESENTATIONS In this problem, you will investigate a property of quadratic equations. a. TABULAR  Copy the table shown and complete the second column.

b. ALGEBRAIC Set each trinomial equal to zero, and solve the equation by completing the square. Complete the last column of the table with the number of roots of each equation.

c. VERBAL Compare the number of roots of each equation to the result in the b2 − 4ac column. Is there a

relationship between these values? If so, describe it.

d. ANALYTICAL Predict how many solutions 2x2 − 9x + 15 = 0 will have.  Verify your prediction by solving the 

equation.

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3    

3x2 + 6x + 9    

x2 − 2x + 7    

x2 + 10x + 25    

x2 + 3x − 12    

SOLUTION:  a.

b.

The trinomial x2 − 8x + 16 has 1 root.

 

The trinomial 2x

2 − 11x + 3 has 2 roots.

 

The square root of a negative number has no real roots. So, the trinomial 3x2 + 6x + 9 has 0 roots.

 

The square root of a negative number has no real roots. So, the trinomial x2 − 2x + 7.

 

The trinomial x2 + 10x + 25 has 1 root.

 

The trinomial x2 + 3x − 12 has 2 roots.

 

c. If b2 − 4ac is negative, the equation has no real solutions. If b

2 − 4ac is zero, the equation has one solution. If b2

− 4ac is positive, the equation has 2 solutions. d.

The equation 2x2 − 9x + 15 = 0 has 0 real solutions because b

2 − 4ac is negative.

 

The equation cannot be solved because the square root of a negative number has no real roots.

Trinomial b2 − 4ac Number

of Roots

x2 − 8x + 16 (–8)

2 – 4(1)(16) = 0 1

2x2 − 11x + 3 (–11)

2 – 4(2)(3) = 97  

3x2 + 6x + 9 (6)

2 – 4(3)(9) = −72  

x2 − 2x + 7 (–2)

2 – 4(1)(7) = −24  

x2 + 10x + 25 (10)

2 – 4(1)(25) = 0  

x2 + 3x − 12 (3)

2 – 4(1)(–12) = 57  

Trinomial b2 − 4ac Number of

Roots

x2 − 8x + 16 0 1

2x2 − 11x + 3 97 2

3x2 + 6x + 9 −72 0

x2 − 2x + 7 −24 0

x2 + 10x + 25 0 1

x2 + 3x − 12 57 2

48. CCSS PERSEVERANCE Given y = ax2 + bx + c with a ≠ 0, derive the equation for the axis of symmetry by 

completing the square and rewriting the equation in the form y = a(x − h)2 + k .

SOLUTION:  

Let    and   , then the equation becomes y = (x - h)2 + k .

For an equation in the form y = ax2 + bx + c,  the equation for the axis of symmetry is   .

So, for an equation in the form  y = (x - h)2 + k, the equation for the axis of symmetry becomes x = h.

49. REASONING Determine the number of solutions x2 + bx = c has if . Explain.

SOLUTION:  

None; Sample answer: If you add to each side of the equation and each side of the inequality, you get

. Since the left side of the last equation is a perfect square, it cannot

equal the negative number . So, there are no real solutions.

50. WHICH ONE DOESN’T BELONG? Identify the expression that does not belong with the other three. Explain your reasoning.

 

SOLUTION:  The first 3 trinomials are perfect squares.

The trinomial  is not a perfect square. So, it does not belong.

51. OPEN ENDED Write a quadratic equation for which the only solution is 4.

SOLUTION:  Find a quadratic that has two roots of 4.

Verify that the solution is 4.

So, the quadratic equation x2 − 8x + 16 = 0 has a solution of 4.

52. WRITING IN MATH Compare and contrast the following strategies for solving x2 − 5x − 7 = 0: completing the

square, graphing, and factoring.

SOLUTION:  Sample answer: Because the leading coefficient is 1, solving the equation by completing the square is simpler and yields exact answers for the solutions.   

 

To solve the equation by graphing, use a graphing calculator to graph the related function y = x2 - 5x - 7. Select the

zero option from the 2nd [CALC] menu to determine the roots. 

   

  The roots are given as decimal approximations. So, for this equation the solutions would have to be given as estimations.   There are no factors of –7 that have a sum of –5, so solving by factoring is not possible.

53. The length of a rectangle is 3 times its width. The area of the rectangle is 75 square feet. Find the length of the rectangle in feet. A 25 B 15 C 10 D 5

SOLUTION:  Let x = the width of the rectangle and let 3x = the length of the rectangle.

The length of the rectangle is 3(5) or 15 feet. Choice B is the correct answer.

54. PROBABILITY At a festival, winners of a game draw a token for a prize. There is one token for each prize. The prizes include 9 movie passes, 8 stuffed animals, 5 hats, 10 jump ropes, and 4 glow necklaces. What is the probabilitythat the first person to draw a token will win a movie pass?

F  

G

H

J  

SOLUTION:  There are 9 + 8 + 5 + 10 + 4 or 36 possible outcomes.

P(movie pass) =  or  . Choice J is the correct answer.

55. GRIDDED RESPONSE The population of a town can be modeled by P = 22,000 + 125t, where P represents the population and t represents the number of years from 2000. How many years after 2000 will the population be 26,000?

SOLUTION:  

In 32 years the population of the town will be 26,000.

56. Percy delivers pizzas for Pizza King. He is paid $6 an hour plus $2.50 for each pizza he delivers. Percy earned $280 last week. If he worked a total of 30 hours, how many pizzas did he deliver? A 250 pizzas B 184 pizzas C 40 pizzas D 34 pizzas

SOLUTION:  Let p = the number of pizzas Percy delivered.

Percy delivered 40 pizzas. Choice C is the correct answer.

Describe how the graph of each function is related to the graph of f (x) = x2.

57. g(x) = −12 + x2

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –12, the translation is

down. So, the graph is shifted down 12 units from the parent function. 

58. h(x) = (x + 2)2

SOLUTION:  

The graph of f (x) = (x – c)2 represents a translation left or right from the parent graph. Since c = –2, the translation

is to the left by 2 units. 

 

59. g(x) = 2x2 + 5

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 2 and c = 5. Since 2 > 0 and  > 1, the graph of y = 2x

2 + 5

is the graph of y = x2 vertically stretched and shifted up 5 units.

60. 

SOLUTION:  

The function can be written f (x) = a(x – b)2, where a = and b = 6. Since > 0 and  < 1, the graph of 

is the graph of y = x2 vertically compressed and shifted right 6 units.

61. g(x) = 6 + x2

SOLUTION:  

The function can be written f (x) = ax2 + c, where a =  and c = 6. Since  > 0 and   > 1, the graph of y = 6 +

x2 is the graph of y = x

2 vertically stretched and shifted up 6 units.

62. h(x) = − 1 −   x2

SOLUTION:  

The function can be written f (x) = –ax2 + c, where a =  and c = –1. Since  < 0 and   > 1, the graph of y

= –1 – x2 is the graph of y = x

2 vertically stretched, shifted down 1 unit and reflected across the x-axis.

63. RIDES A popular amusement park ride whisks riders to the top of a 250-foot tower and drops them. A function for

the height of a rider is h = −16t2 + 250, where h is the height and t is the time in seconds. The ride stops the descent

of the rider 40 feet above the ground. Write an equation that models the drop of the rider. How long does it take to fall from 250 feet to 40 feet?

SOLUTION:  

The ride stops the descent of the rider at 40 feet above ground, so h = 40. Then, the equation 40 = −16t2 + 250

models the drop of the rider.

Time cannot be negative. So, it takes about 3.6 seconds to complete the ride.

Simplify. Assume that no denominator is equal to zero.

64. 

SOLUTION:  

65. 

SOLUTION:  

66. 

SOLUTION:  

67. 

SOLUTION:  

68. 

SOLUTION:  

69. b3(m

−3)(b

−6)

SOLUTION:  

Solve each open sentence.

70. |y − 2| > 7

SOLUTION:  

The solution set is {y |y > 9 or y < −5}.

Case 1 y – 2 is positive.

and Case 2 y – 2 is negative.

 

71. |z + 5| < 3

SOLUTION:  

The solution set is {z |−8 < z < −2}.

Case 1 z +5 is positive.

and Case 2 z +5 is negative.

 

72. |2b + 7| ≤ −6

SOLUTION:  

cannot be negative. So,  cannot be less than or equal to –6. The solution set is empty.

73. |3 − 2y | ≥ 8

SOLUTION:  

The solution set is {y |y  ≥ 5.5 or y  ≤ −2.5}.

Case 1 3 – 2y is positive.

and Case 2 3 – 2y is negative.

 

74. |9 − 4m| < −1

SOLUTION:  

cannot be negative. So, cannot be less than or equal to –1. The solution set is empty.

75. |5c − 2| ≤ 13

SOLUTION:  

The solution set is {c|−2.2 ≤ c ≤ 3}.

Case 1 5c – 2 is positive.

and Case 2 5c – 2 is negative.

 

Evaluate  for each set of values. Round to the nearest tenth if necessary.

76. a = 2, b = −5, c = 2

SOLUTION:  

77. a = 1, b = 12, c = 11

SOLUTION:  

78. a = −9, b = 10, c = −1

SOLUTION:  

79. a = 1, b = 7, c = −3

SOLUTION:  

80. a = 2, b = −4, c = −6

SOLUTION:  

81. a = 3, b = 1, c = 2

SOLUTION:  

This value is not a real number.

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9-4 Solving Quadratic Equations by Completing the Square