chapter solutions key 5 quadratic functions

Solutions Key

Quadratic Functions5CHAPTER

ARE YOU READY? PAGE 311

1. E 2. C

3. A 4. B

5. (3.2)2

= (3.2)(3.2)

= 10.24

6. ( 2__5)

2

= ( 2__5

)( 2__5

)

= 4___25

7. √ �� 121 = 11 8. √ �� 1___16

= 1__4

9. √ �� 72

= √ �� 36 · 2= 6 √ � 2

10. 2( √ �� 144 - 4)

= 2(12 - 4)

= 2(8)

= 16

11. √ �� 33 · √ �� 75

= √ �� 2475

= √ �� 225 · 11

= 15 √ � 11

12. √ �� 54____ √ � 3

= √ �� 18

= √ � 9 · 2= 3 √ � 2

13. (x - 2)(x - 6)

= x2 - 6x - 2x + 12

= x2 - 8x + 12

14. (x + 9)(x - 9)

= x2 - 9x + 9x - 81

= x2 - 81

15. (x + 2)(x + 7)

= x2 + 7x + 2x + 14

= x2 + 9x + 14

16. (2x - 3)(5x + 1)

= 10 x2 + 2x - 15x - 3

= 10 x2 - 13x - 3

17. 2x + 10 = -32

2x = -32 - 10

2x = -42

x = -21

18. 2x - (1 - x ) = 2

2x - 1 + x = 2

3x = 2 + 1

3x = 3

x = 1

19.2__3

(x - 1) = 11

2__3

x - 2__3

= 11

2__3

x = 35___3

x = 105____6

x = 17 1__2

20. 2(x + 5) - 5x = 1

2x + 10 - 5x = 1

-3x = -9

x = 3

21. 22.

23. 24.

5-1 USING TRANSFORMATIONS TO

GRAPH QUADRATIC FUNCTIONS,

PAGES 315-322

CHECK IT OUT!

1. x g(x) = - x2 + 6x - 8 (x, g(x))

1 g(1) = -(1)2 + 6(1) - 8 = -3 (1, -3)

2 g(2) = -(2)2 + 6(2) - 8 = 0 (2, 0)

3 g(3) = -(3)2 + 6(3) - 8 = 1 (3, 1)

4 g(4) = -(4)2 + 6(4) - 8 = 0 (4, 0)

5 g(5) = -(5)2 + 6(5) - 8 = -3 (5, -3)

2a. g is f translated 5 units down.

b. g is f ranslated 3 units left and 2 units down.

3a. g is a horizontal compression of f by a factor of 1__2

.

b. g is f reflected across the x-axis and vertically

compressed by a factor of 1__2

.

153 Holt McDougal Algebra 2

4a. g(x) = 1__3

(x - 2 )2 - 4 b. g(x) = -(x + 5 )

2 + 1

5.dn(v)_____d(v)

= 0.039v

2_______0.045v

2=

13___15

Vertical compression by a factor of 13___15

; the braking

distance wiil be less with optimally inflated new tires

than with tires having more wear.

THINK AND DISCUSS

1. Possible answer: a indicates a reflection, vertical

stretch or vertical compression. h indicates a

horizontal translation (left or right). k indicates a

vertical translation

(up or down).

2. Possible answer: The function for which a is

greater will have a narrower graph.

3.

EXERCISES

GUIDED PRACTICE

1. vertex

2. x -2 -1 0 1 2

f(x) -12 -6 -4 -6 -12

3. x -1 0 1 2 3

g(x) -6 -2 0 0 -2

4.x -2 -1 0 1 2

h(x) 0 -1 0 3 8

5. d is f translated 4 units right.

6. g is f translated 3 units right and 2 units up.

7. h is f translated 1 unit left and 3 units down.


8. g is a vertical stretch of f by a factor of 3.

9. h is a horizontal stretch of f by a factor of 8.

10. p is a vertical compression of f by a factor of 0.25.

11. h is f reflected across the x-axis and horizontally


.

12. g is a vertical stretch of f by a factor of 4.2.

13. d is f reflected across the x-axis and vertically


.

14. g(x) = 2(x + 3 )2

15. h(x) = - x2 - 6

16. Vertical compression by a factor of 415____592

; the safe

working load is less for an old rope than for a newer

rope of the same radius.

PRACTICE AND PROBLEM SOLVING

17. x -2 -1 0 1 2

f(x) 0 3 4 3 0

18. x -2 -1 0 1 2

g(x) 9 4 1 0 1

19. x -2 -1 0 1 2

h(x) -1 -3 -1 5 15

20. g is f translated 2 units down.

21. h is f translated 5 units left.

22. j is f translated 1 unit right.


23. g is f translated 4 units left and 3 units down.

24. h is f translated 2 units up and 2 units left.

25. h is f translated 4 units right and 9 units down.

26. g is a vertical compression of f by a factor of 4__7

.

27. h is f reflected across the x-axis and vertically stretched by a factor of 20.

28. j is a horizontal stretch of f by a factor of 3.

29. g(x) = - 1__2

(x - 1)2 30. h(x) = 2.5(x + 2 )2 + 1

31. Vertical translaton; at any given speed, the gas mileage for an SUV is 18 mi/gal less than for a compact car.

32a. Translation 10 units right and 300 units up.

b. No; the largest pen Keille can build with an 80 ft roll has an area of 400 f t2 , and the largest pen she can build with a 40 ft roll has an area of 100 ft2.Therefore, a roll that is twice as long allows her to build a pen with 4 times the area.

33. p is f reflected across the x-axis and translated 4 units right.

34. g is f vertically stretched by a factor of 8 and translated 2 units left.

35. h is f vertically stretched by a factor of 4 and translated 2 units down.

36. p is f vertically compressed by a factor of 1__4

and translated 2 units up.

37. g is f horizontally compressed by a factor of 1__3

and translated 1 unit up.

38. h is f reflected across the x-axis and horizontally stretched by a factor of 3.

39. C 40. B

41. A

42a. B(r) = 25 - π r2

b. B is A reflected across the x-axis and translated 25 units up.

c. A: D: {r | 0 ≤ r ≤ 2.5}; R: {A | 0 ≤ A ≤ 6.25π}.B: D: {r | 0 ≤ r ≤ 2.5};

R: {B | (25 - 6.25π) ≤ B ≤ 25}. Possible answer: The radius of the circle cannot be less than 0 or greater than half the side length of the square.

43. Horizontal line; linear or constant function.

44. Very narrow parabola opening upward with its vertex at (-5, 5).

45a. Vertical compression by a factor of 0.38 and translation 2.5 units right and 59 units up.

b. y = -6.08(t - 4 )2 + 95

TEST PREP

46. B 47. J

48. C 49. G

50. 5

CHALLENGE AND EXTEND

51. y = -3(x - 3 )2 + 3Translation 6 units up and 6 units right.

52a. f: horizontal compression by a factor of 1__2

and translation 2 units down.g: vertical stretch by a factor of 4 and translation 2 units down.

b.

c. The functions are the same.

d. g(x) = (3x)2

SPIRAL REVIEW

53. Yes; the price is justified because the volume of the large container is more than 3 times the volume of the small container: Vsmall = 28π; V large = 88π.

54. f(x) = x 55. f(x) = √ � x


56. 2y + 5x = 14 2y = -5x + 14

y = - 5__2

x + 7

57. x - 1__2

y + 4 = -1

- 1__2

y = -x - 5

y = 2x + 10

5-2 PROPERTIES OF QUADRATIC

FUNCTIONS IN STANDARD FORM,

PAGES 323-330

CHECK IT OUT!

1. x = 3

2a. (a) downward

(b) x = -b___2a

= -(-4)______2(-2)

= 4___-4

= -1

(c) f(-1) = -2(-1)2 - 4(-1)

= -2(1) + 4 = -2 + 4 = 2 The vertex is (-1, 2).

(d) The y-intercept is 0.

(e)

b. (a) upward

(b) x = -b___2a

= -3____2(1)

= - 3__2

(c) g(- 3__2) = (-

3__2)

2 + 3 (-

3__2) - 1

= 9__4

- 9__2

- 1

= - 13___4

The vertex is (- 3__2

, - 13___4 ) .

(d) The y-intercept is -1.

(e)

3a. x = -b___2a

= -(-6)______2(1)

= 6__2

= 3

f(3) = ( 3)2 - 6(3) + 3 = 9 - 18 + 3 = -6The minimum is -6.

D: �;R: {y | y ≥ -6}.

b. x = -b___2a

= 0_____2(-2)

= 0

g(0) = -2(0)2 - 4

= -4The maximum is -4.

D: �;R: {y | y ≤ -4}.

4. s = -b___a

= -2.45_________2(-0.025)

= -2.45______-0.05

= 49

m(49) = -0.025(49)2 + 2.45(49) - 30

= -0.025(2401) + 120.05 - 30 = 30.025The maximum mileage is 30.0 mi/gal at a speed of 49 mi/h.

THINK AND DISCUSS

1. Possible answer: No; quadratic functions open in 1 direction. If they open upward, they have a minimum value. If they open downward, they have a maximum value.

2. Possible answer: The value of x2 increases faster than the value of 2x decreases.

3.

EXERCISES

GUIDED PRACTICE

1. minimum 2. x = 2

3. x = 0 4. x = -5

5a. downward b. x = -b___2a

= -(-2)______2(-1)

= -1

c. f(-1) = -(-1)2 - 2(-1) - 8

= -1 + 2 - 8 = -7The vertex is (-1, -7).

d. The y-intercept is -8.

e.


6a. upward

b. x = -b___2a

= -(-3)______2(1)

= 3__2

c. g( 3__2) = ( 3__

2)2 - 3 ( 3__

2) + 2

= 9__4

- 9__2

+ 2

= - 1__4

The vertex is ( 3__2

, - 1__4) .

d. The y-intercept is 2.

e.

7a. downward

b. x = -b___2a

= -4_____2(-1)

= 2

c. h(2) = -(2)2 + 4(2) - 1

= -4 + 8 - 1 = 3The vertex is (2, 3).


e.

8. x = -b___2a

= 0____2(1)

= 0

f(0) = -1The minimum is -1.

D: �;R: {y | y ≥ -1}.

9. x = -b___2a

= -3_____2(-1)

= 3__2

g( 3__2) = - ( 3__

2)2

+ 3 ( 3__2) - 2

= - 9__4

+ 9__2

- 2

= 1__4

The maximum is 1__4

.

D: �;R: {y | y ≤ 1__

4}.

10. x = -b___2a

= -32______2(-16)

= 1

h(1) = -16(1)2 + 32(1) + 4

= -16 + 32 + 4 = 20The maximum is 20.

D: �;R: {y | y ≤ 20}.

11. x = -b___2a

= -0.25_________2(-0.005)

= 25

h(25) = -0.005(25)2 + 0.25(25)

= -3.125 + 6.25 = 3.125The maximum height is 3.125 m.


12. x = 0 13. x = 1

14. x = -1

15a. upward

b. x = -b___2a

= -1____2(1)

= - 1__2

c. f(- 1__2) = (-

1__2)

2 + (-

1__2) - 2

= 1__4

- 1__2

- 2

= - 9__4


, - 9__4) .


e.

16a. downward

b. x = -b___2a

= -6_____2(-3)

= 1

c. g(1) = -3(1)2 + 6(1)

= -3 + 6= 3

The vertex is (1, 3).


e.

17a. upward

b. x = -b___2a

= -(-2)______2(0.5)

= 2

c. h(2) = 0.5(2 )2 - 2(2) - 4

= 0.5(4) - 4 - 4= -6

The vertex is (2, -6).


e.


18a. downward

b. x = -b___2a

= -8_____2(-2)

= 2

c. f(2) = -2(2)2 + 8(2) + 5

= -8 + 16 + 5= 13



e.

19a. upward

b. x = -b___2a

= -2____2(3)

= - 1__3

c. g(- 1__3) = 3 (-

1__3)

2 + 2 (-

1__3) - 8

= 1__3

- 2__3

- 8

= - 25___3


, - 25___3 ) .


e.

20a. upward

b. x = -b___2a

= -2____2(1)

= -1

c. h(-1) = 2(-1) - 1 + (-1)2

= -2 -1 + 1= -2

The vertex is (-1, -2).


e.

21a. downward

b. x = -b___2a

= 0_____2(-1)

= 0

c. f(0) = -2 - (0 )2 = -2



e.

22a. upward

b. x = -b___2a

= -3_____2(0.5)

= -3

c. g(-3) = 0.5(-3)2 + 3(-3) - 5= 4.5 - 9 - 5= -9.5

The vertex is (-3, -9.5).


e.

23a. upward

b. x = -b___2a

= -1____2( 1__

4)= -2

c. h(-2) = 1__4

(-2)2 - 2 + 2

= 1 - 2 + 2= 1

The vertex is (-2, 1).


e.

24. x = -b___2a

= -7_____2(-2)

= 7__4

f( 7__4) = -2 ( 7__

4)2 + 7 ( 7__

4) - 3

= -2( 49___16) + 49___

4- 3

= - 49___8

+ 98___8

- 24___8

= 3.125The maximum is 3.125.

D: �;R: {y | y ≤ 3.125}.

25. x = -b___2a

= -6_____2(-1)

= 3

g(3) = 6(3) - (3 )2

= 18 - 9 = 9The maximum is 9.

D: �;R: {y | y ≤ 9}.


26. x = -b___2a

= -(-4)______2(1)

= 2

h(2) = (2 )2 - 4(2) + 3

= 4 - 8 + 3 = -1The minimum is -1.

D: �;R: {y | y ≥ -1}.

27. x = -b___2a

= 0______2(-

1__2)

= 0

f(0) = - 1__2

(0)2 - 4 = -4

The maximum is -4.

D: �;R: {y | y ≤ -4}.

28. x = -b___2a

= -(-6)______2(-1)

= -3

g(-3) = -(- 3)2 -6(-3) + 1

= -9 + 18 + 1 = 10The maximum is 10.

D: �;R: {y | y ≤ 10}.

29. x = -b___2a

= -8____2(1)

= -4

h(-4) = (-4)2 + 8(-4) + 16

= 16 - 32 + 16 = 0The minimum is 0.

D: �;R: {y | y ≥ 0}.

30. d = -b___2a

= -0.657__________ 2(-0.0018)

= 182.5

T = -0.0018(182.5)2 + 0.657(182.5) + 50.95

= -59.95 + 119.9 + 50.95 ≈ 111

The maximum temperature in 2003 is approximately 111°.

31. Maximum height is 64 ft.Possible answer: The axis of symmetry is halfway between any 2 points with the same y-value. Because the points (1, 48) and (3, 48) have the same y-value, the axis of symmetry is x = 2. Because the vertex lies on the axis of symmetry, the vertex of the graph is (2, 64). Therefore, the maximum value of the function is 64.

32a. C(x) = x(32 - 2x)

b. x 0 4 8 12 16

C(x) 0 96 128 96 0

c. D: {0 ≤ x ≤ 16}; R:{y | 0 ≤ y ≤ 128}Neither the width nor the area can be negative.

d. x = 8 cm

33a. t = -b___2a

= -3000________2(-4000)

= 0.375

h(0.375) = -4000(0.375)2 + 3000(0.375)= -562.5 + 1125= 562.5 mm

b. 93.75 to 1. Possible answer: The ratio for spittle bugs is more than 67 times as great as the ratio for humans.

c.x___

1.8= 93.75

x = 168.75 m

34. A(x) = 10x - x2

x = -b___2a

= -10_____2(-1)

= 5

A(5) = 10(5) - (5 )2 = 50 - 25 = 25 yd2

35. min ≈ -3.029771 36. max ≈ 13.178533

37. min ≈ -1.253333 38. max = 5.3715

39. The axis of symmetry is halfway between any 2 points with the same y-value. Halfway between -7 and 3 is -2. Therefore, the axis of symmetry is x = -2.

40. Yes; possible answer: a function such as f(x) = - x

2 - 5 may open downward and have a vertex below the x-axis. A function such as f(x) = x

2 + 2 may open upward and have a vertex above the x-axis.

41a. t = -b___2a

= -50______2(-16)

≈ 1.6 s

b. h(1.5625) = -16(1.5625)2 + 50(1.5625) + 6

= -39.0625 + 78.125 + 6≈ 45 ft

TEST PREP

42. C 43. G

44. B 45. G


46. Because a is negative, the graph will open

downward and have a maximum value. To find the

maximum value, find the x-value of the vertex:

x = -b___2a

= -(-8)______2(-1)

= -4.

Then evaluate the function for x = -4.

f(-4) = (-4)2 - 8(-4) + 4 = 20.

The maximum value is 20.


47. Possible answer: f(x) = x2 + 2x + 4;

g(x) = - x2 - 2x + 2.

48a. (11, 8); possible answer: the point (-5, 8) is

8 units left of the axis of symmetry. The graph of

the quadratic function must also pass through a

point that is 8 units right of the axis of symmetry

and has the same y-value as (-5, 8). This point has

coordinates (11, 8).

b. No; possible answer: you would need to know

the coordinates of at least one other point on the

function’s graph to determine whether it opens

upward or downward.

49. Possible answer: The function has no x-term,

so b = 0. Therefore, the axis of symmetry is x = 0,

and the vertex is (0, c).

50. If the value of the function is the same for different

x-values, the axis of symmetry is halfway between

these x-values. In this case, the axis of symmetry is

x = -1 + 2_______

2= 1__

2.

SPIRAL REVIEW

51. √ �� 40 · √ �� 180

= √ �� 7200

= √ �� 3600 · 2

= 60 √ � 2

52. 2 √ � 8 · 4 √ � 3

= 8 √ �� 24

= 8 √ �� 4 · 6

= 16 √ � 6

53. √ �� 54 ÷ √ �� 30

= √ ��

54___30

= √ �

9__5

= √ ��

9 · 5____5 · 5

= 3 √ � 5____

5

54. √ �� 304

= √ �� 16 · 19

= 4 √ �� 19

55. f(0) = (0 - 3 )2 + 1

= 9 + 1

= 10

f( 1__2) = ( 1__

2- 3)

2 + 1

= 25___4

+ 1

= 29___4

f(-2) = (-2 - 3)2 + 1

= 25 + 1

= 26

56. g(0) = 2 (0 - 1__2)

2

= 2 ( 1__4)

= 1__2

g( 1__2) = 2 ( 1__

2- 1__

2)2

= 2(0)

= 0

g(-2) = 2 (-2 - 1__2)

2

= 2 ( 25___4 )

= 25___2

57. f(0) = -4(0 + 5)

= -4(5)

= -20

f( 1__2) = -4( 1__

2+ 5)

= -4( 11___2 )

= -22

f(-2) = -4(-2 + 5)

= -4(3)

= -12

58. g(0) = (0 )3 - 4(0) + 8

= 0 - 0 + 8

= 8

g( 1__2) = ( 1__

2)3 - 4 ( 1__

2) + 8

= 1__8

- 2 + 8

= 49___8

g(-2) = (-2)3 - 4(-2) + 8

= -8 + 8 + 8

= 8

59. y = mx + b

-4 = 3(1) + b

-4 = 3 + b

b = -7

y = 3x - 7

60. m = -7 - 5_________

-1 - (-3)= -6

y = mx + b

-7 = -6(-1) + b

-7 = 6 + b

b = -13

y = -6x - 13

61. y = mx + b

5 = -2(3) + b

5 = -6 + b

b = 11

y - 5 = - 2 (x - 3) or

y = -2x + 11

62. m = 1 - 6_______

-2 - 4=

5__6

y = mx + b

1 = 5__6

(-2) + b

1 = - 5__3

+ b

b = 8__3

y - 6 = 5__6

(x - 4) or

y = 5__6

x + 8__3


5-3 SOLVING QUADRATIC EQUATIONS

BY GRAPHING AND FACTORING,

PAGES 333-340

CHECK IT OUT!

1. x = -b___2a

= -(-2)______2(-1)

= -1

x -3 -2 -1 0 1

f(x) 0 3 4 3 0

The zeros of the function are x = -3 and 1.

2a. f(x) = x2 - 5x - 6

x 2 - 5x - 6 = 0

(x - 6)(x + 1) = 0

x - 6 = 0 or x + 1 = 0

x = 6 or x = -1

b. g(x) = x2 - 8x

x2 - 8x = 0

x(x - 8) = 0

x = 0 or x - 8 = 0

x = 0 or x = 8

3. h(t) = -16t2 + vt + h

= -16t2 + 48t + 0

= -16t(t - 3)

-16t(t - 3) = 0

-16t = 0 or t - 3 = 0

t = 0 or t = 3

The ball is in the air for 3 s.

4a. x2 - 4x = -4

x 2 - 4x + 4 = 0

(x - 2)(x - 2) = 0

x - 2x = 0

x = 2

b. 25 x2 = 9

25 x2 - 9 = 0

(5x + 3)(5x - 3) = 0

5x + 3 =0 or 5x - 3 = 0

x = - 3__5

or x = 3__5

5. Possible answer:

x = 5 or x = -5

x - 5 = 0 or x + 5 = 0

(x - 5)(x + 5) = 0

x2 - 25 = 0

f(x)= x2 - 25

THINK AND DISCUSS

1. Possible answer: The function has 1 distinct

real zero.

2. Possible answer: Both linear and quadratic functions

have exactly one y-intercept. Linear functions have

at most one x-intercept, and quadratic functions

have at most two x-intercepts.

3. Either the parabola opens downward and the

maximum is less than 0, or the parabola opens

upward and the minimum is greater than zero.

4.

EXERCISES

GUIDED PRACTICE

1. roots

2. x = -b___2a

= -4____2(1)

= -2

x -5 -4 -2 0 1

f(x) 0 -5 -9 -5 0


3. x = -b___2a

= -6_____

2(-1)= 3

x 1 2 3 4 5

f(x) -3 0 1 0 -3

The zeros of the function are x = 2 and 4.

4. x = -b___2a

= 0____

2(1)= 0

x -2 -1 0 1 2

f(x) 3 0 -1 0 3


5. f(x) = x2 - 7x + 6

x2 - 7x + 6 = 0

(x - 6)(x - 1) = 0

x - 6 = 0 or x - 1 = 0

x = 6 or x = 1

6. g(x) = 2 x2 - 5x + 2

2 x2 - 5x + 2 = 0

(2x - 1)(x - 2) = 0

2x - 1 = 0 or x - 2 = 0

x = 1__2

or x = 2

7. h(x) = x2 + 4x

x2 + 4x = 0

x(x + 4) = 0

x = 0 or x + 4 = 0

x = 0 or x = -4

8. f(x) = x2 + 9x + 20

x2 + 9x + 20 = 0

(x + 4)(x + 5) = 0

x + 4 = 0 or x + 5 = 0

x = -4 or x = -5

9. g(x) = x2 - 6x - 16

x2 - 6x - 16 = 0

(x - 8)(x + 2) = 0

x - 8 = 0 or x + 2 = 0

x = 8 or x = -2

10. h(x) = 3 x2 + 13x + 4

3 x2 + 13x + 4 = 0

(3x + 1)(x + 4) = 0

3x + 1 = 0 or x + 4 = 0

x = - 1__3

or x = -4

11. h(t) = -16t2 + 63 t + 4

-16t2 + 63t + 4 = 0

- (16t2 - 63t - 4) = 0

-(16t + 1)(t - 4) = 0

16t + 1 = 0 or t - 4 = 0

t = - 1___16

or t = 4

The arrow is in the air at for 4 s.

12. x2 - 6x = -9

x2 - 6x + 9 = 0

(x - 3)(x - 3) = 0

x - 3 = 0

x = 3

13. 5 x2 + 20 = 20x

5x2 - 20x + 20 = 0

5 (x2 - 4x + 4) = 0

5(x -2)(x - 2) = 0

x - 2 = 0

x = 2

14. x2 = 49

x2 - 49 = 0

(x + 7)(x - 7) = 0

x + 7 = 0 or x - 7 = 0

x = -7 or x = 7

15. x = 3 or x = 4

x - 3 = 0 or x - 4 = 0

(x - 3)(x - 4) = 0

x2 - 7x + 12 = 0

f(x) = x2 - 7x + 12


16. x = -4 or x = -4

x + 4 = 0 or x + 4 = 0

(x + 4)(x + 4) = 0

x2 + 8x + 16 = 0

f(x) = x2 + 8x + 16

17. x = 3 or x = 0

x - 3 = 0 or x = 0

(x - 3)x = 0

x2 - 3x = 0

f(x) = x2 - 3x


18. x = -b___2a

= -4_____2(-1)

= 2

x 0 1 2 3 4

f(x) -3 0 1 0 -3

The zeros of the function are x = 1 and 3.

19. x =-b___2a

=-1____2(1)

= - 1__2

x -3 -1 - 1__2 0 2

f(x) 0 -6 -6.25 -6 0


20. x =-b___2a

=0____

2(1)= 0

x -3 -1 0 1 3

f(x) 0 -8 -9 -8 0


21. f(x) = x2 + 11x + 24

x2 + 11x + 24 = 0

(x + 3)(x + 8) = 0

x + 3 = 0 or x + 8 = 0

x = -3 or x = -8

22. g(x) = 2 x2 + x - 10

2 x2 + x - 10 = 0

(x - 2)(2x + 5) = 0

x - 2 = 0 or 2x + 5 = 0

x = 2 or x = - 5__2

23. h(x) = - x2 + 9x

- x2 + 9x = 0

-x(x - 9) = 0

x = 0 or x - 9 = 0

x = 0 or x = 9

24. f(x) = x2 - 15x + 54

x2 - 15x + 54 = 0

(x - 6)(x - 9) = 0

x - 6 = 0 or x - 9 = 0

x = 6 or x = 9

25. g(x) = x2 + 7x - 8

x2 + 7x - 8 = 0

(x -1)(x + 8) = 0

x - 1 = 0 or x + 8 = 0

x = 1 or x = -8

26. h(x) = 2 x2 - 12x + 18

2x2 - 12x + 18 = 0

2 (x2 - 6x + 9) = 0

2(x - 3 )2 = 0

x - 3 = 0

x = 3 27. h(t) = -16t

2 + 256

-16t2 + 256 = 0

-16(t2 - 16) = 0

-16(t + 4)(t - 4) = 0

t + 4 = 0 or t - 4 = 0

t = -4 or t = 4

It will take 4 s.

28. x2 + 8x = -16

x 2 + 8x + 16 = 0

(x + 4 )2 = 0

x + 4 = 0

x = -4

29. 4 x2 = 81

4 x2 - 81 = 0

(2x + 9)(2x - 9) = 0

2x + 9 = 0 or 2x - 9 = 0

x = - 9__2

or x = 9__2

30. 9 x2 + 12x + 4 = 0

(3x + 2 )2 = 0

3x + 2 = 0

x = - 2__3

31. 36 x2 - 9 = 0

9 (4x2 - 1) = 0

9(2x + 1)(2x - 1) = 0

2x + 1 = 0 or 2x - 1 = 0

x = - 1__2

or x = 1__2

32. x2 - 10x + 25 = 0

(x - 5 )2 = 0

x - 5 = 0

x = 5

33. 49 x2 = 28x - 4

49x2 - 28x + 4 = 0

(7x - 2 )2 = 0

7x - 2 = 0

x = 2__7

34. Possible answer:

x = 5 or x = -1

x - 5 = 0 or x + 1 = 0

(x - 5)(x + 1) = 0

x2 - 4x - 5 = 0

f(x) = x2 - 4x - 5


x = 6 or x = 2

x - 6 = 0 or x - 2 = 0

(x - 6)(x - 2) = 0

x2 - 8x + 12 = 0

f(x) = x2 - 8x + 12


x = 3 or x = 3

x - 3 = 0 or x - 3 = 0

(x - 3)(x - 3) = 0

x2 - 6x + 9 = 0

f(x) = x2 - 6x + 9

37. f(x) = 6x - x2

- x2 + 6x = 0

-x(x - 6) = 0

x = 0 or x - 6 = 0

x = 0 or x = 6

38. g(x) = x2 - 25

x2 - 25 = 0

(x + 5)(x - 5) = 0

x + 5 = 0 or x - 5 = 0

x = -5 or x = 5

39. h(x) = x2 - 12x + 36

x 2 - 12x + 36 = 0

(x - 6 )2 = 0

x - 6 = 0

x = 6

40. f(x) = 3 x2 - 12

3 x2 - 12 = 0

3 (x2 - 4) = 0

3(x + 2)(x - 2) = 0

x + 2 = 0 or x - 2 = 0

x = -2 or x = 2

41. g(x) = x2 - 22x + 121

x 2 - 22x + 121 = 0

(x - 11 )2 = 0

x - 11 = 0

x = 11

42. h(x) = 30 + x - x2

- x2 + x + 30 = 0

- (x2 - x - 30) = 0

-(x - 6)(x + 5) = 0

x - 6 = 0 or x + 5 = 0

x = 6 or x = -5

43. f(x) = x2 - 11x + 30

x2 - 11x + 30 = 0

(x - 5)(x - 6) = 0

x - 5 = 0 or x - 6 = 0

x = 5 or x = 6

44. g(x) = x2 - 8x - 20

x2 - 8x - 20 = 0

(x - 10)(x + 2) = 0

x - 10 = 0 or x + 2 = 0

x = 10 or x = -2

45. h(x) = 2 x2 + 18x + 28

2 x2 + 18x + 28 = 0

2 (x2 + 9x + 14) = 0

2(x + 2)(x + 7) = 0

x + 2 = 0 or x+ 7 = 0

x = -2 or x = -7

46a. h(t) = 73 - 16 t2

-16t2 + 73 = 9

-16t2 + 64 = 0

-16(t2 - 4) = 0

-16(t + 2)(t - 2) = 0

t + 2 = 0 or t - 2 = 0

t = -2 or t = 2

The woman will fall for 2 s.


b. Possible answer: No; the relationship between

the building height and jump time is quadratic, not

linear. Therefore, a jump that is half as high will not

last half as long.

47a. h(t) = -16t2 + 16t + 5

b. -16t2 + 16t + 5 = 0

- (16t2 - 16t - 5) = 0

-(4t + 1)(4t - 5) = 0

4t + 1 = 0 or 4t - 5 = 0

t = - 1__4

or t = 5__4

The juggler has 1.25 s.

48. x2 - 2x + 1 = 0

(x - 1 )2 = 0

x - 1 = 0

x = 1

49. x2 + 6x = -5

x2 + 6x + 5 = 0

(x + 1)(x + 5) = 0

x + 1 = 0 or x + 5 = 0

x = -1 or x = -5

50. 25 x2 + 40x = -16

25x2 + 40x + 16 = 0

(5x + 4)2 = 0

5x + 4 = 0

x = - 4__5

51. 9 x2 + 6x = -1

9x2 + 6x + 1 = 0

(3x + 1)2 = 0

3x + 1 = 0

x = - 1__3

52. 5 x2 = 45

5 x2 - 45 = 0

5 (x2 - 9) = 0

5(x + 3)(x - 3) = 0

x + 3 = 0 or x - 3 = 0

x = -3 or x = 3

53. x2 - 6 = x

x2 - x - 6 = 0

(x - 3)(x + 2 ) = 0

x - 3 = 0 or x + 2 = 0

x = 3 or x = -2

54a. x = -b___2a

= -2____2(1)

= -1

f(-1) = (-1)2 + 2(-1) - 8

= 1 - 2 - 8 = -9


b. The y-intercept is -8.

c. x2 + 2x - 8 = 0

(x + 4)(x - 2) = 0

x + 4 = 0 or x - 2 = 0

x = -4 or x = 2

d.

55a. x = -b___2a

= 0____

2(1)= 0

g(0 ) = (0 )2 - 16

= -16



c. x2 - 16 = 0

(x + 4)(x - 4) = 0

x + 4 = 0 or x - 4 = 0

x = -4 or x = 4

d.

56a. x = -b___2a

= -(-1)______2(1)

= 1__2

h ( 1__2) = ( 1__

2)2 -

1__2

- 12

= 1__4

- 1__2

- 12

= - 49___4

The vertex is

( 1__2

, -12 1__4) .


c. x2 - x - 12 = 0

(x - 4)(x + 3) = 0

x - 4 = 0 or x + 3 = 0

x = 4 or x = -3

d.

57a. x = -b___2a

= -4_____2(-2)

= 1

f(1) = -2(1)2 + 4(1)

= -2(1) + 4

= 2


b. The y-intercept is 0.

c. -2x2 + 4x = 0

-2x(x - 2) = 0

x = 0 or x - 2 = 0

x = 0 or x = 2

d.

58a. x = -b___2a

= -(-5)______2(1)

= 5__2

g( 5__2) = ( 5__

2)2 - 5 ( 5__

2) - 6

= 25___4

- 25___2

- 6

= - 49___4

The vertex is

(2 1__2

, - 12 1__4) .


c. x2 - 5x - 6 = 0

(x - 6)(x + 1) = 0

x - 6 = 0 or x + 1 = 0

x = 6 or x = -1

d.

59a. x = -b___2a

= -1____2(3)

= - 1__6

h( -1___6 ) = 3 (-

1__6)

2 - 1__

6- 4

= 1___12

- 1__6

- 4 = - 49___12


, -4 1___12) .



c. 3 x2 + x - 4 = 0

(x - 1)(3x + 4) = 0

x - 1 = 0 or 3x + 4 = 0

x = 1 or x = -1 1__3

d.

60a. The two legs are (x - 2) and (x - 4).

The equation wil be (x - 2 )2 + (x - 4 )

2 = x

2 .

b. (x - 2 )2 + (x - 4 )

2 = x

2

(x2 - 4x + 4) + (x2

- 8x + 16)= x2

2 x2 - 12x + 20 = x

2

x2 - 12x + 20 = 0

(x - 2)(x - 10) = 0

x - 2 = 0 or x - 10 = 0

x = 2 or x = 10

c. Possible answer: The solutions represent possible

lengths in centimeters of the hypotenuse.If x =

10, the triangle would have side lengths of 10 cm,

8 cm, and 6 cm. If x = 2, the triangle would have

side lengths of 2 cm, 0 cm, and -2 cm. Because

length cannot be negative, only the solution x = 10

is reasonable.

61. x(x + 16) = 80

x 2 + 16x - 80 = 0

(x - 4)(x + 20) = 0

x - 4 = 0 or x + 20 = 0

x = 4 since x > 0

The dimensions are 4 ft by 20 ft.

62. x(x + 1) = 210

x2 + x - 210 = 0

(x - 14)(x + 15) = 0

x - 14 = 0 or x + 15 = 0

x = 14 since x > 0

x + 1 = (14) + 1 = 15

The dimensions are 14 cm by 15 cm.

63. (x + 2)(x - 3) = 50

x2 - x - 6 = 50

x2 - x - 56 = 0

(x - 8)(x + 7) = 0

x - 8 = 0 or x + 7 = 0

x = 8 since x > 0

x + 2 = (8) + 2 = 10; x - 3 = (8) - 3 = 5

The dimensions are 10 m by 5 m.

64. No; Possible anwer: if a function can be factored as

a binomial squared, the two factors are identical.

When each factor is set to zero, each equation will

have the same solution. Therefore, the function has

only one distinct zero.

65. Possible answer: The Zero Product Property states

that if a product equals zero, then at least one of the

factors must equal zero. The zeros of a function can

be found by writing the function rule as a product

of factors and setting it equal to zero. You can then

determine the value that makes each factor equal to

zero. These values are the zeros of the function.

66a. h(t) = -16t2 + 16t + 5

-16t2 + 16t + 5 = 0

- (16t2 - 16t - 5) = 0

-(4t + 1)(4t - 5) = 0

4t + 1 = 0 or 4t - 5 = 0

t = - 1__4

or t = 5__4

The ball will stay in air for 1.25 s.

b. d(1.25) = 85(1.25) = 106.25

The horizontal distance travelled is 106.25 ft.

TEST PREP

67. B 68. J

69. C

70. x2 + 4x - 21 = 0

(x + 7)(x - 3) = 0

x + 7 = 0 or x - 3 = 0

x = -7 or x = 3

The positive root is x = 3.


71. 3 (x2 - x) = x

2

3x2 - 3x = x

2

2 x2 - 3x = 0

x(2x - 3) = 0

x = 0 or 2x - 3 = 0

x = 0 or x = 3__2

72. x2 = 1__

3x

x 2 - 1__

3x = 0

x (x - 1__3) = 0

x = 0 or x - 1__3

= 0

x = 0 or x = 1__3

73. x2 -

3__4

x + 1__8

= 0

8 x2 - 6x + 1 = 0

(4x - 1)(2x - 1) = 0

4x - 1 = 0 or 2x - 1 = 0

x = 1__4

or x = 1__2

74. x2 + x + 0.21 = 0

(x + 0.3)(x + 0.7) = 0

x + 0.3 = 0 or x + 0.7 = 0

x = -0.3 or x = -0.7


75a. (a + b)(a2 - ab + b

2)= a

3 - a

2b + a b

2 + a

2b - a b

2 + b

3

= a3 + b

3

b. 8 x3 + 27

(2x)3 + (3 )

3

(2x + 3) ((2x)2 - (2x)(3) + 3

2) (2x + 3) (4x

2 - 6x + 9)

c. a3 - b

3 = (a - b)(a2

+ ab + b2)

d. x3 - 1

x3 - 1

3

(x - 1) ((x)2 + (x)(1) + 1

2) (x - 1) (x2

+ x + 1)

SPIRAL REVIEW

76. (1.4 × 108) (6.1 × 1 0

-3)= (1.4 · 6.1) × 1 0

5

= 8.54 × 1 05

77. (2.7 × 1 010) (3.2 × 1 0

2)= (2.7 · 3.2) × 1 0

12

= 8.64 × 1 012

78.3.5 × 1 0

6__________1.4 × 1 0

-4

= ( 3.5___1.4) × 1 0

6

= 2.5 × 1 010

79.3.12 × 1 0

-6

___________4.8 × 1 0

3

= ( 3.12____4.8 ) × 1 0

-9

= 0.65 × 1 0-9

= 6.5 × 1 0-10

80. 12___7.5

= n__5

7.5n = 60

n = 8

81. 1.2___4.8

= w___8.8

4.8w = 10.56

w = 2.2

82. 6.8___4.5

= r___90

4.5r = 612

r = 136

83. h is a vertical compression of f by a factor of 0.5.

84. d is f translated 2 units up.

85. g is f translated 1 unit left.

5-4 COMPLETING THE SQUARE, PAGES

341-348

CHECK IT OUT!

1a. 4 x2 - 20 = 5

4 x2 = 25

2x = ± √ �� 25

2x = ±5

x = ± 5__2

b. x2 + 8x + 16 = 49

(x + 4 )2 = 49

x + 4 = ± √ �� 49

x + 4 = ±7

x = 3 or -11

2a. ( b__2)

2

= ( 4__2)

2 = (2 )

2 = 4

x 2 + 4x + 4 = (x + 2 )

2

b. ( b__2)

2

= ( -4___2 )

2 = (-2)

2 = 4

x2 - 4x + 4 = (x - 2 )

2

c. ( b__2)

2

= ( 3__2)

2 =

9__4

x 2 + 3x +

9__4

= x + ( 3__2)

2

3a. x2 -2 = 9x

x2 - 9x = 2

x 2 - 9x + ( 9__

2)2 = 2 + ( 9__

2)2

x2 - 9x +

81___4

= 2 + 81___4

(x - 9__2)

2 =

89___4

x - 9__2

= ± √ ��

89___4

x - 9__2

= ± √ ��

89___2

x = 9 ± √ �� 89________

2

b. 3 x2 - 24x = 27

x2 - 8x = 9

x 2 - 8x + ( -8___

2 )2 = 9 + ( -8___

2 )2

x2 - 8x + 16 = 9 + 16

(x - 4 )2 = 25

x - 4 = ± √ �� 25

x - 4 = ±5

x = 9 or -1

4a. f(x) = x2 + 24x + 145

= (x2 + 24x) + 145

= (x2 + 24x + ( 24___

2 )2) + 145 - ( 24___

2 )2

= (x2 + 24x + 144) + 145 - 144

= (x + 12 )2 + 1


b. g(x) = 5 x2 - 50x + 128

= 5 (x2 - 10x) + 128

= (5x2 - 10x + ( 10___

2 )2) + 128 - 5 ( 10___

2 )2

= 5( x2 - 10x + 25) + 128 - 125

= 5(x - 5)2 + 3



THINK AND DISCUSS

1. Possible answer: Take the square root of each side

by applying the Square Root Property; rewrite the

equation as a difference of squares and factor.

2. Possible answer: Factor the x2 -term and the x-term

so that the coefficient of the x2 -term is 1. Complete

the square of the x2 -term and the x-term. Factor

the perfect square. Multiply the original coefficient

of the x2 -term by the number added to complete

the square. Subtract this product from the constant

term.

3.

EXERCISES

GUIDED PRACTICE

1. ( b__2)

2

2. (x - 2 )2 = 16

x - 2 = ± √ �� 16

x - 2 = ±4

x = 6 or -2

3. x2 - 10x + 25 = 16

(x - 5 )2 = 16

x - 5 = ± √ �� 16

x - 5 = ±4

x = 9 or 1

4. x2 - 2x + 1 = 3

(x - 1 )2 = 3

x - 1 = ± √ � 3

x = 1 ± √ � 3

5. x2 + 14x + ( 14___

2 )2

= x2 + 14x + 49

= (x + 7 )2

6. x2 - 12x + ( -12____

2 )2

= x2 - 12x + 36

= (x - 6 )2

7. x2 - 9x + ( -9___

2 )2

= x2 - 9x +

81___4

= (x - 9__2)

2

8. x2 - 6x = -4

x 2 - 6x + (- 6__

2)2 = -4 + ( -6___

2 )2

x2 - 6x + 9 = -4 + 9

(x - 3 )2 = 5

x - 3 = ± √ � 5

x = 3 ± √ � 5

9. x2 + 8 = 6x

x 2 - 6x = -8

x 2 - 6x + ( -6___

2 )2 = -8 + ( -6___

2 )2

x 2 - 6x + 9 = -8 + 9

(x - 3)2 = 1

x - 3 = ± √ � 1

x - 3 = ±1

x = 4 or 2

10. 2 x2 - 20x = 8

x2 - 10x = 4

x 2 - 10x + ( -10____

2 )2 = 4 + ( -10____

2 )2

x 2 - 10x + 25 = 4 + 25

(x - 5 )2 = 29

x - 5 = ± √ �� 29

x = 5 ± √ �� 29

11. x2 = 24 - 4x

x2 + 4x = 24

x 2 + 4x + ( 4__

2)2 = 24 + ( 4__

2)2

x2 + 4x + 4 = 24 + 4

(x + 2 )2 = 28

x + 2 = ± √ �� 28

x = -2 ±2 √ � 7

12. 10x + x2 = 42

x 2 + 10x + ( 10___

2 )2 = 42 + ( 10___

2 )2

x2 + 10x + 25 = 42 + 25

(x + 5 )2 = 67

x + 5 = ± √ �� 67

x = -5 ± √ �� 67

13. 2 x2 + 8x - 15 = 0

2 x2 + 8x = 15

2 (x2 + 4x) = 15

2(x2 + 4x + ( 4__

2)2) = 15 + 2 ( 4__

2)2

2 (x2 + 4x + 4) = 15 + 8

2(x + 2)2 = 23

(x + 2 )2 =

23___2

x + 2 = ± √ ��

23___2

x = -2 ± √ �� 46____

2

14. f(x) = x2 + 6x - 3

= (x2 + 6x) - 3

= (x2 + 6x + ( 6__

2)2) - 3 - ( 6__

2)2

= (x2 + 6x + 9) - 3 - 9

= (x + 3 )2 - 12


15. g(x) = x2 - 10x + 11

= (x2 - 10x) + 11

= x2 - 10x + ( -10____

2 )2 + 11 - ( -10____

2 )2

= (x2 - 10x + 25) + 11 - 25

= (x - 5 )2 - 14



16. h(x) = 3 x2 - 24x + 53

= (3x2 - 24x) + 53

= 3 (x2 - 8x) + 53

= 3 (x2 - 8x + ( -8___

2 )2) + 53 - 3 ( -10____

2 )2

= 3 (x2 - 8x + 16) + 53 - 48

= 3(x - 4)2 + 5


17. f(x) = x2 + 8x - 10

= (x2 + 8x) - 10

= (x2 + 8x + ( 8__

2)2) - 10 - ( 8__

2)2

= (x2 + 8x + 16) - 10 - 16

= (x + 4)2 - 26


18. g(x) = x2 - 3x + 16

= (x2 - 3x) + 16

= (x2 - 3x + ( -3___

2 )2) + 16 - ( -3___

2 )2

= (x2 - 3x +

9__4) + 16 -

9__4

= (x - 3__2)

2 +

55___4

The vertex is ( 3__2

, 55___4 ) .

19. h(x) = 3 x2 - 12x - 4

= (3x2 - 12x )- 4

= 3 (x2 - 4x) - 4

= 3 (x2 - 4x + ( -4___

2 )2) - 4 - 3 ( -4___

2 )2

= 3 (x2 - 4x + 4) - 4 - 12

= 3(x - 2)2 - 16



20. (x + 2 )2 = 36

x + 2 = ± √ �� 36

x + 2 = ±6

x = 4 or -8

21. x2 - 6x + 9 = 100

(x - 3 )2 = 100

x - 3 = ± √ �� 100

x - 3 = ±10

x = 13 or -7

22. (x - 3 )2 = 5

x - 3 = ± √ � 5

x = 3 ± √ � 5

23. x2 - 18x + ( -18____

2 )2

= x2 - 18x + 81

= (x - 9 )2

24. x2 + 10x + ( 10___

2 )2

= x2 + 10x + 25

= (x + 5 )2

25. x2 - 1__

2x + ( -0.5_____

2 )2

= x2 - 1__

2x + 1___

16

= (x - 1__4)

2

26. x2 + 2x = 7

x 2 + 2x + ( 2__

2)2 = 7 + ( 2__

2)2

x2 + 2x + 1 = 7 + 1

(x + 1 )2 = 8

x + 1 = ± √ � 8

x = -1 ± 2 √ � 2

27. x2 - 4x = -1

x 2 - 4x + ( -4___

2 )2 = -1 + ( -4___

2 )2

x2 - 4x + 4 = -1 + 4

(x - 2)2 = 3

x - 2 = ± √ � 3

x = 2 ± √ � 3

28. 2 x2 - 8x = 22

x 2 - 4x = 11

x 2 - 4x + ( -4___

2 )2 = 11 + ( -4___

2 )2

x2 - 4x + 4 = 11 + 4

(x - 2 )2 = 15

x - 2 = ± √ �� 15

x = 2 ± √ �� 15

29. 8x = x2 + 12

x2 - 8x = -12

x 2 - 8x + ( -8___

2 )2 = -12 + ( -8___

2 )2

x2 - 8x + 16 = -12 + 16

(x - 4 )2 = 4

x - 4 = ± √ � 4

x - 4 = ±2

x = 6 or 2

30. x2 + 3x - 5 = 0

x2 + 3x = 5

x 2 + 3x + ( 3__

2)2 = 5 + ( 3__

2)2

x2 + 3x +

9__4

= 5 + 9__4

(x + 3__2)

2 =

29___4

x + 3__2

= ± √ ��

29___4

x = -3 ± √ �� 29_________

2

31. 3 x2 + 6x = 1

x2 + 2x = 1__

3

x 2 + 2x + ( 2__

2)2 = 1__

3+ 1

x2 + 2x + 1 = 1__

3+ 1

(x + 1 )2 = 4__

3

x + 1 = ± √ �

4__3

x = -1 ± 2 √ � 3____

3


32. f(x) = x2 - 4x + 13

= (x2 - 4x) + 13

= (x2 - 4x + ( -4___

2 )2) + 13 - ( -4___

2 )2

= (x2 - 4x + 4) + 13 - 4

= (x - 2)2 + 9


33. g(x) = x2 + 14x + 71

= (x2 + 14x) + 71

= (x2 + 14x + ( 14___

2 )2) + 71 - ( 14___

2 )2

= (x2 + 14x + 49) + 71 - 49

= (x + 7 )2 + 22


34. h(x) = 9 x2 + 18x - 3

= (9x2 + 18x) - 3

= 9 (x2 + 2x) - 3

= 9 (x2 + 2x + ( 2__

2)2) - 3 - 9 ( 2__

2)2

= 9 (x2 + 2x + 1) - 3 - 9

= 9(x + 1 )2 - 12


35. f(x) = x2 + 4x - 7

= (x2 + 4x )- 7

= (x2 + 4x + ( 4__

2)2) - 7 - ( 4__

2)2

= (x2 + 4x + 4) - 7 - 4

= (x + 2 )2 - 11


36. g(x) = x2 - 16x + 2

= (x2 - 16x) + 2

= (x2 - 16x + ( -16____

2 )2) + 2 - ( -16____

2 )2

= (x2 - 16x + 64) + 2 - 64

= (x - 8 )2 - 62


37. h(x) = 2 x2 + 6x + 25

= (2x2 + 6x) + 25

= 2 (x2 + 3x) + 25

= 2 (x2 + 3x + ( 3__

2)2) + 25 - 2 ( 3__

2)2

= 2 (x2 + 3x +

9__4) + 25 -

9__2

= 2 (x + 1.5)2 + 20.5

The vertex is (-1.5, 20.5).

38a. h(x) = 1_____9000

x2 - 7___

15x + 500

= 1_____9000

(x2 - 4200x) + 500

= 1_____9000(x2

- 4200x + ( -4200______2 )

2) + 500

- 1_____9000

( -4200______2 )

2

= 1_____9000

(x2 - 4200x + 4,410,000)

+ 500 - 490

= 1_____9000

(x - 2100 )2 + 10

b. The vertex is (2100, 10). The vertex represents the

distance from the left tower (2100 ft) at which the

height of the main cable reaches its lowest point

(10 ft above the roadway).

c. The distance is 2100 × 2 = 4200 ft.

39a. h(t) = -16t2 + 2421

-16t2 + 2421 = 0

-16(t2 - 151.3) ≈ 0

-16(t + 12.3)(t - 12.3) ≈ 0

t + 12.3 = 0 or t - 12.3 = 0

t = -12.3 or t = 12.3

It takes about 12.3 s.

b. h(t) = - 16t2 + 1612

-16t2 + 1612 = 0

-16(t2 - 100.75) = 0

-16(t + 10.0)(t - 10.0) ≈ 0

t ≈ -10.0 or t ≈ 10.0

It takes 12.3 - 10.0 ≈ 2.3 s longer.

40. h(t) = -16t2 + 24t + 6

= - (16t2 + 24t)+ 6

= - (16t2 - 1.5t) + 6

= -16(t2 - 1.5t + ( -1.5_____2 )

2) + 6 + 16 ( -1.5_____2 )

2

= -16(t2 - 1.5t + 0.5625) + 6 + 9

= -16(t - 0.75 )2 + 15

The maximum height of 15 ft occurs 0.75 s after the

ball is shot.

41. x2 - 1 = 2

x2 = 3

x = ± √ � 3

42. 25 x2 = 0

x2 = 0

x = 0

43. 8 x2 - 200 = 0

x2 = 25

x = ± √ �� 25

x = ±5

44. -3x2 + 6 = -1

-3x2 = -7

x2 = 7__

3 x = ± √

� 7__

3

x = ± √ �� 21____

3


45. (x + 13 )2 = 7

x + 13 = ± √ � 7

x = -13 ± √ � 7

46. (x + 1__4)

2 -

9___16

= 0

(x + 1__4)

2=

9___16

x + 1__4

= ± √ ��

9___

16

x + 1__4

= ± 3__4

x = 1__2

or -1

47. (x + 3__2)

2 =

25___2

x + 3__2

= ± √ ��

25___2

x + 3__2

= ± 5 √ � 2____

2

x = -3 ± 5 √ � 2_________

2

48. x2 + 14x + 49 = 64

(x + 7 )2 = 64

x + 7 = ± √ �� 64

x + 7 = ±8

x = 1 or -15

49. 9 x2 + 18x + 9 = 5

9(x2 + 2x + 1) = 5

9(x + 1)2 = 5

(x + 1 )2 =

5__9

x + 1 = ± √ �

5__9

x + 1 = ± √ � 5___3

x = -1 ± √ � 5___3

50. A is incorrect; possible answer: in the third step,

the number 4 is added inside the parentheses.

Because the expression in parentheses is multiplied

by 2, the total number added to the function rule is

8. Therefore, the number subtracted from the rule

should be 8 instead of 4.

51. x2 + 8x = -15

x 2 + 8x + ( 8__

2)2 = -15 + ( 8__

2)2

x 2 + 8x + 16 = -15 + 16

(x + 4 )2 = 1

x + 4 = ± √ � 1

x = -5 or -3

52. x2 + 22x = -21

x 2 + 22x + ( 22___

2 )2 = -21 + ( 22___

2 )2

x2 + 22x + 121 = -21 + 121

(x + 11 )2 = 100

x + 11 = ± √ �� 100

x = -1 or - 21

53. 3 x2 + 4x = 1

3 (x2 + 4__

3x) = 1

3(x2 + 4__

3x + ( 2__

3)2) = 1 + 3 ( 2__

3)2

3 (x2 + 4__

3x + 4__

9) = 1 + 4__3

3 (x + 2__3)

2 = 7__

3

(x + 2__3)

2 = 7__

9

x + 2__3

= ± √ �

7__9

x = -2 ± √ � 7________

3

54. 2 x2 = 5x + 12

2 x2 - 5x = 12

x2 -

5__2

x = 6

x2 -

5__2

x + ( -5___4 )

2 = 6 + ( -5___

4 )2

x2 -

5__2

x + 25___16

= 6 + 25___16

(x - 5__4)

2 = 121____

16

x - 5__4

= ± √ ��

121____16

x = 4 or -3___2

55. x2 - 7x - 2 = 0

x2 - 7x = 2

x2 - 7x + ( -7___

2 )2 = 2 + ( -7___

2 )2

x2 - 7x +

49___4

= 2 + 49___4

(x - 7__2)

2 =

57___4

x - 7__2

= ± √ ��

57___4

x = 7 ± √ �� 57________

2

56. x2 = 4x + 11

x2 - 4x = 11

x2 - 4x + ( -4___

2 )2 = 11 + ( -4___

2 )2

x2 - 4x + 4 = 11 + 4

(x - 2 )2 = 15

x - 2 = ± √ �� 15

x = 2 ± √ �� 15

57. x2 + 6x + 4 = 0

x2 + 6x = -4

x2 + 6x + ( 6__

2)2 = -4 + ( 6__

2)2

x2 + 6x + 9 = -4 + 9

(x + 3 )2 = 5

x + 3 = ± √ � 5

x = -3 ± √ � 5


58. 5 x2 + 10x - 7 = 0

5 x2 + 10x = 7

x2 + 2x = 7__

5

x2 + 2x + ( 2__

2)2 = 7__

5+ ( 2__

2)2

x2 + 2x + 1 = 7__

5+ 1

(x + 1 )2 = 12___

5

x + 1 = ± √ ��

12___5

x + 1 = ± 2 √ �� 15_____

5

x = -1 ± 2 √ �� 15_____

5

59. x2 - 8x = 24

x2 - 8x + ( -8___

2 )2 = 24 + ( -8___

2 )2

x2 - 8x + 16 = 24 + 16

(x - 4 )2 = 40

x - 4 = ± √ �� 40

x = 4 ± 2 √ �� 10

60a. h = 5 m

h(t) = 5 - 5 t2

5 t2 - 5 = 0

5 (t2 - 1) = 0

5(t + 1)(t - 1) = 0

t + 1 = 0 or t - 1 = 0

t = -1 or t = 1

when h = 5 m, t = 1 s;

h = 10 m

h(t) = 10 - 5 t2

5 t2 - 10 = 0

5 (t2 - 2) = 0

5 (t + √ � 2) (t - √ � 2) = 0

t + √ � 2 = 0 or t- √ � 2 = 0

t = - √ � 2 or t = √ � 2

when h = 10 m, t ≈ 1.41 s;

h = 20 m

h(t) = 20 - 5 t2

5 t2 - 20 = 0

5 (t2 - 4) = 0

5(t + 2)(t - 2) = 0

t + 2 = 0 or t - 2 = 0

t = -2 or t = 2

when h = 20 m, t = 2 s;

h = 30

h(t) = 30 - 5 t2

5 t2 - 30 = 0

5 (t2 - 6) = 0

5 (t + √ � 6) (t - √ � 6) = 0

t + √ � 6 = 0 or t - √ � 6 = 0

t = - √ � 6 or t = √ � 6

when h = 30 m, t ≈ 2.45 s.

b. The height will be h = 20 m.

c. h = 5 m, s = 18t = 18(1) = 18 km/h;

h = 10 m, s = 18t = 18(1.41) = 25.38 km/h;

h = 20 m, s = 18t = 18(2) = 36 km/h;

h = 30 m, s = 18t = 18(2.45) = 44.1 km/h.

d. The dive has to be 4 times as high.

61a. 7.5 + 6 = -16t2 + 32t + 5.5

-16t2 + 32t = 8

t2 - 2t = - 1__

2

t2 - 2t + ( 2__

2)2 = - 1__

2+ ( 2__

2)2

(t - 1 )2 = 1__

2

t - 1 = ± √ �

1__2

t = 1 ± √ � 2___2

t ≈ 0.3 or t ≈ 1.7

The ball ascends at t = 0.3 s and descends over

the player’s head at t = 1.7 s.

b. v = 120 ft_____1.7 s

≈ 71 ft/s

62. 525 × 20 = 10500 ft2

√ �� 10500 = 102.5 ft

The side length is about 100 ft.

63. f(x) = x2 + 2x - 2

f(x) = (x + 1 )2 - 3

f(x) = g(x)

64a. h(t) = -16t2 + 20t + 4

b. h(t) = (-16t2 + 20t) + 4

= -16(t2 - 5__4

t) + 4

= -16(t2 - 5__4

t + ( -5___8 )

2) + 4 + 16 ( -5___8 )

2

= -16(t2 - 5__4

t + 25___64) + 4 +

25___4

= -16 (t - 5__8)

2 + 10 1__

4

c. The maximum height is 41___4

ft.

d. h(t) = -16t2 + 32t + 4

= (-16t2 + 32t) + 4

= -16(t2 - 2t) + 4

= -16(t2 - 2t + ( 2__2)

2) + 4 + 16 ( 2__2)

2

= -16(t2 - 2t + 1) + 4 + 16

= -16(t - 1)2 + 20

20 - 41___4

= 39___4

The ball will go 9 3__4

ft higher.

65. ±7.416 66. ±1.688

67. ±4.192 68. -3.320, 2.120

69. ±1.528 70. -1.163, 0.663

71. The square root of a negative number is not a real

number.


72. Possible answer: Factoring is useful for solving quadratic equations with integer roots when the coefficient of the x

2 term is not a large number. Completing the square is useful for solving quadratic equations that cannot be factored easily. It involves rewriting part of an equation so that it can be factored as a perfect square binomial.

TEST PREP

73. B 74. F

75. A 76. H

77. 2x2 - x = 10

x2 - 0.5x = 5

x 2 - 0.5x + ( -0.5_____

2 )2 = 5 + ( -0.5_____

2 )2

x 2 - 0.5x + 0.0625 = 5 + 0.0625

(x - 0.25 )2 = 5.0625

x - 0.25 = ± √ �� 5.0625 x = 2.5 or -2

78. x2 - 6x = 16

x 2 - 6x + ( -6___

2 )2 = 16 + ( -6___

2 )2

x2 - 6x + 9 = 16 + 9

(x - 3 )2 = 25

x - 3 = ± √ �� 25 x = 8 or -2

add ( b__2)

2

to both sides

simplify both sides

create a square

binomial on the left

side

square root both sides

solve for both values

of x


79. ( b__2)

2= 144

b__2

= ±12

b = ±24

80. 4 x2 - bx + 16

4(x2 - b__4

x + 4)

( b__4__2)

2

= 4

b__8

= ±2

b = ±16

81. 3 x2 + bx + 27

3(x2 + b__3

x + 9)

( b__3__2)

2

= 9

b__6

= ±3

b = ±18

82. ( b__2)

2 = ac

b__2

= ± √ � ac

b = ±2 √ � ac

83. f(x) = x2 - 4x √ � 5 + 19

x2 - 4 √ � 5x + 19 = 0

x2 - 4 √ � 5x = -19

x 2 - 4 √ � 5x + ( -4 √ � 5______

2 )2

= -19 + ( -4 √ � 5______2 )

2

x2 - 4 √ � 5x + 20 = -19 + 20

(x - 2 √ � 5)2 = 1

x - 2 √ � 5 = ± √ � 1

x = 2 √ � 5 ± 1

84. f(x) = x2 + 6x √ � 3 + 23

x2 + 6 √ � 3x + 23 = 0

x2 + 6 √ � 3x = -23

x 2 + 6 √ � 3x + ( 6 √ � 3____

2 )2

= -23 + ( 6 √ � 3____2 )

2

x2 + 6 √ � 3x + 27 = -23 + 27

(x + 3 √ � 3)2 = 4

x + 3 √ � 3 = ± √ � 4

x = -3 √ � 3 ± 2

85a. Let � be the length, and w be the width of the field. 2� + 3w = 1800 2� = 1800 - 3w

� = 900 - 1.5w

A =� · w

= (900 - 1.5w) · w

= -1.5w2 + 900w

= -1.5(w2 - 600w)

= -1.5(w2 - 600w + ( -600_____2 )

2) + 1.5 ( -600_____2 )

2

= -1.5(w2 - 600w + 90000) + 135000

= -1.5(w - 300 )2 + 135000 The largest area of the field is 135,000 f t2 .

b. w = 300 ft;� = 900 - 1.5w = 900 - 1.5(300) = 450 ft.

The dimensions of the largest field are 450 ft by 300 ft.

c. Let x be the side length of the square field. 5x = 1800

x = 360

A = x2 = (360 )2 = 129600

The largest area of the square field is 129,600 f t2 .

SPIRAL REVIEW

86. {x | x > 72} 87. {x | -6 ≤ x ≤ 14}

88. {x | x = 4n for n ∈ N} 89. {x | -1 ≤ x ≤ 5}

90.

B =

352 895 426

675 1368 642

�

185 615 295

91. The matrix is 3 × 3.

92. The entry is b31 .

93. The value is 1368; it represents the amount in dollarsthat the Hernandez family budgeted for housing.

94. x = 2; (2, 0) 95. x = 0; (0, -1)

96. x = 0; (0, 2.5)


5-5 COMPLEX NUMBERS AND ROOTS,

PAGES 350-355

CHECK IT OUT!

1a. √ �� -12

= √ �� (12)(-1)

= √ �� 12 √ �� -1

= √ � 4 √ � 3 √ �� -1

= 2 √ � 3 √ �� -1

= 2i √ � 3

b. 2 √ �� -36

= 2 √ �� (36)(-1)

= 2 √ �� 36 √ �� -1

= 2(6) √ �� -1

= 12i

c. - 1__3

√ �� -63

= - 1__3

√ �� (63)(-1)

= - 1__3

√ �� 63 √ �� -1

= - 1__3

√ � 9 √ � 7 √ �� -1

= - 1__3

(3) √ � 7 √ �� -1

= -i √ � 7

2a. x2 = -36

x = ± √ �� -36

x = ±6i

b. x2 + 48 = 0

x 2 = -48

x = ± √ �� -48

x = ±4i √ � 3

c. 9 x2 + 25 = 0

9 x2 = -25

x2 = - 25___

9

x = ± √ ��

- 25___9

x = ± 5__3

i

3a. 2x - 6i = -8 + (20y)i

2x = -8

x = -4

-6i = (20y)i

-6 = 20y

- 3___10

= y

b. -8 + (6y)i = 5x - i √ � 6

-8 = 5x

- 8__5

= x

(6y)i = -i√ � 6

6y = - √ � 6

y = - √ � 6___6

4a. f(x) = x2 + 4x + 13

x 2 + 4x + 13 = 0

x2 + 4x = -13

x2 + 4x + 4 = -13 + 4

(x + 2 )2 = -9

x + 2 = ± √ �� -9

x = -2 ± 3i

b. g(x) = x2 - 8x + 18

x 2 - 8x + 18 = 0

x 2 - 8x = -18

x 2 - 8x + 16 = -18 + 16

(x - 4 )2 = -2

x - 4 = ± √ �� -2

x = 4 ± i √ � 2

5a. 9 - i

9 + i

b. i + √ � 3

√ � 3 + i

√ � 3 - i

c. 0 - 8i

0 + 8i

8i

THINK AND DISCUSS

1. Possible answer: If a quadratic equation has nonreal

roots, the roots are complex conjugates. Because

3 + i is nonreal, the other solution is its complex

conjugate, 3 - i.

2. Possible answer: A real number equal to a; an

imaginary number equal to bi; yes; both are

complex, because real numbers and imaginary

numbers are both sunsets of complex numbers.

3.

EXERCISES

GUIDED PRACTICE

1. imaginary

2. 5 √ �� -100

= 5 √ �� (100)(-1)

= 5 √ �� 100 √ �� -1

= 5(10) √ �� -1

= 50i

3.1__2

√ �� -16

= 1__2

√ �� (16)(-1)

= 1__2

√ �� 26 √ �� -1

= 1__2

(4) √ �� -1

= 2i

4. - √ �� -32

= - √ �� (32)(-1)

= - √ �� 32 √ �� -1

= - √ �� 16 √ � 2 √ �� -1

= -4i √ � 2

5. √ �� -144

= √ �� (144)(-1)

= √ �� 144 √ �� -1

= 12i

6. x2 = -9

x = ± √ �� -9

x = ±3i

7. 2 x2 + 72 = 0

2 x2 = -72

x2 = -36

x = ± √ �� -36

x = ±6i

8. 4 x2 = -16

x2 = -4

x = ± √ �� -4

x = ±2i

9. x2 + 121 = 0

x2 = -121

x = ± √ �� 121

x = ±11i

10. -2x + 6i = (-24y)i - 14

-2x = -14

x = 7

6i = (-24y)i

6 = -24y

- 1__4

= y

11. -4 + (y)i = -12x - i + 8

-4 = -12x + 8

12x = 12

x = 1

(y)i = -i

y = -1


12. f(x) = x2 - 12x + 45

x2 - 12x + 45 = 0

x2 - 12x = -45

x2 - 12x + 36 = -45 + 36

(x - 6 )2 = -9

x - 6 = ± √ �� -9

x = 6 ± 3i

13. g(x) = x2 + 6x + 34

x2 + 6x = -34

x2 + 6x + 9 = -34 + 9

(x + 3 )2 = -25

x + 3 = ± √ �� -25

x = -3 ± 5i

14. - 9i

9i

15. √ � 5 + 5i

√ � 5 - 5i

16. 8i - 3

-3 + 8i

-3 - 8i

17. 6 + i √ � 2

6 - i √ � 2


18. 8 √ �� -4

= 8 √ �� (4)(-1)

= 8 √ � 4 √ �� -1

= 8(2) √ �� -1

= 16i

19. - 1__3

√ �� -90

= - 1__3

√ �� (90)(-1)

= - 1__3

√ �� 90 √ �� -1

= - 1__3

√ � 9 √ �� 10 √ �� -1

= - 1__3

(3) √ �� 10 √ �� -1

= -i √ �� 10

20. 6 √ �� -12

= 6 √ �� (12)(-1)

= 6 √ �� 12 √ �� -1

= 6 √ � 4 √ � 3 √ �� -1

= 6(2) √ � 3 √ �� -1

= 12i √ � 3

21. √ �� -50

= √ �� (50)(-1)

= √ �� 50 √ �� -1

= √ �� 25 √ � 2 √ �� -1

= 5i √ � 2

22. x2 + 49 = 0

x2 = -49

x = ± √ �� -49

x = ±7i

23. 5 x2 = -80

x2 = -16

x = ± √ �� -16

x = ±4i

24. 3 x2 + 27 = 0

3 x2 = -27

x2 = -9

x = ± √ �� -9

x = ±3i

25.1__2

x2 = -32

x2 = -64

x = ± √ �� -64

x = ±8i

26. 9x + (y)i - 5 = -12i + 4

9x - 5 = 4

9x = 9

x = 1

(y)i = -12i

y = -12

27. 5(x - 1) + (3y)i = -15i - 20

5(x - 1) = -20

x - 1 = -4

x = -3

(3y)i = -15i

3y = -15

y = -5

28. f(x) = x2 + 2x + 3

x 2 + 2x + 3 = 0

x2 + 2x = -3

x 2 + 2x + 1 = -3 + 1

(x + 1 )2 = -2

x + 1 = ± √ �� -2

x = -1 ± i √ � 2

29. g(x) = 4 x2 - 3x + 1

4 x2 - 3x + 1 = 0

4 x2 - 3x = -1

x2 -

3__4

x = - 1__4

x 2 -

3__4

x + 9___

64= - 1__

4+

9___64

(x - 3__8)

2 = - 7___

64

x - 3__8

= ± √ ��

- 7___64

x = 3 ± i√ � 7_______

8

30. f(x) = x2 + 4x + 8

x 2 + 4x + 8 = 0

x 2 + 4x = -8

x 2 + 4x + 4 = -8 + 4

(x + 2 )2 = -4

x + 2 = ± √ �� -4

x = -2 ± 2i

31. g(x) = 3 x2 - 6x + 10

3x2 - 6x + 10 = 0

3 x2 - 6x = -10

x2 - 2x = - 10___

3

x2 - 2x + 1 = - 10___

3+ 1

(x - 1 )2 = - 7__

3

x - 1 = ± √ ��

-7___3

x = 1 ± i √ �� 21_____

3

32. i

-i

33. - √ �

3__2

- 2i

- √ �

3__2

+ 2i

34. -2.5i + 1

1 - 2.5i

1 + 2.5i

35.i___

10- 1

-1 + i___10

-1 - i___10

36. No; the participant will not win a prize.

Possible answer: The solutions to the equation

16x2 - 32x + 18 = 0 are imaginary so the distance

between the puck and the bell never reaches 0.

37. 1 - 14i 38.-5___7

i

39. -2 √ � 5 - 4i 40. -12 + i

41. 9 + i √ � 2 42.17___3

i


43. 2ci + 1 = -d + 6 - ci

1 = -d + 6

d = 5

2ci = -ci

3ci = 0

c = 0

44. c + 3ci = 4 + di

c = 4

3ci = di

3c = d

12 = d

45. c2 + 4i = d + di

4i = di

4 = d

c 2 = d

c = ± √ � d

c = ±2

46. 8 x2 = -8

x 2 = -1

x = ± √ �� -1

x = ±i

47.1__3

x2 = -27

x2 = -81

x = ± √ �� -81

x = ±9i

48. 2 x2 + 12.5 = 0

2 x2 = -12.5

x2 = -6.25

x = ± √ �� -6.25

x = ±2.5i

49.1__2

x2 + 72 = 0

1__2

x2 = -72

x2 = -144

x = ± √ �� -144

x = ±12i

50. x2 = -30

x = ± √ �� -30

x = ±i √ �� 30

51. 2 x2 + 16 = 0

2 x2 = -16

x2 = -8

x = ± √ �� -8

x = ±2i √ � 2

52. x2 - 4x + 8 = 0

x 2 - 4x = -8

x 2 - 4x + 4 = -8 + 4

(x - 2) 2 = -4

x - 2 = ± √ �� -4

x = 2 ± 2i

53. x2 + 10x + 29 = 0

x 2 + 10x = -29

x 2 + 10x + 25 = -29 + 25

(x + 5 )2 = -4

x + 5 = ± √ �� -4

x = -5 ± 2i

54. x2 - 12x + 44 = 0

x2 - 12x = -44

x 2 - 12x + 36 = -44 + 36

(x - 6 )2 = -8

x - 6 = ± √ �� -8

x = 6 ± 2i√ � 2

55. x2 + 2x = -5

x 2 + 2x + 1 = -5 + 1

(x + 1 )2 = -4

x + 1 = ± √ �� -4

x = -1 ± 2i

56. x2 + 18 = -6x

x2 + 6x = -18

x 2 + 6x + 9 = -18 + 9

(x + 3 )2 = -9

x + 3 = ± √ �� -9

x = -3 ± 3i

57. -149 = x2 - 24x

-149 + 144 = x2 - 24x + 144

-5 = (x - 12 )2

± √ �� -5 = x - 12

12 ± i √ � 5 = x

58. Never true. 59. Always true.

60. Always true.

61. Sometimes true.

Possible answer: 3i is a complex number that

is imaginary; 3 is a complex number that is not

imaginary.

62. Always true.

63. Sometimes true.

Possible answer: The quadratic equation x2 = -4

has no real solutions; the quadratic equation x2 = 4

has 2 real solutions.

64. Sometimes true.

Possible answer: The quadratic equation x2 = 4 has

2 real, complex roots,; the quadratic equation x2

= -4 has 2 complex roots, but these roots are not

real.

65. Sometimes true.

Possible answer: The quadratic equation x2 = -4

has 2 roots that form a conjugate pair; the quadratic

equation x2 = 4 has 2 roots that do not form a

conjugate pair.

66. f(X) = x2 - 10x + 26

x 2 - 10x + 26 = 0

x2 - 10x = -26

x 2 - 10x + 25 = -26 + 25

(x - 5 )2 = -1

x - 5 = ± √ �� -1

x = 5 ± i

67. g(x) = x2 + 2x + 17

x 2 + 2x + 17 = 0

x 2 + 2x = -17

x2 + 2x + 1 = -17 + 1

(x + 1 )2 = -16

x + 1 = ± √ �� -16

x = -1 ± 4i

68. h(x) = x2 - 10x + 50

x 2 - 10x + 50 = 0

x2 - 10x = -50

x 2 - 10x + 25 = -50 + 25

(x - 5 )2 = -25

x - 5 = ± √ �� -25

x = 5 ± 5i

69. f(x) = x2 + 16x + 73

x 2 + 16x + 73 = 0

x2 + 16x = -73

x 2 + 16x + 64 = -73 + 64

(x + 8 )2 = -9

x + 8 = ± √ �� -9

x = -8 ± 3i

70. g(x) = x2 - 10x + 37

x 2 - 10x + 37 = 0

x2 - 10x = -37

x 2 - 10x + 25 = -37 + 25

(x - 5 )2 = -12

x - 5 = ± √ �� -12

x = 5 ± 2i √ � 3


71. h(x) = x2 - 16x + 68

x 2 - 16x + 68 = 0

x2 - 16x = 68

x 2 - 16x + 64 = -68 + 64

(x - 8 )2 = -4

x - 8 = ± √ �� -4

x = 8 ± 2i

72. Possible answer: No; the graph of the function

does not cross the x-axis. Therefore, it has nonreal,

complex zeros. Algebra must be used to determine

these zeros.

73. The complex conjugate of a real number “a” is the

number “a”.

74. Possible answer: You can use the Square Root

Property or complete the square to solve for

complex roots with imaginary parts.

75a. 208 = -16t2 + 112t

208 = -16(t2 - 7t)

-13 = t2 - 7t

-13 + 49___4

= t2 - 7t +

49___4

- 3__4

= (t - 7__2)

2

± √ ��

-3___4

= t- 7__2

± i √ � 3____

2= t - 7__

2

7__2

± √ � 3___2

i = t

b. Based on the solution to part a, there are no real

values of t for which the height of the ball is 208 ft.

Therefore, the ball does not hit the roof.

c. h(t) = -16t2 + 112t

= -16(t2 - 7t)

= -16(t2 - 7t + 49___4 ) + 16 ( 49___

4 )= -16 (t - 7__

2)2 + 196

The maximum height that the ball can hit is 196 ft.

TEST PREP

76. D 77. F

78. A;

f(x) = x2 - 2x + 17

x 2 - 2x + 17 = 0

x2 - 2x = -17

x2 - 2x + 1 = -16

(x - 1 )2 = -16

x - 1 = ± √ �� -16

x = 1 ± 4i

79. G;

3 - 4i - 5 = 9 + ci - 11

-4i = ci

-4 = c

80. A;

- 1__6

x2 = 6

x 2 = -36

x = ± √ �� -36

x = ±6i

81. When a < 0, the solutions are imaginary and

complex. When a > 0, the 2 solutions are real and

complex.


82. 5a + 3b = 1

5a = 1 - 3b

a = 1 - 3b______

5 -5b = 7 + 4a

-5b = 7 + 4 ( 1 - 3b______5 )

-5b = 7 + 4__5

- 12b____5

- 13___5

b = 39___5

b = -3

a = 1 - 3b______

5=

1 - 3(-3)_________5

= 2

a + bi = 2 - 3i

83. Possible answer: A quadratic equation can have a

single distinct root. For example, the only distinct

root of (x - 1 )2 = 0 is x = 1. A quadratic equation

cannot have a single imaginary root because

imaginary roots occur only as conjugate pairs.

Because a quadratic equation can have single

distinct real number root and because real numbers

are complex, a quadratic equation can have a single

distinct complex root.

84a. If b = 0 and c ≤ 0, the equation has 2 real

solutions. If b = 0 and c > 0, the equation has 2

nonreal complex solutions.

b. If c ≤ 0, the equation has 2 real solutions.

c. If c > 0 and c ≤ ( b__2)

2 , the equation has 2 real

solutions. If c > 0 and c > ( b__2)

2

, the equation has

2 nonreal complex solutions.

d. The solutions have imaginary pairs if c > ( b__2)

2 .

SPIRAL REVIEW

85.

T2 =

12 -3 5

2 7 -1

-4 4 -2

row 1 (-4 · -4) + (1 · 0) + (-2 · 2)

= 16 + 0 - 4 = 12

(-4 · 1) + (1 · -3) + (-2 · -2)

= -4 - 3 + 4 = -3

(-4 · -2) + (1 · 1) + (-2 · 2)

= 8 + 1 - 4 = 5

row 2 (0 · -4) + (-3 · 0) + (1 · 2)

= 0 + 0 + 2 = 2

(0 · 1) + (-3 · -3) + (1 · -2)

= 0 + 9 - 2 = 7

(0 · -2) + (-3 · 1) + (1 · 2)

= 0 - 3 + 2 = -1

row 3 (2 · -4) + (0 · -2) + (2 · 2)

= -8 + 0 + 4 = -4

(2 · 1) + (-2 · -3) + (2 · -2)

= 2 + 6 - 4 = 4

(2 · -2) + (-2 · 1) + (2 · 2)

= -4 - 2 + 4 = -2


86.

TV =

-30 -15

-5 8

�

10 14

row 1 (-4 · 10) + (1 · 0) + (-2 · -5)

= -40 + 0 + 10 = -30

(-4 · 1) + (1 · -1) + (-2 · 5)

= -4 - 1 - 10 = -15

row 2 (0 · 10) + (-3 · 0) + (1 · -5)

= 0 + 0 - 5 = -5

(0 · 1) + (-3 · -1) + (1 · 5)

= 0 + 3 + 5 = 8

row 3 (2 · 10) + (0 · -2) + (2 · -5)

= 20 + 0 - 10 = 10

(2 · 1) + (-2 · -1) + (2 · 5)

= 2 + 2 + 10 = 14

87. ST is not defined.

88.

S2 =

11 -5

�

-2 10

row 1 (1 · 1) + (-2 · -5)

= 1 + 10 = -11

(1 · -5) + (-5 · 0)

= -5 + 0 = -5

row 2 (-2 · 1) + (-2 · 0)

= -2 + 0 = -2

(-2 · -5) + (0 · 0)

= 10 + 0 = 10

89a. upward

b. x = -b___2a

= -1____

2( 1__5)

= -2.5

c. f(-2.5) = 1__5

(-2.5)2 + (-2.5) - 10

= 1.25 - 2.5 - 10

= -11.25

The vertex is (-2.5, -11.25).


e.

90a. downward

b. x = -b___2a

= 0_____

2(-1)= 0

c. f(0) = -(0)2 + 3 = 3



e.

91a. upward

b. x = -b___2a

= -4____2(2)

= -1

c. f(-1) = 2(-1)2 + 4(-1) - 3

= 2 - 4 - 3

= -5



e.

92a. downward

b. x = -b___2a

= -3______

2(- 1__2

)

= 3

c. f(3) = - 1__2

(3)2 + 3(3) + 1

= - 9__2

+ 9 + 1

= 11___2

The vertex is (3, 5.5).


e.

93. x2 + 5x = 14

x2 + 5x - 14 = 0

(x - 2)(x + 7) = 0

x - 2 = 0 or x + 7 = 0

x = 2 or x = -7

94. 6 x2 = -x + 2

6 x2 + x - 2 = 0

(2x - 1)(3x + 2) = 0

2x - 1 = 0 or 3x + 2 = 0

x = 1__2

or x = - 2__3

95. 4 x2 + 9 = 15x

4 x2 - 15x + 9 = 0

(x - 3)(4x - 3) = 0

x - 3 = 0 or 4x - 3 = 0

x = 3 or x = 3__4


96. 4 x2 = 1

4 x2 - 1 = 0

(2x + 1)(2x - 1) = 0

2x + 1 = 0 or 2x - 1 = 0

x = - 1__2

or x = 1__2

97. x2 + 11x = -24

x2 + 11x + 24 = 0

(x + 3)(x + 8) = 0

x + 3 = 0 or x + 8 = 0

x = -3 or x = -8

98. x2 = -7x

x2 + 7x = 0

x(x + 7) = 0

x = 0 or x + 7 = 0

x = 0 or x = -7

5-6 THE QUADRATIC FORMULA,

PAGES 356-363

CHECK IT OUT!

1a. x = -b ± √ �� b

2 - 4ac

______________

2a

= -3 ± √ �� 3

2 - 4(1)(-7)

__________________ 2(1)

= -3 ± √ �� 9 + 28

_____________2

= -3 ± √ �� 37_________

2

1b. x = -b ± √ �� b

2 - 4ac

______________

2a

= -(-8) ± √ �� (-8)

2 - 4(1)(10)

_______________________ 2(1)

= 8 ± √ �� 64 - 40

____________2

= 8 ± 2 √ � 6________

2 = 4 ± √ � 6

2. x = -b ± √ �� b

2 - 4ac

______________

2a

= -(-1) ± √ �� (-1)

2 - 4(3)(8)

______________________ 2(3)

= 1 ± √ �� 1 - 96

___________6

= 1 ± √ �� -95_________

6

3a. b2 - 4ac

(-4)2 - 4(1)(4)

16 - 16 = 0

b 2 - 4ac = 0;

the equation has 1

distinct solution.

b. b2 - 4ac

(-4)2 - 4(1)(8)

16 - 32 = -16

b 2 - 4ac < 0;

the equation has 2 distinct

nonreal complex solutions.

c. b2 - 4ac

(-4)2 - 4(1)(-2)

16 + 8 = 24

b 2 - 4ac > 0;

the equation has 2 distinct real solutions.

4. 0 = -16t2 - 2t + 400

t = -b ± √ �� b

2 -4ac

______________2a

= -(-2) ± √ �� (-2)

2 - 4(-16)(400)

___________________________ 2(-16)

= 2 ± √ �� 4 + 25600

______________-32

= 2 ± 2 √ �� 6401

___________-32

= 1 ± √ �� 6401

__________-16

t ≈ -5.0625 or t ≈ 4.9375

The time cannot be negative, so the water lands on

the target about 4.9375 seconds after it is released.

x = 91t = 91(4.9375) ≈ 449

The water will have traveled a horizontal distance of

about 449 ft.

THINK AND DISCUSS

1. Possible answer: If the function has 2 x-intercepts.

there are 2 real zeros. If there is 1 x-intercept, the

function has 1 distinct real zero. If there are no

x-intercepts, the function has 2 nonreal complex

zeros.

2. Possible: For c < 16, the equation has 2 real

solutions. For c = 16, the equation has 1 distinct

real solution. For c > 16, the equation has 2 nonreal

complex solutions.

3.

EXERCISES

GUIDED PRACTICE

1. Possible answer: The value of the discriminant

indicates the number and types of roots.

2. x = -b ± √ �� b

2 - 4ac

______________

2a

= -7 ± √ �� 7

2 - 4(1)(10)

__________________ 2(1)

= -7 ± √ �� 49 - 40

______________2

= -7 ± 3_______

2 = -2 or -5


3. x = -b ± √ �� b

2 - 4ac

______________

2a

= -(-4) ± √ �� (-4)

2 - 4(3)(-1)

________________________ 2(3)

= 4 ± √ �� 16 + 12

____________6

= 4 ± 2 √ � 7________

6

= 2 ± √ � 7_______

3

4. x = -b ± √ �� b

2 - 4ac

______________

2a

= -(-5) ± √ �� (-5)

2 - 4(3)(0)

______________________ 2(3)

= 5 ± √ �� 25________

6

= 5 ± 5_____

6

= 5__3

or 0

5. x = -b ± √ �� b

2 - 4ac

______________

2a

= -(-5) ± √ �� (-5)

2 - 4(-1)(6)

________________________ 2(-1)

= 5 ± √ �� 25 + 24

____________-2

= 5 ± 7_____

-2

= -6 or 1

6. x = -b ± √ �� b

2 - 4ac

______________

2a

= -(-5) ± √ �� (-5)

2 - 4(4)(-6)

________________________ 2(4)

= 5 ± √ �� 25 + 96

____________8

= 5 ± 11______

8

= 2 or - 3__4

7. x = -b ± √ �� b

2 - 4ac

______________

2a

= -0 ± √ �� 0

2 - 4(2)(-19)

___________________ 2(2)

= 0 ± √ �� 152_________

4

= ±2 √ �� 38_______

4

= ± √ �� 38____2

8. x = -b ± √ �� b

2 - 4ac

______________

2a

= -(-2) ± √ �� (-2)

2 - 4(2)(3)

______________________ 2(2)

= 2 ± √ �� 4 - 24

___________4

= 2 ± 2i √ � 5________

4

= 1 ± i √ � 5_______

2

9. x = -b ± √ �� b

2 - 4ac

______________

2a

= -6 ± √ �� 6

2 - 4(1)(12)

__________________ 2(1)

= -6 ± √ �� 36 - 48

______________2

= -6 ± 2i √ � 3__________

2

= -3 ± i √ � 3

10. x = -b ± √ �� b

2 - 4ac

______________

2a

= -4 ± √ �� 4

2 - 4(3)(3)

_________________ 2(3)

= -4 ± √ �� 16 - 36

______________6

= -4 ± 2i √ � 5__________

6

= -2 ± i √ � 5_________

3

11. x = -b ± √ �� b

2 - 4ac

______________

2a

= -4 ± √ �� 4

2 - 4(1)(10)

__________________ 2(1)

= -4 ± √ �� 16 - 40

______________2

= -4 ± 2i

√ � 6 __________

2

= -2 ± i √ � 6

12. x = -b ± √ �� b

2 - 4ac

______________

2a

= -7 ± √ �� 7

2 - 4(-5)(-3)

____________________ 2(-5)

= -7 ± √ �� 49 - 60

______________-10

= -7 ± i √ � 11_________

-10

13. x = -b ± √ �� b

2 - 4ac

______________

2a

= -7 ± √ �� 7

2 - 4(10)(4)

__________________ 2(10)

= -7 ± √ �� 49 - 160

_______________20

= -7 ± i √ �� 111

___________20


14. 4 x2 - 4x + 1 = 0

b 2 - 4ac

(-4)2 - 4(4)(1)

16 - 16 = 0

b 2 - 4ac = 0;

the equation has

1 distinct real solution.

15. x2 + 2x - 10 = 0

b 2 - 4ac

22 - 4(1)(-10)

4 + 40 = 44

b 2 - 4ac > 0;

the equation has

2 distinct real solutions.

16. - x2 + 2x - 4 = 0

b 2 - 4ac

22 - 4(-1)(-4)

4 - 16 = -12

b 2 - 4ac < 0;

the equation has 2 distinct nonreal complex

solutions.

17. Let x be the length of the shorter leg, then (x + 6) is

the length of the longer leg.

x 2 + (x + 6 )

2 = 2 5

2

x 2 + x

2 + 12x + 36 = 625

2x2 + 12x - 589 = 0

x = -b ± √ �� b

2 - 4ac

______________

2a

= -12 ± √ �� 12

2 - 4(2)(-589)

______________________ 2(2)

= -12 ± √ �� 4856

____________4

= -3 ± √ �� 1214

___________2

x ≈ 14 or x ≈ -20

The length of the shorter leg is 14 in., and the length

of the longer leg is 14 + 6 = 20 in.


18. x = -b ± √ �� b

2 - 4ac

______________

2a

= -(-10) ± √ �� (-10)

2 - 4(3)(3)

________________________ 2(3)

= 10 ± √ �� 100 - 36

______________6

= 10 ± 8______

6

= 3 or 1__3

19. x = -b ± √ �� b

2 - 4ac

______________

2a

= -6 ± √ �� 6

2 - 4(1)(0)

_________________ 2(1)

= -6 ± √ �� 36 - 0

_____________2

= -6 ± 6_______

2 = 0 or -6

20. x2 - 3x - 4 = 0

x = -b ± √ �� b

2 - 4ac

______________

2a

= -(-3) ± √ �� (-3)

2 - 4(1)(-4)

________________________ 2(1)

= 3 ± √ �� 9 + 16

___________2

= 3 ± 5_____

2 = 4 or -1

21. x = -b ± √ �� b

2 - 4ac

______________

2a

= -(-2) ± √ �� (-2)

2 - 4(-1)(9)

________________________ 2(-1)

= 2 ± √ �� 4 + 36

___________-2

= 2 ± 2 √ �� 10_________

-2 = -1 ± √ �� 10

22. x = -b ± √ �� b

2 - 4ac

______________

2a

= -(-7) ± √ �� (-7)

2 - 4(2)(-8)

________________________ 2(2)

= 7 ± √ �� 49 + 64

____________4

= 7 ± √ �� 113_________

4

23. x = -b ± √ �� b

2 - 4ac

______________

2a

= -0 ± √ �� 0

2 - 4(7)(-3)

__________________ 2(7)

= 0 ± √ �� 84________

14

= ±2 √ �� 21_______

14

= ± √ �� 21____

7

24. x = -b ± √ �� b

2 - 4ac

______________

2a

= -1 ± √ �� 1

2 - 4(1)(1)

_________________ 2(1)

= -1 ± √ �� 1 - 4

____________2

= -1 ± i √ � 3_________

2

25. x = -b ± √ �� b

2 - 4ac

______________

2a

= -(-1) ± √ �� (-1)

2 - 4(-1)(-1)

_________________________ 2(-1)

= 1 ± √ �� 1 - 4

__________-2

= -1 ± i √ � 3_________

2


26. x = -b ± √ �� b

2 - 4ac

______________

2a

= -0 ± √ �� 0

2 - 4(2)(8)

_________________ 2(2)

= 0 ± √ �� -64_________

4

= ±8i___4

= ±2i

27. x = -b ± √ �� b

2 - 4ac

______________

2a

= -7 ± √ �� 7

2 - 4(2)(-13)

___________________ 2(2)

= -7 ± √ �� 49 + 104

_______________4

= -7 ± 3 √ �� 17

__________4

28. x = -b ± √ �� b

2 - 4ac

______________

2a

= -(-1) ± √ �� (-1)

2 - 4(1)(-5)

________________________ 2(1)

= 1 ± √ �� 1 + 20

___________2

= 1 ± √ �� 21________

2

29. x = -b ± √ �� b

2 - 4ac

______________

2a

= -4 ± √ �� 4

2 - 4(-3)(-4)

____________________ 2(-3)

= -4 ± √ �� 16 - 48

______________-6

= -4 ± 4i √ � 2__________

-6

= 2 ± 2i √ � 2________

3

30. b2 - 4ac

(-2)2 - 4(2)(5)

4 - 40 = -36

b 2 - 4ac < 0;

the equation has

2 distinct nonreal

complex solutions.

31. b2 - 4ac

(-3)2 - 4(2)(-8)

9 + 64 = 73

b 2 - 4ac > 0;

the equation has


32 . b2 - 4ac

(-16)2 - 4(2)(32)

256 - 256 = 0

b 2 - 4ac = 0;

the equation has


33. b2 - 4ac

(-28)2 - 4(4)(49)

784 - 784 = 0

b 2 - 4ac = 0;

the equation has


34. b2 - 4ac

(-8)2 - 4(3)(8)

64 - 96 = -32

b 2 - 4ac < 0;

the equation has

2 distinct nonreal

complex solutions.

35. b2 - 4ac

(-8.5)2 - 4(3.2)(1.3)

72.25 - 16.64 = 55.61

b2 - 4ac > 0;

the equation has


36. -16t2 + (60 - 11) = 0

-16t2 + 49 = 0

t = -b ± √ �� b

2 - 4ac

______________

2a

= -0 ± √ �� 0

2 - 4(-16)(49)

____________________ 2(-16)

= 0 ± √ �� 3136

__________-32

= ±56____-32

= -1.75 or 1.75

The walker will fall for 1.75 s.

37a. 5 t2 + 20t = 585

5 t2 + 20t - 585 = 0

5 (t2 + 4t - 117) = 0

t = -b ± √ �� b

2 - 4ac

______________

2a

= -4 ± √ �� 4

2 - 4(1)(-117)

____________________ 2(1)

= -4 ± √ �� 484

__________2

= -4 ± 22________

2= 9 or -14

It will take her 9 s to reach the bottom.

b. 5 t2 + 20t = 292.5

5 t2 + 20t - 292.5 = 0

5 (t2 + 4t - 58.5) = 0

t = -b ± √ �� b

2 - 4ac

______________

2a

= -4 ± √ �� 4

2 - 4(1)(-58.5)

_____________________ 2(1)

= -4 ± √ �� 250

__________2

= -2 ± 5 √ �� 10_____

2t ≈ 6 or t ≈ -10

It will take about 6 s to reach the bottom.

38. x = -b ± √ �� b

2 - 4ac

______________

2a

= -(-4) ± √ �� (-4)

2 - 4(3)(-2)

________________________ 2(3)

= 4 ± √ �� 16 + 24

____________6

= 4 ± 2 √ �� 10_________

6

= 2 ± √ �� 10________

3


39. x = -b ± √ �� b

2 - 4ac

______________

2a

= -(-2) ±

√ �� (-2)2 - 4(2)(-1)

________________________ 2(2)

= 2 ± √ �� 4 + 8

__________4

= 2 ± 2 √ � 3________

4

= 1 ± √ � 3_______

2

40. x = -b ± √ �� b 2 - 4ac

______________

2a

= -6 ± √ �� 6

2 - 4(2)(5)

_________________ 2(2)

= -6 ± √ �� 36 - 40

______________4

= -6 ± 2i_______

4

= -3 ± i______

2

41. x = -b ± √ �� b

2 - 4ac

______________

2a

= -3 ± √ �� (-3)

2 - 4(2)(-1)

_____________________ 2(2)

= -3 ± √ �� 9 + 8

____________4

= -3 ± √ �� 17_________

4

42. x = -b ± √ �� b 2 - 4ac

______________

2a

= -(-5) ± √ �� (-5)

2 - 4(3)(-4)

________________________ 2(3)

= 5 ± √ �� 25 + 48

____________6

= 5 ± √ �� 73________

6

43. x = -b ± √ �� b

2 - 4ac

______________

2a

= -(-1) ± √ �� (-1)

2 - 4(1)(22)

_______________________ 2(1)

= 1 ± √ �� 1 - 88

___________2

= 1 ± i √ �� 87________

2

44a. -0.17t2 + 187t + 61000 = 0

t = -b ± √ �� b 2 - 4ac

______________

2a

= -187 ± √ �� 187

2 - 4(-0.17)(61000)

_____________________________ 2(-0.17)

= -187 ± √ �� 76449

______________ -0.34t ≈ -261.7 or t ≈ 1363

The flight took about 1363 s.

b. axis of symmetry: t = -b___2a

= -187________

2(-0.17)= 550

h(550) = -0.17(550)2 + 187(550) + 61,000

= -51425 + 102850 + 61,000

= 112425

The highest altitude is about 112,000 m.


c. -0.17t2 + 187t + 61000 = 5000

-0.17t2 + 187t + 56000 = 0

t = -b ± √ �� b

2 - 4ac

______________

2a

= -187 ± √ �� 187

2 - 4(-0.17)(56000)

_____________________________ 2(-0.17)

= -187 ± √ �� 18649

______________ -0.34

t ≈ 148 or x ≈ 951.6

The ship entered the thermosphere first at

t ≈ 149 s then the mesosphere at t ≈ 952 s.

-0.17t2 + 187t + 61000 = 50000

-0.17t2 + 187t + 11000 = 0

t = -b ± √ �� b

2 - 4ac

______________

2a

= -187 ± √ �� 187

2 - 4(-0.17)(11000)

_____________________________ 2(-0.17)

= -187 ± √ �� 42449

______________ -0.34

t ≈ 1156

After the mesosphere, the shop entered the

stratosphere at t ≈ 1156 s.

-0.17t2 + 187t + 61000 = 10000

-0.17t2 + 187t + 51000 = 0

t = -b ± √ �� b

2 - 4ac

______________

2a

= -187 ± √ �� 187

2 - 4(-0.17)(51000)

_____________________________ 2(-0.17)

= -187 ± √ �� 69649

______________ -0.34

t ≈ 1326.17

The ship entered the troposphere at t ≈ 1326 s.

45. x2 - 3x - 10 = 0

(x - 5)(x + 2) = 0

x - 5 = 0 or x + 2 = 0

x = 5 or x = -2

46. x2 - 16 = 0

(x - 4)(x + 4) = 0

x - 4 = 0 or x + 4 = 0

x = 4 or x = -4

47. 4 x2 + 4x - 15 = 0

(2x - 3)(2x + 5) = 0

2x - 3 = 0 or 2x + 5 = 0

x = 1.5 or x = -2.5

48. x2 + 2x - 2 = 0

x = -b ± √ �� b

2 - 4ac

______________

2a

= -(2) ± √ �� (2)

2 - 4(1)(-2)

_____________________ 2(1)

= -2 ± √ �� 4 + 8

____________2

= -2 ± 2 √ � 3_________

2 = -1 ± √ � 3

49. x2 - 4x - 21 = 0

(x - 7)(x + 3) = 0

x - 7 = 0 or x + 3 = 0

x = 7 or x = -3

50. 4 x2 - 4x - 1 = 0

x = -b ± √ �� b

2 - 4ac

______________

2a

= -(-4) ± √ �� (-4)

2 - 4(4)(-1)

________________________ 2(4)

= 4 ± √ �� 16 + 16

____________8

= 4 ± 4 √ � 2________

8

= 1 ± √ � 2_______

2

51. 6 x2 = 150

x2 = 25

x = ± √ �� 25

x = ±5

52. x2 = 7

x = ± √ � 7

53. x2 - 16x + 64 = 0

(x - 8)(x - 8) = 0

x -8 = 0

x = 8

54. The problem does not have a meaningful solution

because the roots of the quadratic equation are

imaginary.

55. (25 - 2w)(20 - 2w) = 266

500 - 50w - 40w + 4 w2 = 266

4 w2 - 90w + 234 = 0

2 (2w2 - 45w + 117) = 0

2(w - 3)(2w - 39) = 0

w - 3 = 0 or 2w - 39 = 0

w = 3 or w = 19.5

The width of the frame is 3 in.

56. b2 - 4ac = 0

82 - 4(1)(c) = 0

64 - 4c = 0

-4c = -64

c = 16

57. x2 + 12x - c = 0

b2 - 4ac = 0

122 - 4(1)(-c) = 0

144 + 4c = 0

4c = -144

c = -36

58. b2 - 4ac = 0

(2c)2 - 4(1)(49) = 0

4 c2 = 196

c2 = 49

c = ± √ �� 49

c = ±7

59. Possible answer: The Quadratic Formula; this

method would be easier because the absolute value

of a is great and the value of c is not an integer.

Therefore, it would be difficult to solve the equation

by factoring or by completing the square.


60a. -16t2 + 19t + 5 = 4

-16t2 + 19t + 1 = 0

t = -b ± √ �� b 2 - 4ac

______________

2a

= -19 ± √ �� 19

2 - 4(-16)(1)

_____________________

2(-16)

= -19 ± √ �� 425

___________

-32

= 19 ± 5 √ �� 17

__________

32 t ≈ -0.05 or t ≈ 1.2

The ball is in the air for about 1.2 s.

b. No, the ball will reach third base before the runner.

Possible answer: When the ball reaches third base,

the runner will have been running for about 2.4 s.

The runner will have traveled only about 65 ft in

this amount of time.

TEST PREP

61. B 62. H;

b 2 - 4ac = (-8)

2 - 4(2)(-14)

= 64 + 112 = 176

63. C;

x = -b ± √ �� b 2 - 4ac

______________

2a

= -(-6) ± √ �� (-6)

2 - 4(1)(10)

_______________________ 2(1)

= 6 ± √ �� 36 - 40

____________2

= 6 ± 2i_____

2 = 3 ± i

64. F

CHALLENGE AND EXTENDED

65. Let x be the length of one leg, then the length of the

other leg is 40 - 17 -x = 23 - x.

x2 + (23 - x)

2 = 1 7

2

x 2 + 529 - 46x + x

2 = 289

2 x2 - 46x + 240 = 0

2 (x2 - 23x + 120) = 0

2(x - 8)(x - 15) = 0

x - 8 = 0 or x - 15 = 0

x = 8 or x = 15

The length of the legs are 8 cm and 15 cm.

66a. Let x be the length of the rectangle, then the width

of the rectangle will be 88___2

- x = 44 - x.

d2 = x

2 + (44 - x)

2

= x2 + 1936 - 88x + x

2

= 2 x2 - 88x + 1936

= 2 (x2 - 44x) + 1936

= 2 (x2 - 44x + 484) + 1936 - 2(484)

= 2(x - 22 )2 + 968

Therefore, the least possible value of the length of

the diagonal is √ �� 968 ≈ 31.1 cm.

b. The dimensions of the rectangle are 22 cm by

22 cm.

67. Possible answer: x2 = 4

68. Possibele anwer: x2 = 1__

2

69. Possible answer: x2 = -2

70a. x1 + x2 = -b + √ �� b

2 - 4ac

______________2a

+ -b - √ �� b

2 - 4ac

______________2a

= -b + √ �� b

2 - 4ac - b - √ �� b

2 - 4ac _____________________________

2a

= -2b____2a

= - b__a

x1 · x2 = -b + √ �� b

2 - 4ac

______________2a

· -b - √ �� b 2 - 4ac

______________2a

= (-b + √ �� b

2 - 4ac )· (-b -

√ �� b 2 - 4ac) _________________________________

(2a) · (2a)

= (-b)

2 - ( √ �� b

2 - 4ac)

2

___________________ 4a

2

= b

2 - (b2

- 4ac) ______________

4a2

= 4ac____4a

2=

c__a

b. x2 - 2x - 15 = 0

71. The solutions are nonreal and complex. Regardless

of the values of a, b and c, the roots are -1 ± i √ � 3_________

2.

SPIRAL REVIEW

72.2.1 cm_______60 days

= 0.035 cm/day

h(x) = 0.035x

73. 2x - 3y = -7

x - 6y = 1

2 -3 -7

�

1 -6 -1

R 1 = R2

1 -6 1

�

2 -3 -7

R 2 = R2 - 2 R1

1 -6 1

�

0 9 -9

R 2 = 1__9

R2

1 -6 1

�

0 1 -1

R 1 = R1 + 6 R2

1 0 -5

�

0 1 -1

solution (-5, -1)

74. 2x + 3y = 12

x + y = 14

1 1 14

�

2 3 12

R 2 = R2 - 2 R1

1 1 14

�

0 1 -16

R 1 = R1 - R2

1 0 30

�

0 1 -16

solution (30, -16)


75. 4x + 5y = -1

2x - 7y = 9

4 5 -1

�

2 -7 9

R 1 = R2

2 -7 9

�

4 5 -1

R 1 = 1__2

R2

1 - 7__

2

9__2

�

4 5 -1

R 2 = R2 - 4 R1

1 - 7__

2

9__2

�

0 19 -19

R 2 = 1___19

R2

1 - 7__

2

9__2

�

0 1 -1

R 1 = R1 + 7__2

R2

1 0 0

�

0 1 -1

solution (1, -1)

76. x2 - 5x = 1

x 2 - 5x +

25___4

= 1 + 25___4

(x - 25___2 )

2 =

29___4

x - 5__2

= ± √ 29______

4

x = 5 ± √ 29________

2

77. 2 x2 = 16x - 4

2 x2 - 16x = -4

x 2 - 8x = -2

x 2 - 8x + 16 = -2 + 16

(x - 4 )2 = 14

x - 4 = ± √ 14

x = 4 ± √ 14

78. 3x = 5 x2 - 12

-5x2 + 3x = -12

x2 - 0.6 = 2.4

x2 - 0.6 + 0.09 = 2.4 + 0.09

(x - 0.3 )2 = 2.49

x - 0.3 = ± √ 2.49

x = 0.3 ± √ 2.49

x = 3 ± √ 249_________

10

READY TO GO ON? PAGE 365

1. g is f translated 2 units left and 4 units down.

2. g is f reflected across the x-axis, vertically stretched

by a factor of 4, and translated 1 unit right.

3. g is f vertically compressed by a factor of 1__2

and

translated 1 unit up.

4. g(x) = 9(x + 2 )2

5. g(x) = - x2 + 4

6a. upward

b. x = -b___2a

= -(-4)______2(1)

= 2

c. f(2) = (2 )2 - 4(2) + 3

= 4 - 8 + 3

= -1



e.

7a. downward

b. x = -b___2a

= -2_____2(-1)

= 1

c. g(1) = -(1)2 + 2(1) - 1

= -1 + 2 - 1

= 0



e.

8a. upward

b. x = -b___2a

= -(-6)______2(1)

= 3

c. h(3) = 32 - 6(3)

= 9 - 18 = -9



e.


9. x = -b___2a

= -0.5__________

2(-0.0075)=

-0.5_______-0.015

= 33.33

h(33.33) = -0.0075(33.33)2 + 0.5(33.33) + 5

= -8.33 + 16.665 + 5

= 13.3

The maximum height of the ball is about 13 ft.

10. x2 - 100 = 0

(x - 10)(x + 10) = 0

x - 10 = 0 or x + 10 = 0

x = 10 or x = -10

11. x2 + 5x - 24 = 0

(x - 3)(x + 8) = 0

x - 3 = 0 or x + 8 = 0

x = 3 or x = -8

12. 4 x2 + 8x = 0

4x(x + 2) = 0

x = 0 or x + 2 = 0

x = 0 or x = -2

13. x2 - 6x = 40

x 2 - 6x + 9 = 40 + 9

(x - 3 )2 = 49

x - 3 = ± √ �� 49

x = -4 or 10

14. x2 + 18x = 15

x 2 + 18x + 81 = 15 + 81

(x + 9 )2 = 96

x + 9 = ± √ �� 96

x = -9 ± 4 √ � 6

15. x2 + 14x = 8

x 2 + 14x + 49 = 8 + 49

(x + 7 )2 = 57

x + 7 = ± √ �� 57

x = -7 ± √ �� 57

16. f(x) = x2 + 24x + 138

= (x2 + 24x) + 138

= (x2 + 24x + 144) + 138 - 144

= (x + 12 )2 - 6


17. g(x) = x2 - 12x + 39

= (x2 - 12x) + 39

= (x2 - 12x + 36) + 39 - 36

= (x - 6 )2 + 3


18. h(x) = 5 x2 - 20x + 9

= (5x2 - 20x) + 9

= 5 (x2 - 4x) + 9

= 5 (x2 - 4x + 4) + 9 - 5(4)

= 5(x - 2 )2 - 11


19. 3 x2 = -48

x 2 = -16

x = ± √ �� -16

x = ±4i

20. x2 - 20x = -125

x 2 - 20x + 100 = -125 + 100

(x - 10 )2 = -25

x - 10 = √ �� -25

x = 10 ± 5i

21. x2 - 8x + 30 = 0

x = -b ± √ �� b

2 - 4ac

______________2a

= -(-8) ± √ �� (-8)

2 - 4(1)(30)

_______________________ 2(1)

= 8 ± √ �� -56_________

2

= 8 ± 2i √ �� 14_________

2 = 4 ± i √ �� 14

22. f(x) = x2 + 12x + 36 + 2

= x2 + 12x + 38

x = -b ± √ �� b

2 - 4ac

______________2a

= -12 ± √ �� 12

2 - 4(1)(38)

____________________ 2(1)

= -12 ± √ �� -8

___________2

= -12 ± 2i √ � 2

___________2

= -6 ± i √ � 2

23. x = -b ± √ �� b

2 - 4ac

______________2a

= -7 ± √ �� 7

2 - 4(1)(15)

__________________ 2(1)

= -7 ± √ �� 49 - 60

______________2

= - 7__2

± i √ � 11____

2

24. x = -b ± √ �� b

2 - 4ac

______________2a

= -(-5) ± √ �� (-5)

2 - 4(2)(3)

______________________ 2(2)

= 5 ± √ �� 25 - 24

____________4

= 5 ± 1_____

4 = 1.5 or 1

25. 400 = 4 t2 + 18t

0 = 4 t2 + 18t - 400

t = - 18 ± √ �� (18)

2 - 4(4)(- 400)

________________________ 2 (4)

t = 8 s


5-7 SOLVING QUADRATIC INEQUALITIES,

PAGES 366-373

CHECK IT OUT!

1a. b.

2a. -1 < x < 2 b. x ≤ 0 or x ≥ 2.5

3a. x2 - 6x + 10 = 2

x 2 - 6x + 8 = 0

(x - 4)(x - 2) = 0

x - 4 = 0 or x - 2 = 0

x = 4 or x = 2

x-value test: 02 - 6(0) + 10 ≥ 2 �

32 - 6(3) + 10 ≥ 2 �

52 - 6(5) + 10 ≥ 2 �

Therefore x ≤ 2 or x ≥ 4.

b. -2x2 + 3x + 7 = 2

-2x2 + 3x + 5 = 0

-1(x + 1)(2x - 5) = 0

x + 1 = 0 or 2x - 5 = 0

x = -1 or x = 5__2

x-value test: -2(-2)2 + 3(-2) + 7 < 2 �

-2(0)2 + 3(0) + 7 < 2 �

-2(3)2 + 3(3) + 7 < 2 �

Therefore x < -1 or x > 2.5.

4. -25x2 + 1250x - 5000 ≥ 7500

-25x2 + 1250x - 5000 = 7500

-25x2 + 1250x - 12500 = 0

-25(x2 - 50 + 500) = 0

x = -b ± √ �� b

2 - 4ac

______________2a

= -(-50) ± √ �� (-50)

2 - 4(1)(500)

__________________________ 2(1)

= 50 ± √ �� 500

__________2

= 25 ± 5 √ � 5

x ≈ 36.2 or x ≈ 13.8

x-value test: -25(0)2 + 1250(0) - 5000 > 7500 �

-25(20)2 + 1250(20) - 5000 > 7500 �

-25(40)2 + 1250(40) - 5000 > 7500 �

Therefore there should be fewer than 14 peope or

more than 36 people.

THINK AND DISCUSS

1. Possible answer: Graphing a quadratic inequality

is essentially the same as graphing a linear

inequality: plot points on a boundary, connect them

with a solid or dashed curve, and shade above or

below accordingly. For a quadratic inequality, the

boundary is a parabola, and for a linear inequality,

the boundary is a line.

2. Possible answer: The intersection points are

included when the inequality symbol is ≥ or ≤.

The intersection points are not included when the

inequality symbol is > or <.

3.

EXERCISES

GUIDED PRACTICE

1. Possible answer: y < x2 + 4x + 4

2. 3.

4.

5. 0 ≤ x ≤ 5 6. x < 0 or x > 1

7. 1.5 ≤ x ≤ 3

8. x2 + 10x + 1 = 12

x2 + 10x - 11 = 0

(x - 1)(x + 11) = 0

x - 1 = 0 or x + 11 = 0

x = 1 or x = -11

x-value test: (-15)2 + 10(15) + 1 ≥ 12 �

(0)2 + 10(0) + 1 ≥ 12 �

(2 )2 + 10(2) + 1 ≥ 12 �

Therefore x ≤ -11 or x ≥ 1.


9. x2 + 13x + 45 = 5

x 2 + 12x + 40 = 0

(x + 5)(x + 8) = 0x + 5 = 0 or x + 8 = 0 x = -5 or x = -8x-value test: (-10)

2 + 13(-10) + 45 < 5 �

(-6)2 + 13(-6) + 45 < 5 �

02 + 13(0) + 45 < 5 �

Therefore -8 < x < -5.

10. -2x2 + 3x + 12 = 10

-2x2 + 3x + 2 = 0

-12(x2 - 3x - 2) = 0

-1(x - 2)(2x + 1) = 0x - 2 = 0 or 2x + 1 = 0

x = 2 or x = - 1__2

x-value test: -2(-1)2 + 3(-1) + 12 > 10 �

-2(1)2 + 3(1) + 12 > 10 �

-2(3)2 + 3(3) + 12 > 10 �

Therefore -0.5 < x < 2.

11. -50x2 + 3500x - 2500 > 50000

-50x2 + 3500x - 2500 = 50000

-50x2 + 3500x - 52500 = 0

-50(x2 - 70x + 1050) = 0

x = -b ± √ �� b

2 - 4ac

______________

2a

= -(-70) ± √ �� (-70)

2 - 4(1)(1050)

___________________________ 2(1)

= 70 ± √ �� 700

__________2

= 35 ± 5 √ � 7x ≈ 48.22 or x ≈ 21.78x-value test: -50(0)

2 + 3500(0) - 2500 > 50000 �

-50(30)2 + 3500(30) - 2500 > 50000 �

-50(50)2 + 3500(50) - 2500 > 50000 �

Therefore the costs must be in the range of $21.78 < x < $48.22.


12. 13.

14. 15.

16. 17.

18. -2 < x < 3 19. x ≤ -1 or x ≥ -0.5

20. x < 2 or x > 3 21. x < -2 or x > 4

22. -7 ≤ x ≤ 0 23. 2 < x < 10

24. x2 - 11x + 13 = 25

x2 - 11x - 12 = 0

(x + 1)(x - 12) = 0x + 1 = 0 or x - 12 = 0 x = -1 or x = 12

x-value test: (-5)2 - 11(-5) + 13 ≤ 25 �

(0 )2 - 11(0) + 13 ≤ 25 �

(15)2 - 11(15) + 13 ≤ 25 �

Therefore -1 ≤ x ≤ 12.

25. -2x2 + 3x + 4 = -1

-2x2 + 3x + 5 = 0

-12(x2 - 3x - 5) = 0

-1(x + 1)(2x - 5) = 0x + 1 = 0 or 2x - 5 = 0

x = -1 or x = 5__2

x-value test: -2(-5)2 + 3(-5) + 4 ≥ -1 �

-2(0)2 + 3(0) + 4 ≥ -1 �

-2(5)2 + 3(5) + 4 ≥ -1 �

Therefore -1 ≤ x ≤ 2.5.

26. x2 - 5x - 4 = -9

x 2 - 5x + 5 = 0

x = -b ± √ �� b

2 - 4ac

______________

2a

= -(-5) ± √ �� (-5)

2 - 4(1)(5)

______________________ 2(1)

= 5 ± √ � 5_______

2x ≈ 3.62 or x ≈ 1.38x-value test: (0 )

2 - 5(0) - 4 < -9 �

(2 )2 - 5(2) - 4 < -9 �

(4 )2 - 5(4) - 4 < -9 �

Therefore 5 - √ � 5______

2< x <

5 + √ � 5______2

.


27. -0.0095x2 + x + 7 > 10

-0.0095x2 + x + 7 = 10

-0.0095x2 + x - 3 = 0

x = -b ± √ �� b

2 - 4ac

______________

2a

= -1 ± √ �� 1

2 - 4(-0.0095)(-3)

________________________ 2(-0.0095)

= -1 ± √ �� 0.886

____________-0.019

x ≈ 3.12 or x ≈ 102.1

The ball wil be knocked down at distances less than

3 ft and greater than 102 ft.

28. 29.

30. 31.

32. 33.

34a. -0.007x2 + x + 20 ≥ 25

-0.007x2 + x + 20 = 25

-0.007x2 + x - 5 = 0

x = -b ± √ �� b

2 - 4ac

______________

2a

= -1 ± √ �� 1

2 - 4(-0.007)(-5)

_______________________ 2(-0.007)

= -1 ± √ �� 0.86

___________-0.014

x ≈ 5 or x ≈ 138

axis of symmetry: x = x1 + x2_______

2= 71.5

h(71.5) = -0.007(71.5)2 + 71.5 + 20 ≈ 55.7 ft

The height of the tent can be no more than

55.7 - 5 ≈ 51 ft.

b. The plot should be placed at the axis of symmetry,

or at 71.5 ft.

35. x2 - 5x - 24 = 0

(x - 8)(x + 3) = 0

x - 8 = 0 or x + 3 = 0

x = 8 or x = -3

x-value test: (-5)2 - 5(-5) - 24 ≤ 0 �

(0 )2 - 5(0) - 24 ≤ 0 �

(10)2 - 5(10) - 24 ≤ 0 �


36. x2 - 14 = 2

x2 - 16 = 0

(x - 4)(x + 4) = 0

x - 4 = 0 or x + 4 = 0

x = 4 or x = -4

x-value test: (-5)2 - 14 ≥ 2 �

(0 )2 - 14 ≥ 2 �

(5 )2 - 14 ≥ 2 �

Therefore x ≤ -4 or x ≥ 4.

37. -2x2 - x + 8 = 6

-2x2 - x + 2 = 0

x = -b ± √ �� b

2 - 4ac

______________

2a

= -(-1) ± √ �� (-1)

2 - 4(-2)(2)

________________________ 2(-2)

= 1 ± √ �� 17________

-4x ≈ 0.78 or x ≈ -1.28

x-value test: -2(-2)2 - (-2) + 8 > 6 �

-2(0)2 - (0) + 8 > 6 �

-2(1)2 - (1) + 8 > 6 �

Therefore 1 + √ �� 17________

-4< x <

1 - √ �� 17________-4

.

38. x2 - 4x - 5 = -9

x 2 - 4x + 4 = 0

(x - 2 )2 = 0

x - 2 = 0

x = 2

39. 3 x2 + 6x + 11 = 10

3 x2 + 6x + 1 = 0

x = -b ± √ �� b

2 - 4ac

______________

2a

= -6 ± √ �� 6

2 - 4(3)(1)

_________________ 2(3)

= -6 ± √ �� 24_________

6

= -1 ± √ � 6___3

x ≈ -0.18 or x ≈ -1.81

x-value test: 3(-2)2 + 6(-2) + 11 < 10 �

3(-1)2 + 6(-1) + 11 < 10 �

3(0)2 + 6(0) + 11 < 10 �

Therefore -1 - √ � 6___3

< x < -1 + √ � 6___3

.


40. 4x2 - 9 = 0

(2x - 3)(2x + 3) = 0

2x - 3 = 0 or 2x + 3 = 0

x = 3__2

or x = - 3__2

x-value test: 4(-2)2 - 9 > 0 �

4(0 )2 - 9 > 0 �

4(2 )2 - 9 > 0 �

Therefore x < -1.5 or x > 1.5.

41. 3 x2 + 5x + 13 = 16

3 x2 + 5x - 3 = 0

x = -b ± √ �� b

2 - 4ac

______________

2a

= -5 ± √ �� 5

2 - 4(3)(-3)

__________________ 2(3)

= -5 ± √ �� 61_________

6x ≈ -2.14 or x ≈ 0.47

x-value test: 3(-3)2 + 5(-3) + 13 ≤ 16 �

3(0 )2 + 5(0) + 13 ≤ 16 �

3(1 )2 + 5(1) + 13 ≤ 16 �

Therefore -5 - √ �� 61_________

6≤ x ≤

-5 + √ �� 61_________6

.

42. -2x2 + 3x + 17 = 11

-2x2 + 3x + 6 = 0

x = -b ± √ �� b

2 - 4ac

______________

2a

= -3 ± √ �� 3

2 - 4(-2)(6)

__________________ 2(-2)

= -3 ± √ �� 57_________

-4x ≈ 2.6 or x ≈ -1.125

x-value test: -2(-2)2 + 3(-2) + 17 ≥ 11 �

-2(0)2 + 3(0) + 17 ≥ 11 �

-2(3)2 + 3(3) + 17 ≥ 11 �

Therefore -3 + √ �� 57_________

-4≤ x ≤

-3 - √ �� 57_________-4

.

43. 5 x2 - 2x - 1 = 0

x = -b ± √ �� b

2 - 4ac

______________

2a

= -(-2) ± √ �� (-2)

2 - 4(5)(-1)

________________________ 2(5)

= 2 ± √ �� 24________

10

= 1 ± √ � 6_______

5x ≈ 0.69 or x ≈ -0.3

x-value test: 5(-1)2 - 2(-1) - 1 ≥ 0 �

5(0)2 - 2(0) - 1 ≥ 0 �

5(1 )2 - 2(1) - 1 ≥ 0 �

Therefore x ≤ 1 - √ � 6_______

5 or x ≥

1 + √ � 6_______5

.

44. (x - 2)(x + 11) = 2

x2 + 9x - 22 = 2

x2 + 9x - 24 = 0

x = -b ± √ �� b

2 - 4ac

______________

2a

= -9 ± √ �� 9

2 - 4(1)(-24)

___________________ 2(1)

= -9 ± √ �� 177

__________2

x ≈ 2.15 or x ≈ -11.15

x-value test: (-12 - 2)(-12 + 11) ≥ 2 �

(0 - 2)(0 + 11) ≥ 2 �

(3 - 2)(3 + 11) ≥ 2 �

Therefore x ≤ -9 - √ �� 177

__________2

or x ≥ -9 + √ �� 177

__________2

.

45. x2 + 27 = 12x

x2 - 12x + 27 = 0

(x - 3)(x - 9) = 0

x - 3 = 0 or x - 9 = 0

x = 3 or x = 9

x-value test: (0 )2 + 27 > 12(0) �

(5 )2 + 27 > 12(5) �

(10 )2 + 27 > 12(10) �

Therefore x < 3 or x > 9.

46. -2x2 + 3x + 6 = 0

x = -b ± √ �� b

2 - 4ac

______________

2a

= -3 ± √ �� 3

2 - 4(-2)(6)

__________________ 2(-2)

= -3 ± √ �� 57_________

-4x ≈ 2.6 or x ≈ -1.14

x-value test: -2(-2)2 + 3(-2) + 6 > 0 �

-2(0)2 + 3(0) + 6 > 0 �

-2(3)2 + 3(3) + 6 > 0 �

Therefore -3 + √ �� 57_________

-4< x <

-3 - √ �� 57_________-4

.

47. (200 + x)(120 + x) ≥ 35000

(200 + x)(120 + x) = 35000

x 2 + 320x + 24000 = 35000

x 2 + 320x - 11000 = 0

x = -b ± √ �� b

2 - 4ac

______________

2a

= -320 ± √ �� 320

2 - 4(1)(-11000)

__________________________ 2(1)

= -320 ± √ �� 146400

_______________2

x ≈ 31.3 or x ≈ -335.6

The lot can be expanded by a distance between 0 ft

and about 31 ft.

48. B 49. A

50. C


51a. A(x) = b × h_____

2

A(x) = x · (20 - x)

__________2

A(x) = - 1__2

x2 + 10x

b. - 1__2

x2 + 10x ≥ 30

- 1__2

x2 + 10x = 30

- 1__2

x2 + 10x - 30 = 0

x = -b ± √ �� b

2 - 4ac

______________

2a

=

-10 ± √ ��

102 - 4(- 1__

2)(-30)

_______________________ 2(- 1__

2)

= -10 ± √ �� 40

__________-1

= 10 ± √ �� 40x ≈ 16.32 or x ≈ 3.68

The range of the x value is 3.7 ≤ x ≤ 16.3.

c. - 1__2

x2 + 10x > 40

- 1__2

x2 + 10x = 40

- 1__2

x2 + 10x - 40 = 0

x = -b ± √ �� b

2 - 4ac

______________

2a

=

-10 ± √ ��

102 - 4(- 1__

2)(-40)

_______________________ 2(- 1__

2)

= -10 ± √ �� 20

__________-1

= 10 ± √ �� 20x ≈ 14.47 or x ≈ 5.53

The range of the x value is 0 < x ≤ 5.5 or 14.5 ≤ x ≤ 20.

52. Freemont Park: -200t

2 + 2500t - 900 ≥ 1000

-200t2 + 2500t - 900 = 1000

-200t2 + 2500t - 1900 = 0

-100(2t2 - 25t + 19) = 0

t = -b ± √ �� b

2 - 4ac

______________

2a

= -(-25) ± √ �� (-25)

2 - 4(2)(19)

_________________________ 2(2)

= 25 ± √ �� 473

__________4

t ≈ 0.82 or t ≈ 11.68 The tickets should be between $0.82 and $11.68.Saltillo Plaza:-200t

2 + 2500t - 1500 ≥ 1000

-200t2 + 2500t - 1500 = 1000

-200t2 + 2500t - 2500 = 0

-100(2t2 - 25t + 25) = 0

t = -b ± √ �� b

2 - 4ac

______________

2a

= -(-25) ± √ �� (-25)

2 - 4(2)(25)

_________________________ 2(2)

= 25 ± √ �� 425

__________4

t ≈ 1.1 or t ≈ 11.4The tickets should be between $1.10 and $11.40.Riverside Walk:-200t

2 + 2500t - 2500 ≥ 1000

-200t2 + 2500t - 2500 = 1000

-200t2 + 2500t - 3500 = 0

-100(2t2 - 25t + 35) = 0

t = -b ± √ �� b

2 - 4ac

______________

2a

= -(-25) ± √ �� (-25)

2 - 4(2)(35)

_________________________ 2(2)

= 25 ± √ �� 345

__________4

t ≈ 1.61 or t ≈ 10.89The tickets should be between $1.61 and $10.89.

53a. Possible answer: A(x) = -2x2 + 40x

b. -2x2 + 40x ≥ 150

-2x2 + 40x = 150

-2x2 + 40x - 150 = 0

-2(x2 - 20x + 75) = 0

-2(x - 5)(x - 15) = 0x = 5 or x = 15

A width between 5 ft to 15 ft will give the required area.

c. -2x2 + 40x ≥ 200

-2x2 + 40x = 200

-2x2 + 40x - 200 = 0

-2(x2 - 20x + 100) = 0

-2(x - 10 )2 = 0

x = 10 A width of 10 ft will give the required area.


54. x < -8.1 or x > 2.1 55. 0.9 ≤ x ≤ 14.1

56. -7.2 < x < 7.2 57. x ≤ -2.2 or x ≥ 0.7

58a. - x2 + 125x = 0

-x(x - 125) = 0x = 0 or x - 125 = 0x = 0 or x = 125

axis of symmetry : 0 + 125_______

2= 62.5

r(62.5) = -(62.5)2 + 125(62.5) = 3906.25

The maximum revenue is about $3906.

b. - x2 + 125x ≥ 1500

- x2 + 125x = 1500

- x2 + 125x - 1500 = 0

x = -b ± √ �� b

2 - 4ac

______________

2a

= -125 ± √ �� (125)

2 - 4(-1)(-1500)

____________________________ 2(-1)

= -125 ± √ �� 9625

_____________-2

x ≈ 111.55 or x ≈ 13.45 The snow borads must be purchased between $14 and $111 inclusive.

59. It is not; the solution can be a single value, all real numbers, or the empty set.

60. Yes; possible answer: 4 x2 - x + 3 > 0.

61. The solutions of x2 - 3x - 4 ≤ 6 include a range of

x-values, - 2 ≤ x ≤ 5, while the solutions of

x 2 - 3x - 4 = 6 consist of only two x-values,

x = - 2 and x = 5.

TEST PREP

62. A 63. J

64. B

65. x2 + 4x + 4 = 1

x2 + 4x + 3 = 0

(x + 1)(x + 3) = 0x + 1 = 0 or x + 3 = 0 x = -1 or x = -3x-value test: (-5)

2 + 4(-5) + 4 > 1 �

(-2)2 + 4(-1) + 4 > 1 �

(0 )2 + 4(0) + 4 > 1 �

Therefore x < -3 or x > -1. Check students’ work.


66.

4

6

2

-2

x

y

0-2 2 4-4

67.

8

12

4

-4

x

y

0-4 4 8-8

68.

8

12

4

-4

x

y

0-4 4 8-8

69. x2 + 5x - 6 = 8

x 2 + 5x - 14 = 0

(x - 2)(x + 7) = 0x - 2 = 0 or x + 7 = 0 x = 2 or x = -7base = x2 - x1 = -7 - 2 = -9 = 9 units

axis of symmetry: 2 + (-7)________

2= -5___

2= -2.5

h = f(-2.5) = (-2.5)2 + 5(-2.5) - 14

= -20.25 = 20.25 units

A = 2__3

bh = 2__3

(9)(20.25) = 121.5 square units

70. -2x2 + 3x + 9 = -5

-2x2 + 3x + 14 = 0

-1(2x2 - 3x - 14) = 0

-1(x + 2)(2x - 7) = 0x + 2 = 0 or 2x - 7 = 0 x = -2 or x = 3.5base = x2 - x1 = 3.5 - (-2) = 5.5 units

axis of symmetry: -2 + 3.5________

2= 0.75

h = f(0.75) = -2(0.75)2 + 3(0.75) + 14

= 15.125 units

A = 2__3

bh = 2__3

(5.5)(15.125) = 55.46 square units

SPIRAL REVIEW

71. Reed Home: Laundry, Shopping;Brown Home: Lawn, Laundry;Sandi Home: Shopping, Cleaning, Lawn;Clem Home: Laundry, Cleaning, Shopping.

72. 73.

74.

75. 4 - 2c = -14 -2c = -18 x = 9


76. 4c + 2 - 3i + 2(i - 5) = 4(2i - 6) - 9i

4c + 2 - 3i + 2i - 10 = 8i - 24 - 9i

4c - 8 - i = -i - 24

4c - 8 = -24

4c = -16

c = -4

5-8 CURVE FITTING WITH QUADRATIC

MODELS, PAGES 374-381

CHECK IT OUT!

1a. Quadratic; second differences are constant for

equally spaced x-values.

b. Not quadratic; first differences are constant so the

function is linear.

2. (x, y) f(x) = a x 2 + bx + c System in a, b, c

(0, -3) -3 = a(0)2 + b(0) + c

c = -3 �

a + b + c = 0 �

4a + 2b + c = 1 �

(1, 0) 0 = a(1)2 + b(1) + c

(2, 1) 1 = a(2)2 + b(2) + c

Substitute c = -3 from equation 1 into both

equation 2 and equation 3.

� a + b - 3 = 0

a + b = 3

�

� 4a + 2b - 3 = 1

4a + 2b = 4 �

Solve equation 4 and equation 5 for a and b using

elimination.

� 4(a + b) = 4(3) → 4a + 4b = 12

� -(4a + 2b) = -4 → -4a - 2b = -4

2b = 8

b = 4

Substitute 4 for b into equation 4 to find a.

� a + b = 3 → a + (4) = 3

a = -1

f(x) = a x 2 + bx + c = -1x

2 + 4x - 3

= - x2 + 4x - 3

3. L(d) ≈ 14.3 d2 - 112.4d + 430.1

L(8) ≈ 14.3(8 )2 - 112.4(8) + 430.1 ≈ 446

The reel length for an 8-inch-diameter film is about

446 ft.

THINK AND DISCUSS

1. Possible answer: If the data set has equally

spaced x-values and the second differences of the

corresponding function values are constant and

nonzero, then the data set is quadratic.

2. Possible answer: No, a quadratic function would

not be a good model, because a quadratic function

could only model one of the plane’s changes in

direction.

3.

EXERCISES

GUIDED PRACTICE

1. Possible answer: A linear model is used to fit a set

of data whose points lie on or close to a line.

A quadratic model is used to fit a set of data whose

points lie on or close to a parabola.

2. Not quadratic; first differences are constant so the

function is linear.

3. Not quadratic; second differences are not constant

for equally spaced x-values.

4. Quadratic; second differences are constant for

equally spaced x-values.

5. (x, y) y = a x 2 + bx + c System in a, b, c

(0, -3) -3 = a(0)2 + b(0) + c

c = -3 �

9a + 3b + c = 0 �

4a - 2b + c = 5 �

(3, 0) 0 = a(3)2 + b(3) + c

(-2, 5) 5 = a(-2)2 + b(-2) + c



� 9a + 3b - 3 = 0

9a + 3b = 3

3a + b = 1 �

� 4a - 2b - 3 = 5

4a - 2b = 8 �


elimination.

� 2(3a + b) = 2(1) → 6a + 2b = 2

� 4a - 2b = 8 → -4a - 2b = -4

2a = -2

a = 1

Substitute 1 for a into equation 4 to find b.

� 3a + b = 1 → 3(1) + b = 1

b = -2

y = a x 2 + bx + c = 1 x

2 + (-2)x - 3

= x2 - 2x - 3



(0, 1) 1 = a(0)2 + b(0) + c

c = 1 �

4a + 2b + c = -1 �

9a + 3b + c = -8 �

(2, -1) -1 = a(2)2 + b(2) + c

(3, -8) -8 = a(3)2 + b(3) + c

Substitute c = 1 from equation 1 into both equation 2 and equation 3.� 4a + 2b + 1 = -1 4a + 2b = -2 2a + b = -1 �

� 9a + 3b + 1 = -8 9a + 3b = -9 3a + b = -3 �

Solve equation 4 and equation 5 for a and b using elimination.� 2a + b = -1 → 2a + b = -1� -(3a + b) = -(-3) → -3a - b = 3 -a = 2 a = -2Substitute -2 for a into equation 4 to find b.� 2a + b = -2 → 2(-2) + b = -1

b = 3

y = a x 2 + bx + c = -2x

2 + 3x + 1

= -2x2 + 3x + 1


(0, 4) 4 = a(0)2 + b(0) + c

c = 4 �

a - b + c = 8 �

4a + 2b + c = 2 �

(-1, 8) 8 = a(-1)2 + b(-1) + c

(2, 2) 2 = a(2)2 + b(2) + c

Substitute c = 4 from equation 1 into both equation 2 and equation 3.� a - b + 4 = 8

a - b = 4 �

� 4a + 2b + 4 = 2 4a + 2b = -2 2a + b = -1 �

Solve equation 4 and equation 5 for a and b using elimination.� a - b = 4 → a - b = 4� 2a + b = -1 → 2a + b = -1 3a = 3 a = 1Substitute 1 for a into equation 4 to find b.� a - b = 4 → (1) - b = 4

b = -3

y = a x 2 + bx + c = 1 x

2 + (-3)x + 4

= x2 - 3x + 4


(0, -7) -7 = a(0)2 + b(0) + c

c = -7 �

16a - 4b + c = 9 �a + b + c = -1 �

(-4, 9)9 = a(-4)2 + b(-4) + c

(1, -1) -1 = a(1)2 + b(1) + c

Substitute c = -7 from equation 1 into both equation 2 and equation 3.� 16a - 4b - 7 = 9 16a - 4b = 16 4a - b = 4 4

� a + b - 7 = -1a + b = 6 �

Solve equation 4 and equation 5 for a and b using elimination.� 4a - b = 4 → 4a - b = 4� a + b = 6 → a + b = 6 5a = 10 a = 2Substitute 2 for a into equation 5 to find b.� a + b = 6 → (2) + b = 6

b = 4

y = a x 2 + bx + c = 2 x

2 + 4x - 7

9. y = a x 2 + bx + c System in a, b, c

3 = a(2)2 + b(2) + c

4a + 2b + c = 336a + 6b + c = 3 64a + 8b + c = -1

3 = a(6)2 + b(6) + c

-3 = a(8)2 + b(8) + c

→

4 2 1

�

a

�

=

3

�

→

a

�

=

-0.5

�

36 6 1 b 3 b 4

64 8 1 c -1 c -3

y = a x 2 + bx + c = - 1__

2x

2 + 4x - 3


-12 = a(-1)2 + b(-1) + c

a - b + c = -12a + b + c = 04a + 2b + c = 9

0 = a(1)2 + b(1) + c

9 = a(2)2 + b(2) + c

→

1 -1 1

�

a

�

=

-12

�

→

a

�

=

1

�

1 1 1 b 0 b 6

4 2 1 c 9 c -7

y = a x 2 + bx + c = x

2 + 6x - 7

11. C(x) ≈ 0.0098 x2 + 0.62x + 3.8

The average side length is 24 + 36_______

2= 30 in.

C(30) ≈ 0.0098(30 )2 + 0.62(30) + 3.8 ≈ 31.20

The cost of mounting a 24 in. × 36 in. photo is about $31.20.



12. Quadratic; second differences are constant for equally spaced x-values.

13. Quadratic; second differences are constant for equally spaced x-values.

14. Not quadratic; first differences are constant so the function is linear.


5 = a(-2)2 + b(-2) + c

4a - 2b + c = 5a - b + c = 0a + b + c = -2

0 = a(-1)2 + b(-1) + c

-2 = a(1)2 + b(1) + c

→

4 -2 1

�

a

�

=

5

�

→

a

�

=

4__3

�

1 -1 1 b 0 b -1

1 1 1 c -2 c - 7__3

y = a x 2 + bx + c = 4__

3x

2 - x - 7__

3


2 = a(1)2 + b(2) + c

a + b + c = 24a + 2b + c = -125a + 5b + c = 2

-1 = a(2)2 + b(2) + c

2 = a(5)2 + b(5) + c

→

1 1 1

�

a

�

=

2

�

→

a

�

=

1

�

4 2 1 b -1 b -6

25 5 1 c 2 c 7

y = a x 2 + bx + c = x

2 - 6x + 7


12 = a(-4)2 + b(-4) + c

16a - 4b + c = 124a - 2b + c = 04a + 2b + c = -12

0 = a(-2)2 + b(-2) + c

-12 = a(2)2 + b(2) + c

→

16 -4 1

�

a

�

=

12

�

→

a

�

=

0.5

�

4 -2 1 b 0 b -3

4 2 1 c -12 c -8

y = a x 2 + bx + c = 0.5 x

2 - 3x - 8


2.6 = a(-1)2 + b(-1) + c

a - b + c = 2.6a + b + c = 4.24a + 2b + c = 14

4.2 = a(1)2 + b(1) + c

14 = a(2)2 + b(2) + c

→

1 -1 1

�

a

�

=

2.6

�

→

a

�

=

3

�

1 1 1 b 4.2 b 0.8

4 2 1 c 14 c 0.4

y = a x 2 + bx + c = 3x

2 + 0.8x + 0.4

19. y ≈ -3.7x2 + 216x + 781

x = 2015 - 1999 = 16

y ≈ -3.7(16)2 + 216(16) + 781 ≈ 3290

The amount spent will be about $3290 million or $3.29 billion.

20. The function is C(r) = 2πr, which is linear.

21. The function is A(b) = ( 1__2

h) b, which is linear.

22. The function is P(t) = 2t, which is neither linear nor

quadratic.

23. The function is A(s) = s2 , which is quadratic.

24a. Galileo’s rule

b. Aristotle’s rule: linear; da Vinci’s rule: quadratic; Galileo’s rule: quadratic.

c. In da Vinci’s pattern, the first differences increase by 1. In Galileo’s pattern, the first differences increase by 2.

d. Galileo’s rule shows the odd-number pattern.

25. -3 26. -2

27. -1

28a. f(x) = a x 2 + bx + c System in a, b, c

28 = a(4)2 + b(4) + c

16a + 4b + c = 2836a + 6b + c = 5464a + 8b + c = 88

54 = a(6)2 + b(6) + c

88 = a(8)2 + b(8) + c

→

16 4 1

�

a

�

=

28

�

→

a

�

=

1

�

36 6 1 b 54 b 3

64 8 1 c 88 c 0

f(x) = a x 2 + bx + c = x

2 + 3x

b. f(9) = (9 )2 + 3(9) = 108

The area is 108 i n2 .

29a.f(x) = a x

2 + bx + c System in a, b, c

9.95 = a(12)2 + b(12) + c

144a + 12b + c = 9.95196a + 14b + c = 13.95256a + 16b + c = 16.95

13.95 = a(14)2 + b(14) + c

16.95 = a(16)2 + b(16) + c

→

144 12 1

�

a

�

=

9.95

�

→

a

�

=

-0.125

�

196 14 1 b 13.95 b 5.25

256 16 1 c 16.95 c -35.05

f(x) = a x 2 + bx + c = -0.125x

2 + 5.25x - 35.05

b. f(18) = -0.125(18)2 + 5.25(18) - 35.05 = 18.95

The pizza is $18.95.


c.

a maximum point; the price and size of the most expensive pizza.

d. f(30) = -0.125(30)2 + 5.25(30) - 35.05 = 9.95

f(8) = -0.125(8)2 + 5.25(8) - 35.05 = -1.05

A 30 in. pizza will cost $9.95, and an 8 in. one will cost -$1.05.

e. Possible answer: No, the function is not a good model. It gives negative prices for very small pizzas and low (or negative) prices for very large pizzas.


(0, -1) -1 = a(0)2 + b(0) + c

c = -1 �

4a + 2b + c = -1 �

16a + 4b + c =-9�

(2, -1) -1 = a(2)2 + b(2) + c

(4, -9) -9 = a(4)2 + b(4) + c

Substitute c = -1 from equation 1 into both equation 2 and equation 3.� 4a + 2b - 1 = -1 4a + 2b = 0 2a + b = 0 �

16a + 4b - 1 = -9 16a + 4b = -8 4a + b = -2 �

Solve equation 4 and equation 5 for a and b using elimination.� 2a + b = 0 → 2a + b = 0� -(4a + b) = -(-2) → -4a - b = 2

-2a = 2 a = -1Substitute -1 for a into equation 4 to find b.� 2a + b = 0 → 2(-1) + b = 0

b = 2

y = a x 2 + bx + c = -1x

2 + 2x - 1

= - x2 + 2x - 1

31. Not quadratic 32. Not quadratic


(0, 0) 0 = a(0)2 + b(0) + c

c = 0 �

a - b + c = 3 �

a + b + c = 7 �

(-1, 3) 3 = a(-1)2 + b(-1) + c

(1, 7) 7 = a(1)2 + b(1) + c

Substitute c = 0 from equation 1 into both equation 2 and equation 3.� a - b + 0 = 3

a - b = 3 4 v a + b + 0 = 7

a + b = 7 �

Solve equation 4 and equation 5 for a and b using elimination.� a - b = 3 → a - b = 3� a + b = 7 → a + b = 7 2a = 10

a = 5Substitute 5 for a into equation 4 to find b.� a - b = 3 → (5) - b = 3

b = 2

f(x) = a x 2 + bx + c = 5 x

2 + 2x

34. Not quadratic 35. Not quadratic

36.f(x) = a x


13.2 = a(0)2 + b(0) + c

c = 13.2 �

1.21a + 1.1b + c =18.7�

9a + 3b + c = 0 �

18.7 = a(1.1)2 + b(1.1) + c

0 = a(3)2 + b(3) + c

Substitute c = 13.2 from equation 1 into both equation 2 and equation 3.1.21a + 1.1b + 13.2 = 18.7 1.21a + 1.1b = 5.5 1.1a + b = 5�

9a + 3b + 13.2 = 0 9a + 3b = -13.2 3a + b = -4.4�

Solve equation 4 and equation 5 for a and b using elimination.� 1.1a + b = 5 → 1.1a + b = 5� -(3a + b) = -(-4.4) → -3a - b = 4.4

-1.9a = 9.4 a ≈ -4.95Substitute -4.95 for a into equation 5 to find b.� 3a + b = -4.4 → 3(-4.95) + b = -4.4

b ≈ 10.44

h(t) = a t 2 + bt + c ≈ -4.95 t

2 + 10.44t + 13.2

h(2) ≈ -4.95(2)2 + 10.44(2) + 13.2 ≈ 14.3

The skater’s height is about 14.3 m after 2 seconds.

37a. Answer will vary. Possible answer: 0.75 m.

b. Answer will vary. Possible answer: h(t) ≈ -4.7t

2 + 18.1t - 16.8


38a. y ≈ 0.187 x2 + 84.3x - 863.6

b. y ≈ 119x - 2159

c. Quadratic:

y ≈ 0.187(200 )2 + 84.3(200) - 863.6 = 23,476

Linear: 119(200) - 2159 = 21,641

The light output is 23,476 lumens using the

quadratic model, and 21,641 lumens using the

linear model.

d. Possible answer: The quadratic model would

probably give a better estimate of the light output

for higher values of energy usage because the

data points lie along a slight curve.

39a. y ≈ -10.7x + 208.1

b. No, the relationship is linear because the

coefficient of the x2 term is 0.

c. linear; y = 4x + 8

40a. Possible answer: The last 2 columns appear to

be quadratic because the successive terms grow

faster than a linear function would.

b. Max. length: y ≈ 102 x2 + 114x + 66;

max. weight: y ≈ 1063 x2 + 103x - 16;

yes, the models fit the data well because the

coefficient of determination is very close to 1.

c. max. length: about 178 in.; max. weight: about

464lb

41. t(n) = 1__2

n2 + 1__

2n

42. Three points define a unique parabola. The 3 points

cannot lie on the same line.

43a.

b. y ≈ 0.5x + 3 c. y ≈ -0.13x2 + 2.8x - 6

d. Possible answer: The quadratic model best

describes the data set because the points appear

to lie on a parabola rather than a straight line.

44. The model is a linear model of the form y = bx + c.

TEST PREP

45. B 46. G;

x 1 + x2_______

2=

7 + 3_____2

= 5

47. D

48. H;

f(x) = a x 2 + bx + c System in a, b, c

-8 = a(0)2 + b(0) + c

c = -8 �

9a + 3b + c = 10 �

36a + 6b + c = 34 �

10 = a(3)2 + b(3) + c

34 = a(6)2 + b(6) + c



� 9a + 3b - 8 = 10

9a + 3b = 18

3a + b = 6 �

� 36a + 6b - 8 = 34

36a + 6b = 42

6a + b = 7 �


elimination.

� 3a + b = 6 → 3a + b = 6

� -(6a + b) = -7 → -6a - b = -7

-6a = -1

a = 1__3

Substitute 1__3

for a into equation 4 to find b.

� 3a + b = 6 → 3( 1__3

) + b = 6

b = 5

f(x) = a x 2 + bx + c = 1__

3x

2 + 5x - 8

f(-3) = 1__3

(-3)2 + 5(-3) - 8 = -20


(0, -5) -5 = a(0)2 + b(0) + c

c = -5 �

a + b + c = -3 �

4a + 2b + c = 3 �

(1, -3) -3 = a(1)2 + b(1) + c

(2, 3) 3 = a(2)2 + b(2) + c



� a + b - 5 = -3

a + b = 2 �

� 4a + 2b - 5 = 3

4a + 2b = 8

2a + b = 4 �


elimination.

� -(a + b) = -2 → -a - b = -2

� 2a + b = 4 → 2a + b = 4

a = 2

Substitute 2 for a into equation 4 to find b.

� a + b = 2 → (2) + b = 2

b = 0

f(x) = a x 2 + bx + c = 2 x

2 + (0)x - 5

= 2x2 - 5



50.f(x) = a x


2 = a(1)2 + b(1) + c

a + b + c = 2

16a + 4b + c = 6

49a + 7b + c = w

6 = a(4)2 + b(4) + c

w = a(7)2 + b(7) + c

50a.

→

1 1 1

a

=

2

→

a

=

- 1___18

16 4 1 b 6 b 29___18

49 7 1 c 9 c 4__9

f(x) = a x 2 + bx + c = - 1___

18x

2 +

29___18

x + 4__9

x = -b___2a

= - 29___

18_______

2(- 1___18)

= 29___2

= 14.5

f(14.5) = - 1___18

(14.5)2 +

29___18

(14.5) + 4__9

= 12.125

It has a maximum value, and the vertex is

(14 1__2

, 12 1__8) .

b.

→

1 1 1

a

=

2

→

a

=

1___18

16 4 1 b 6 b 19___18

49 7 1 c 11 c 8__9

f(x) = a x 2 + bx + c =

1___18

x2 +

19___18

x + 8__9

x = -b___2a

= - 19___

18_____2( 1___

18)= - 19___

2= -9.5

f(-9.5) = 1___18

(-9.5)2 +

19___18

(-9.5) + 8__9

= -4.125

It has a minimum value, and the vertex is

(-9 1__2

, -4 1__8) .

c.

→

1 1 1

a

=

2

→

a

=

0

16 4 1 b 6 b 4__3

49 7 1 c 10 c 2__3

f(x) = a x 2 + bx + c = 4__

3x +

2__3

51. Possible answer: Draw a line connecting the point

with the least x-value to the point with the greatest

x-value. If the third point lies above this line, the

parabola opens downward. If the third point lies

below this line, the parabola opens upward.

SPIRAL REVIEW

52. No 53. Yes

54. detA = 1__3

(1) - 0(-4) = 1__3

A -1

= 1__ 1__3

1 0

= 3

1 0

=

3 0

4 1__3 12 14 1__

3

55. No, A-1

does not exist because

detA = 2(-1) - 1(-2)

= 0 and row 1 is a multiple of row 2.

56.

[A I] =

-2 0 1 1 0 0

=

1 0 0 -0.5 0.5 0

0 0 1 0 1 0 0 1 0 1 -2 0.5

4 2 2 0 0 1 0 0 1 0 1 0

A -1

=

- 1__2

1__2

0

1 -2 1__2

0 1 0

57. detA = - 1__2

(3) - 0(-4) = - 3__2

A -1

= 1___- 3__

2

- 1__

2 4

= - 2__

3

- 1__

2 4

=

1__3

- 8__3

0 3 0 -20 3

58. x = -b ± √ �� b

2 - 4ac

______________

2a

= -(-4) ± √ �� (-4)

2 - 4(2)(1)

______________________ 2(2)

= 4 ± √ � 8_______

4

= 4 ± 2 √ � 2________

4

= 2 ± √ � 2_______

2

59. x = -b ± √ �� b

2 - 4ac

______________

2a

= -0 ± √ �� 0

2 - 4(1)(9)

_________________ 2(1)

= 0 ± √ �� -36_________

2

= ±6i___2

= ±3i

60. x = -b ± √ �� b

2 - 4ac

______________

2a

= -10 ± √ �� 10

2 - 4(-3)(12)

_____________________ 2(-3)

= -10 ± √ �� 244

___________-6

= -10 ± 2 √ �� 61

___________-6

= 5 ± √ �� 61________

3


5-9 OPERATIONS WITH COMPLEX

NUMBERS, PAGES 382-389

CHECK IT OUT!

1a-d.

2a. 1 - 2i

√ �� (1)2 + (-2)

2

√ �� 1 + 4

√ � 5

b. - 1__2

- 1__2

+ 0i √

�� (- 1__

2)2 + (0 )

2

√ �

1__4

1__2

c. 23i0 + 23i

√ �� (0)2 + (23 )

2

√ �� 529

23

3a. (-3 + 5i) + (-6i)

-3 + (5i - 6i)

-3 - i

b. 2i - 3 - 5i

-3 + (2i - 5i)

-3 - 3i

c. (4 + 3i) + (4 - 3i)

(4 + 4) + (3i - 3i)

8

4a. b.

5a. 2i(3 - 5i)

6i - 10 i2

6i - 10(-1)

6i + 10

10 + 6i

b. (4 - 4i)(6 - i)

24 - 4i - 24i + 4 i2

24 - 28i + 4(-1)

20 - 28i

c. (3 + 2i)(3 - 2i)

9 - 6i + 6i - 4 i2

9 - 4(-1)

13

6a.1__2

i7

1__2

i4 · i

3

1__2

(1) · -i

- 1__2

i

b. i42

(i6)7

(i4 · i2)7

(1 · (-1))7

(-1)7

-1

7a.3 + 8i_____

-i· i_

i

3i + 8 i

2_______- i2

3i+ 8(-1)_________

-(-1)-8 + 3i

b.3 - i____2 - i

· 2 + i____2 + i

6 + 3i - 2i - i

_____________ 4 + 2i - 2i - i

2

6 + i - (-1)

___________4 - (-1)

7 + i____

5

7__5

+ 1__5

i

THINK AND DISCUSS

1. Possible answer: A complex number a + bi and its

conjugate a - bi can be equal only when b = 0.

2. Possible answer: The real terms are ac and -bd.

The imaginary terms are adi and bci.

3.

EXERCISE

GUIDED PRACTICE

1. real, imaginary

2-5.

6. 4 - 5i

= √ �� (4)2 + (-5)

2

= √ �� 41

7. -33.3= -33.3 + 0i

= √ �� (-33.3)2 + (0 )

2

= √ �� 33.32

= 33.3

8. -9i=0 + (-9i)

= √ �� (0)2 + (-9)

2

= √ �� 81

= 9

9. 5 + 12i

= √ �� (5)2 + (12 )

2

= √ �� 169

= 13

10. -1 + i

= √ �� (-1)2 + (1 )

2

= √ � 2

11. 15i= 0 + 15i

= √ �� (0)2 + (15 )

2

= √ �� 225

= 15

12. (2 + 5i) + (-2 + 5i)

= (2 - 2) + (5i + 5i)

= 10i

13. (-1 - 8i) + (4 + 3i)

= (-1 + 4) + (-8i + 3i)

= 3 - 5i


14. (1 - 3i) - (7 + i)

= (1 - 7) + (-3i - i)

= -6 - 4i

15. (4 - 8i) + (-13 + 23i)

= (4 - 13) + (-8i + 23i)

= -9 + 15i

16. (6 + 17i) - (18 - 9i)

= (6 - 18) + (17i + 9i)

= -12 + 26i

17. (-30 + i) - (-2 + 20i)

= (-30 + 2) + (i - 20i)

= -28 - 19i

18. 1 19. -3 - i

20. -2i 21. (1 - 2i)(1 + 2i)

= 1 - 2i + 2i - 4 i2

= 1 - 4(-1)

= 5

22. 3i(5 + 2i)

= 15i + 6 i2

= 15i + 6(-1)

= -6 + 5i

23. (9 + i)(4 - i)

= 36 - 9i + 4i - i2

= 36 - 5i - (-1)

= 37 - 5i

24. (6 + 8i)(5 - 4i)

= 30 - 24i + 40i - 32 i2

= 30 + 16i -32(-1)

= 62 + 16i

25. (3 + i)2

= 9 + 3i + 3i + i2

= 9 + 6i + (-1)

= 8 + 6i

26. (-4 - 5i)(2 + 10i)

= -8 - 40i - 10i - 50 i2

= -8 - 50i - 50(-1)

= 42 - 50i

27. - i9

=- ( i2)4 i

= - i

28. 2 i15

= 2 (i5)3

= 2(i)3

= 2(-i)

= -2i

29. i30

= (i6)5

= (-1)5

= -1

30.5 - 4i_____

i· -i__

-i

= -5i + 4 i2________

- i2

= -5i + 4(-1)

___________-(-1)

= -5i - 4_______1

= -4 - 5i

31.11 - 5i______2 - 4i

· 2 + 4i_____2 + 4i

= 22 + 44i - 10i - 20 i2

__________________ 4 + 8i - 8i - 16 i

2

= 22 + 34i - 20(-1)

________________ 4 - 16(-1)

= 42 + 34i_______20

= 21___10

+ 17___10

i

32.8 + 2i_____5 + i

· 5 - i____5 - i

= 40 - 8i + 10i - 2 i2

________________ 25 - 5i + 5i - i

2

= 40 + 2i - 2(-1)

______________

25 - (-1)

= 42 + 2i______

26

= 21___13

+ 1___13

i

33.17___4i

· 4 - i____4 - i

= 68 - 17i

______________ 16 - 4i + 4i - i

2

= 68 - 17i_________16 - (-1)

= 68 - 17i_______17

= 4 - i

34.45 - 3i______7 - 8i

· 7 + 8i_____7 + 8i

= 315 + 360i - 21i - 24 i2

____________________ 49 + 56i - 56i - 64 i

2

= 315 + 339i - 24(-1)

__________________ 49 - 64(-1)

= 339 + 339i_________113

= 3 + 3i

35.-3 - 12i________

6i· -6i___

-6i

= 18i + 72 i2_________

-36i2

= 18i + 72(-1)

___________-36(-1)

= 18i - 72_______36

= -2 + 1__2

i


36-39.

40. 2 + 3i

√ �� (2)2 + (3 )

2

√ �� 4 + 9

√ �� 13

41. -1818

42. 4__5

i √

�� ( 4__

5)2

4__5

43. 6 - 8i

√ �� (6)2 + (-8)

2

√ �� 36 + 64

10

44. -0.5i

√ �� (-0.5)2

0.5

45. 10 - 4i

√ �� (10)2 + (-4)

2

√ �� 100 + 16

2 √ �� 29

46. (8 - 9i) - (-2 - i)

(8 + 2) + (i - 9i)

10 - 8i

47. 4i - (11 - 3i)

4i - 11 + 3i

-11 + 7i

48. (4 - 2i) + (-9 - 5i)

(4 - 9) + (-2i - 5i)

-5 - 7i

49. (13 + 6i) + (15 + 35i)

(13 + 15) + (6i + 35i)

28 + 41i

50. (3 - i) - (-3 + i)

(3 + 3) + (-i - i)

6 - 2i

51. -16 + (12 + 9i)

-16 + 12 + 9i

-4 + 9i


52. 4 - 2i 53. 4 + 6i

54. 1 - 6i 55. -12i(-1 + 4i)

12i - 48 i2

-48(-1) + 12i

48 + 12i

56. (3 - 5i)(2 + 9i)

6 - 10i + 27i - 45 i2

6 + 45 + 17i

51 + 17i

57. (7 + 2i)(7 - 2i)

49 - 14i + 14i - 4 i2

49 - 4(-1)

53

58. (5 + 6i)2

25 + 30i + 30i + 36 i2

25 + 60i + 36(-1)

-11 + 60i

59. (7 - 5i)(-3 + 9i)

-21 + 63i + 15i - 45 i2

-21 -45(-1) + 78i

24 + 78i

60. -4(8 + 12i)

-32 - 48i

61. i27

(i9)3

(i)3

-i

62. -i11

-(-i)

i

63. 5 i10

5(-1)

-5

64.2 - 3i_____

i· -i__

-i

-2i + 3 i2________

-i2

-2i + 3(-1)

___________-(-1)

-2i - 3_______1

-3 - 2i

65.5 - 2i_____3 + i

· 3 - i____3 - i

15 - 5i - 6i + 2 i

2

_______________ 9 - 3i + 3i - i

2

15 - 11i + 2(-1)

______________ 9 - (-1)

13 - 11i_______

10

13___10

- 11___10

i

66.3_______

-1 - 5i· -1 + 5i_______

-1 + 5i

-3 + 15i

_______________ 1 - 5i + 5i - 25 i

2

- 3 + 15i__________

1 - 25(-1)

-3 + 15i________26

- 3___26

+ 15___26

i

67.19 + 9i______5 + i

· 5 - i____5 - i

95 - 19i + 45i - 9 i

2

_________________ 25 - 5i + 5i - i

2

95 + 26i - 9(-1)

_______________ 25 - (-1)

104 + 26i________

264 + i

68.8 + 4i_____7 + i

· 7 - i____7 - i

56 - 8i + 28i - 4 i

2

________________ 49 - 7i + 7i - i

2

56 + 20i - 4(-1)

_______________ 49 - (-1)

60 + 20i_______

50

6__5

+ 2__5

i

69.6 + 3i_____2 - 2i

· 2 + 2i_____2 + 2i

12 + 12i + 6i + 6 i

2

________________ 4 + 4i - 4i - 4 i

2

12 + 18i + 6(-1)

_______________ 4 - 4(-1)

6 + 18i______

8

3__4

+ 9__4

i

70. -3 + 3i

71. 3i 72. 2 + i

73. -2 - i 74. 3- 2i

75. 3 - i

√ �� 32 + (-1)

2

√ �� 9 + 1

√ �� 10

76. 7i

√ � 72

7

77. -2 - 6i

√ �� (-2)2 + (-6)

2

√ �� 40

2 √ �� 10

78. -1 - 8i

√ �� (-1)2 + (-8)

2

√ �� 65

79. 00

80. 5 + 4i

√ �� (5)2 + (4 )

2

√ �� 41

81. 3__2

- 1__2

i √

�� (

3__2

)2 + (- 1__

2)2

√ ��

10___4

1__2

√ �� 10

82. 5 - i √ � 3 √ �� (5)

2 + (- √ � 3)

2

√ �� 28

2 √ � 7

83. 2 √ � 2 - i √ � 3 √ �� (2 √ � 2)

2 + (- √ � 3)

2

√ �� 8 + 3

√ � 11


84a. Z2 = (0.5 + 0.6i)2 + 0.25

= (0.5 + 0.6i)(0.5 + 0.6i) + 0.25

= 0.25 + 0.3i + 0.3i + 0.36 i2 + 0.25

= 0.5 + 0.6i + 0.36(-1)

= 0.14 + 0.6i

b. Z3 = (Z2)2 + 0.25

= (0.14 + 0.6i)2 + 0.25

= 0.0196 + 0.168i + 0.36 i2 + 0.25

= 0.2696 + 0.168i + 0.36(-1)

= -0.0904 + 0.168i

c. Z4 = (Z3)2 + 0.25

= (-0.0904 + 0.168i)2 + 0.25

= 0.00817216 - 0.0303744i + 0.028224 i2 +

0.25

= 0.25817216 - 0.0303744i + 0.028224(-1)

= 0.22994816 - 0.0303744i

85. (3.5 + 5.2i) + (6 - 2.3i)

(3.5 + 6) + (5.2i - 2.3i)

9.5 + 2.9i

86. 6i - (4 + 5i)

(6i - 5i) - 4

i - 4

87. (-2.3 + i) - (7.4 - 0.3i)

(-2.3 - 7.4) + (i + 0.3i)

-9.7 + 1.3i

88. (-8 - 11i) + (-1 + i)

(-8 - 1) + (-11i + i)

-9 - 10i

89. i(4 + i)

4i + i2

4i + (-1)

-1 + 4i

90. (6 - 5i)2

36 - 30i - 30i + 25 i2

36 - 60i + 25(-1)

11 - 60i

91. (-2 - 3i)2

4 + 6i + 6i + 9 i2

4 + 12i + 9(-1)

-5 + 12i

92. (5 + 7i)(5 - 7i)

25 - 35i + 35i - 49 i2

25 - 49(-1)

74

93. (2 - i)(2 + i)(2 - i)

(4 - i2) (2 - i)

(4 - (-1)) (2 - i)

5(2 - i)

10 - 5i

94. 3 - i11

3 - (-i)

3 + i

95. i52

- i48

(i4)13

- (i6)8

(1)13

- (-1)8

1 - 1

0

96. i35

- i24

+ i18

(i5)7 - (i4)

6 + (i2)

9

(i)7 - (1 )

6 + (-1)

9

-i - 1 + (-1)

-2 - i

97.12 + i_____

i· -i__

-i

-12i - i2________

- i2

-12i - (-1)

___________-(-1)

-12i + 1________

11 - 12i

98.18 - 3i______

i· -i__

-i

-18i + i

2________- i2

3(-1) - 18i

__________-(-1)

-3 - 18i________1

-3 - 18i

99.4 + 2i_____6 + i

· 6 - i____6 -i

24 - 4i + 12i - 2 i

2

________________ 36 - i

2

24 + 8i - 2(-1)

______________ 36 - (-1)

26 + 8i______

37

26___37

+ 8___37

i

100.1 + i_______

-2 + 4i· -2 - 4i_______

-2 - 4i

-2 - 4i - 2i - 4 i2

_______________ 4 - 16 i

2

-2 - 4(-1) - 6i ______________

4 - 16(-1)2 - 6i_____

201___10

- 3___10

i

101.4_____

2 - 3i·

2 + 3i_____2 + 3i

8 + 12i______4 - 9 i

2

8 + 12i_________4 - 9(-1)

8 + 12i______13

8___13

+ 12___13

i

102.6______

√ � 2 - i·

√ � 2 + i______ √ � 2 + i

6 √ � 2 + 6i________2 - i

2

6 √ � 2 + 6i________2 - (-1)

6 √ � 2 + 6i________3

2 √ � 2 + 2i

103. Zeq = Z1 Z2_______

Z1 + Z2

= (3 + 2i)(1 - 2i)

_______________ (3 + 2i) + (1 - 2i)

= 3 - 6i + 2i - 4 i

2

_______________ (3 + 1) + (2i - 2i)

= 3 - 4i - 4(-1)

_____________4

= 7 - 4i_____4

= 7__4

- i

104. Zeq = Z1 Z2_______

Z1 + Z2

= (2 + 2i)(4 - i)

______________ (2 + 2i) + (4 - i)

= 8 - 2i + 8i - 2 i

2

______________ (2 + 4) + (2i - i)

= 8 + 6i - 2(-1)

_____________6 + i

= 10 + 6i______

6 + i

= 10 + 6i

______________ 6 + i · 6 - i/6 -i

= 60 - 10i + 36i - 6 i

2

_________________ 36 - i

2

= 60 + 26i - 6(-1)

_______________ 36 - (-1)

= 66 + 26i_______

37

= 66___37

+ 26___37

i 105. Always true.

106. Never true; the difference between any complex

number a + bi and its complex conjugate is an

imaginary number: (a + bi) - (a - bi) = 2bi.

107. Always true.

108. Sometimes true; possible answer: true when

b = -2 and d = 4 because (bi)(di) = (-2i)(4i) = 8;

false when b = 2 and d = 4 because

(bi)(di) = (2i)(4i) = -8.


109. A is incorrect. The product (2 + i)(2 + i) is equal

to 4 + 4i + 4 i2 , not 4 + i

2 . Students may also note

that multiplying the numerator and denominator

by 2 + i will not rationalize the denominator of the

original expression.

110. Possible answer: The values are equal because

a complex number and its complex conjugate are

the same distance from the origin in the complex

number plane. The graph shows that the number

3 + 4i and its complex conjugate 3 - 4i are the

same distance from the origin. Therefore, their

absolute values are equal.

111. Possible answer: The general form for the product

of a complex number and its conjugate is

(a + bi)(a - bi) = a2 + b

2 . This equation is nearly

identical to the equation for a difference of squares

except that the factors on the left side of the

equation are nonreal complex numbers and the

operation on the right side is addition rather than

subtraction.

112a. Possible answer: 4 and 4i; 2 + 5i and 2 - i; 5 + 2i

and -1 + 2i.

b. Possible answer:

c. Each sum can be represented by a parallelogram

with one pair of opposite vertices at 0 + 0i and

4 + 4i in the complex plane.

TEST PREP

113. D 114. F

115. C 116. G


117a. i1 = i, i

0 = 1, i

-1 = -i, i

-2 = -1, i

-3 = i, i

-4 = 1,

i-5

= -i

b.The resulting pattern is very similar to the pattern

for positive power of i. The only possible values of

negative powers of i are 1, -1, i and -i.

c. i -12

(i-4)3

(1)3

1

i-37

i -36

· i-1

(i-4)9 · i

-1

(1)9 · (-i)

-i

i -90

(i-9)10

(i)10

-1

118. (a + bi)(c + di)

ac + adi + bci + bd i 2

ac + bd(-1) + (adi + bci)

(ac - bd) + (ad + bc)i

119.a + bi_____c + di

· c - di_____c - di

ac - bd i 2

_________________ c

2 - cdi + cdi - d

2 i2

ac - bd(-1) ___________

c2 - d

2 (-1)

ac + bd_______c

2 + d

2+

(bc - ad)________c

2 + d

2 i

SPIRAL REVIEW

120. y ≈ 1.16x + 3.88

121. 3 x2 - 6x = 0

3x(x - 2) =0

x = 0 or x - 2 = 0

x = 0 or x = 2

Therefore 0 ≤ x ≤ 2.

122. x2 - 4x - 11 = 10

x2 - 4x - 21 = 0

(x - 7)(x + 3) = 0

x - 7 = 0 or x + 3 = 0

x = 7 or x = -3

Therefore, x < -3 or x > 7.

123. 2 x2 + 7x - 21 = -6

2 x2 + 7x - 15 = 0

(x + 5)(2x - 3) = 0

x + 5 = 0 or 2x - 3 = 0

x = -5 or x = 3__2

Therefore, -5 ≤ x ≤ 3__2

.


124. 3 - x2 = 7 - 5x

x2 - 5x + 4 = 0

(x - 1)(x - 4) = 0x - 1 = 0 or x - 4 = 0x = 1 or x = 4

Therefore, x < 1 or x > 4.

125. Yes, the function has constant second difference for equally spaced x-valued.

126. No; the function does not have constant second differences for equally spaced x-values.

READY TO GO ON? PAGE 391

1. 2.

3. x < -1 or x > 5 4. -3 ≤ x ≤ 2

5. x2 + 4x - 7 = 5

x2 + 4x - 12 = 0

(x - 2)(x + 6) = 0x - 2 = 0 or x + 6 = 0 x = 2 or x = -6Therefore x ≤ -6 or x

≥ 2.

6. x2 - 8x = 0

x(x - 8) = 0x = 0 or x - 8 = 0x = 0 or x = 8Therefore 0 < x < 8.

7. -1000r2 + 6400r - 4400 ≥ 5000

-1000r2 + 6400r - 4400 = 5000

-1000r2 + 6400r - 9400 = 0

-200(5r2 - 32 + 47) =0

r = -b ± √ �� b

2 - 4ac

______________2a

= -(-32) ± √ �� (-32)

2 - 4(5)(47)

_________________________ 2(5)

= 32 ± √ �� 84_________

10

= 32 ± 2 √ �� 21

__________10

= 16 ± √ �� 21_________

5r ≈ 2.29 or r ≈ 4.11The rental price is between $2.29 and $4.11.

8. Quadratic function; the second differences of the function values are constant for equally spaced x-values.

9. Not a quadratic function; the second differences of the function values are not constant for equally spaced x-values.


(0, 4) 4 = a(0)2 + b(0) + c

c = 4 �

4a + 2b + c = 0 �

9a + 3b + c = 1 �

(2, 0) 0 = a(2)2 + b(2) + c

(3, 1) 1 = a(3)2 + b(3) + c

Substitute c = 4 from equation 1 into both equation 2 and equation 3.� 4a + 2b + 4 = 0 4a + 2b = -4 2a + b = -2 �

� 9a + 3b + 4 = 1 9a + 3b = -3 3a + b = -1 �

Solve equation 4 and equation 5 for a and b using elimination.� -(2a + b) = -(-2) → -2a - b = 2� 3a + b = 4 → 3a + b = -1 a = 1Substitute 1 for a into equation 4 to find b.� 2a + b = -2 → 2(1) + b = -2

b = -4

f(x) = a x 2 + bx + c = x

2 - 4x + 4


(1, 3) 3 = a(1)2 + b(1) + c

a + b + c = 34a + 2b + c = 516a + 4b + c = 1

(2, 5) 5 = a(2)2 + b(2) + c

(4, 3) 3 = a(4)2 + b(4) + c

→

1 1 1

�

a

�

=

3

�

→

a

�

=

-1

�

4 2 1 b 5 b 5

16 4 1 c 3 c -1

f(x) = a x 2 + bx + c = -1x

2 + 5x - 1

= - x2 + 5x - 1

12. y ≈ -127.5x2 + 961.5x + 5474.5

13. y ≈ -127.5(6.5)2 + 961.5(6.5) + 5474.5 ≈ 6337

The maximum load allowed is about 6337 lb.

14. y ≈ -127.5(8)2 + 961.5(8) + 5474.5 ≈ 5007

The maximum load allowed is about 5007 lb.

15. -6i

√ �� 02 + (-6)

2

√ �� 366

16. 3 + 4i

√ �� (3)2 + (4 )

2

√ �� 255

17. 2 - i

√ �� (2)2 + (-1)

2

√ � 5

18. (3 - 5i) - (6 - i)(3 - 6) + (i - 5i)-3 - 4i

19. (-6 + 4i) + (7 - 2i)(-6 + 7) + (4i - 2i)1 + 2i

20. 3i(4 + i)

12i + 3 i2

3(-1) + 12i

-3 + 12i

21. (3 + i)(5 - i)15 - 3i + 5i - i

2

15 + 2i - (-1)16 + 2i

22. (1 - 4i)(1 + 4i)

12 - (4i)

2

1 - 16 i2

1 - 16(-1)17

23. 3 i15

3 (i5)3

3(i)3

3(-i)-3i


24. 2 - 7i_____-i

· i_i

2i - 7 i2_______

- i2

2i - 7(-1)_________

-(-1)

7 + 2i_____

17 + 2i

25.3 - i_____

4 - 2i·

4 + 2i_____4 + 2i

12 + 6i - 4i - 2 i

2

_______________ 16 - 4 i

2

12 + 2i - 2(-1)

______________ 16 - 4(-1)

14 + 2i______

20

7___10

+ 1___10

i

STUDY GUIDE: REVIEW, PAGES 392-395

1. imaginary number; complex number

2. zero of a function 3. vertex of a parabola

4. discriminant 5. minimum value

LESSON 5-1

6. 7.

8. g is f vertically stretched by a factor of 4 and

translated 2 units right.

9. g is f reflected across the x-axis, vertically stretched

by a factor of 2, and translated 1 unit left.

10. g is f vertically compressed by a factor of 1__3

and

translated 3 units down.

11. g is f reflected across the x-axis and translated 2

units left and 6 units up.

12. Possible answer: g(x) = - x2 - 3

13. Possible answer: g(x) = 2(x - 4 )2

14. Possibe answer: g(x) = 1__4

(x + 1 )2

LESSON 5-2

15a. upward

b. x = -b___2a

= -(-4)______2(1)

= 2

c. f(2) = (2 )2 - 4(2) + 3

= 4 - 8 + 3

= -1


d. The y-intercept is 3

e.

16a. upward

b. x = -b___2a

= -2____2(1)

= -1

c. g(-1) = (-1)2 + 2(-1) + 3

= 1 - 2 + 3

= 2



e.

17a. upward

b. x = -b___2a

= -3____2(1)

= 1.5

c. h(1.5) = (1.5 )2 - 3(1.5)

= -2.25

The vertex is (1.5, -2.25).


e.


18a. upward

b. x = -b___2a

= -(-2)______

2( 1__2)

= 2

c. j(2) = 1__2

(2)2 - 2(2) + 4

= 2 - 4 + 4

= 2



e.

19. x = -b___2a

= -2____2(1)

= -1

f(-1) = (-1)2 + 2(-1) + 6

= 1 - 2 + 6

= 5

The minimum value is 5.

20. x = -b___2a

= -6_____

2(-2)= 1.5

g(1.5) = 6 (1.5) - 2 (1.5)2

= 9 - 4.5

= 4.5

The maximum value is 4.5.

21. x = -b___2a

= -(-5)______2(1)

= 2.5

f(2.5) = (2.5 )2 - 5(2.5) + 1

= 6.25 - 12.5 + 1

= -5.25

The minimum value is -5.25.

22. x = -b___2a

= -(-8)______2(-2)

= -2

f(-2) = -2(-2)2 - 8(-2) + 10

= -8 + 16 + 10

= 18


23. x = -b___2a

= -(-4)______2(-1)

= -2

f(-2) = -(-2)2 - 4(-2) + 8

= -4 + 8 + 8

= 12


24. x = -b___2a

= 0____

2(3)= 0

g(0) = 3(0 )2 + 7 = 7

The minimum value is 7.

LESSON 5-3

25. x2 - 7x - 8 = 0

(x - 8)(x + 1) = 0

x - 8 = 0 or x + 1 = 0

x = 8 or x = -1

26. x2 - 5x + 6 = 0

(x - 2)(x - 3) = 0

x - 2 = 0 or x - 3 = 0

x = 2 or x = 3

27. x2 = 144

x = ± √ �� 144

x = ±12

28. x2 - 21x = 0

x(x - 21) = 0

x = 0 or x - 21 = 0

x = 0 or x = 21

29. 4 x2 - 16x + 16x = 0

4 (x2 - 4x + 4) = 0

4(x - 2 )2 = 0

x - 2 = 0

x = 2

30. 2 x2 + 8x + 6 = 0

2 (x2 + 4x + 3) = 0

2(x + 1)(x + 3) = 0

x + 1 = 0 or x + 3 = 0

x = -1 or x = -3

31. x2 + 14x = 32

x2 + 14x - 32 = 0

(x - 2)(x + 16) = 0

x - 2 = 0 or x + 16 = 0

x = 2 or x = -16

32. 9 x2 + 6x + 1 = 0

(3x)2 + 2(3x)(1) + 1

2 = 0

(3x + 1 )2 = 0

3x + 1 = 0

x = - 1__3


f(x) = (x - 2)(x + 3)

= x2 + 3x - 2x - 6

= x2 + x - 6


f(x) = (x - 1)(x + 1)

= x2 + x - x - 1

= x2 - 1


f(x) = (x - 4)(x - 5)

= x2 - 5x - 4x + 20

= x2 - 9x + 20


f(x) = (x + 2)(x + 3)

= x2 + 3x + 2x + 6

= x2 + 5x + 6


f(x) = (x + 5)(x + 5)

= x2 + 5x + 5x + 25

= x2 + 10x + 25


f(x) = (x - 9)(x + 0)

= x(x - 9)

= x2 - 9x

LESSON 5-4

39. x2 - 16x + 48 = 0

x2 - 16x = -48

x 2 - 16x + 64 = -48 + 64

(x - 8 )2 = 16

x - 8 = ± √ �� 16

x = 12 or 4

40. x2 + 20x + 84 = 0

x 2 + 20x = -84

x 2 + 20x + 100 = -84 + 100

(x + 10 )2 = 16

x + 10 = ± √ �� 16

x = -6 or -14


41. x2 - 6x = 16

x 2 - 6x + 9 = 16 + 9

(x - 3 )2 = 25

x - 3 = ± √ �� 25

x = 8 or -2

42. x2 - 14x = 13

x 2 - 14x + 49 = 13 + 49

(x - 7 )2 = 62

x - 7 = ± √ �� 62

x = 7 ± √ �� 62

43. f(x) = x2 - 4x + 9

= (x2 - 4x) + 9

= (x2 - 4x + 4) + 9 - 4

= (x - 2 )2 + 5


44. g(x) = x2 + 2x - 7

= (x2 + 2x) - 7

= (x2 + 2x + 1) - 7 - 1

= (x + 1 )2 - 8


LESSON 5-5

45. x2 = -81

x = ± √ �� -81

x = ±9i

46. 6 x2 + 150 = 0

6 x2 = -150

x2 = -25

x = ± √ �� -25

x = ±5i

47. x2 + 6x + 10 = 0

x 2 + 6x = -10

x 2 + 6x + 9 = -10 + 9

(x + 3 )2 = -1

x + 3 = ± √ �� -1

x = -3 ± i

48. x2 + 12x + 45 = 0

x2 + 12x = -45

x 2 + 12x + 36 = -45 + 36

(x + 6 )2 = -9

x + 6 = ± √ �� -9

x = -6 ± 3i

49. x2 - 14x + 75 = 0

x2 - 14x = -75

x 2 - 14x + 49 = -75 + 49

(x - 7 )2 = -26

x - 7 = ± √ �� -26

x = 7 ± i√ �� 26

50. x2 - 22x + 133 = 0

x2 - 22x = -133

x 2 - 22x + 121 = -133 + 121

(x - 11 )2 = -12

x - 11 = ± √ �� -12

x = 11 ± 2i√ � 3

51. -5i - 4 52. 3 - i √ � 5

LESSON 5-6

53. x = -b ± √ �� b

2 - 4ac

______________2a

= -(-3) ± √ �� (-3)

2 - 4(1)(-8)

________________________ 2(1)

= 3 ± √ �� 9 + 32

___________2

= 3 ± √ �� 41________

2

54. (x - 5 )2 + 12 = 0

x 2 - 10x + 25 + 12 = 0

x 2 - 10x + 37 = 0

x = -b ± √ �� b

2 - 4ac

______________2a

= -(-10) ± √ �� (-10)

2 - 4(1)(37)

_________________________ 2(1)

= 10 ± √ �� -48

__________2

= 10 ± 4i √ � 3_________

2

= 5 ± 2i √ � 3

55. x = -b ± √ �� b

2 - 4ac

______________2a

= -(-10) ± √ �� (-10)

2 - 4(2)(18)

_________________________ 2(2)

= 10 ± √ �� -44

__________4

= 10 ± 2i √ � 11

__________4

= 5__2

± i √ � 11_____

2

56. x = -b ± √ �� b

2 - 4ac

______________2a

= -3 ± √ �� 3

2 - 4(1)(3)

_________________ 2(1)

= -3 ± √ �� -3__________

2

= -3___2

± i √ � 3____

2

57.x = -b ± √ �� b

2 - 4ac

______________

2a

= -(-5) ± √ �� (-5)

2 - 4(1)(10)

_______________________ 2(1)

= 5 ± √ �� -15_________

2

= 5__2

± i √ �� 15_____

2


58. b2 - 4ac

(-16)2 - 4(2)(32)

256 - 256 = 0

b 2 - 4ac = 0;

the equation has


59. b2 - 4ac

(-6)2 - 4(1)(5)

36 - 20 = 16

b 2 - 4ac > 0;

the equation has


60. b2 - 4ac

32 - 4(1)(8)

9 - 32 = -23

b 2 - 4ac < 0;

the equation has

2 nonreal complex

solutions.

61. b2 - 4ac

(-246)2 - 4(1)(144)

60526 - 576 = 59940

b 2 - 4ac > 0;

the equation has


62. b2 - 4ac

52 - 4(1)(12)

25 - 48 = -23

b 2 - 4ac < 0;

the equation has

2 nonreal complex

solutions.

63. b2 - 4ac

(-5)2 - 4(3)(3)

25 - 36 = -11

b 2 - 4ac < 0;

the equation has

2 nonreal complex

solutions.

LESSON 5-7

64. 65.

66. x ≤ -3 or x ≥ 1 67. -4 < x < -1

68. - x2 + 6x - 5 = 0

- (x2 - 6x + 5) = 0

-(x - 1)(x - 5) = 0

x - 1 = 0 or x - 5 = 0

x = 1 or x = 5

Therefore x < 1 or x > 5.

69. 3 x2 - 25 = 2

3 x2 = 27

x 2 = 9

x = ± √ � 9

x = ±3


70. x2 - 3 = 0

x2 = 3

x = ± √ � 3

Therefore - √ � 3 < x < √ � 3.

71. 3 x2 + 4x - 3 = 1

3 x2 + 4x - 4 = 0

(x + 2)(3x - 2) = 0

x + 2 = 0 or 3x - 2 = 0

x = -2 or x = 2__3

Therefore -2 ≤ x ≤ 2__3

.

LESSON 5-8


(0, 6) 6 = a(0)2 + b(0) + c

c = 6 �

a - b + c = 8 �

a + b + c = 2 �

(-1, 8) 8 = a(-1)2 + b(-1) + c

(1, 2) 2 = a(1)2 + b(1) + c

Substitute c = 6 from equation 1 into both equation

2 and equation 3.

� a - b + 6 = 8

a - b = 2 �

3 a + b + 6 = 2

a + b = -4 �


elimination.

� a - b = 2 → a - b = 2

� a + b = -4 → a + b = -4

2a = -2

a = -1

Substitute -1 for a into equation 4 to find b.

� a - b = 2 → (-1) - b = 2

b = -3

f(x) = a x 2 + bx + c = -1x

2 + (-3)x + 6

= -x2 - 3x + 6


(0, 0) 0 = a(0)2 + b(0) + c

c = 0 �

a + b + c = -1 �

4a + 2b + c =-6�

(1, -1) -1 = a(1)2 + b(1) + c

(2, -6) -6 = a(2)2 + b(2) + c

Substitute c = 0 from equation 1 into both equation

2 and equation 3.

� a + b + 0 = -1

a + b = -1 �

3 4a + 2b + 0 = -6

4a + 2b = -6

2a + b = -3 �


elimination.

� a + b = -1 → a + b = -1

� -(2a + b) = -(-3) → -2a - b = 3

-a = 2

a = -2

Substitute -2 for a into equation 4 to find b.

� a + b = -1 → (-2) + b = -1

b = 1

f(x) = a x 2 + bx + c = -2x

2 + 1x + 0

= -2x2 + x

74. y ≈ 0.000188 x2 - 0.0112x + 0.182

75. y ≈ 0.000188(12 )2 - 0.0112(12) + 0.182 ≈ 0.0747

The diameter is about 0.0747 in.

76. y ≈ 0.360 x2 - 11.9x + 105

77. y ≈ 0.360(26 )2 - 11.9(26) + 105 ≈ 37.8

The resistance is about 37.8 ohms.


LESSON 5-9

78. -3i

√ �� 02 + (-3)

2

√ � 9

3

79. 4 - 2i

√ �� (4)2 + (-2)

2

√ �� 20

2 √ � 5

80. 12 - 16i

√ �� (12)2 + (-16)

2

√ �� 400

20

81. 7i

√ �� 02 + (7 )

2

√ �� 49

7

82. (1 + 5i) + (6 - i)

(1 + 6) + (5i - i)

7 + 4i

83. (9 + 4i) - (3 + 2i)

(9 - 3) + (4i - 2i)

6 + 2i

84. (5 - i) - (11 - i)

(5 - 11) - (i - i)

-6

85. -5i(3 - 4i)

-15i + 20 i2

20(-1) - 15i

-20 - 15i

86. (5 - 2i)(6 + 8i)

30 + 40i - 12i - 16 i2

30 + 28i - 16(-1)

46 + 28i

87. (3 + 2i)(3 - 2i)

32 - (2i)

2

9 - 4 i2

9 - 4(-1)

13

88. (4 + i)(1 - 5i)

4 - 20i + i - 5 i2

4 - 19i - 5(-1)

9 - 19i

89. (-7 + 4i)(3 + 9i)

-21 - 63i + 12i + 36 i2

-21 - 51i + 36(-1)

-57 - 51i

90. i32

( i 2)16

(-1)16

1

91. -5i21

-5 (i3)7

-5(-i)7

-5(i)

-5i

92.2 + 9i_____

-2i· 2i__

2i

4i + 18 i

2________-4i

2

4i + 18(-1)

__________-4(-1)

-18 + 4i________4

- 9__2

+ i

93.5 + 2i_____3 - 4i

· 3 + 4i_____3 + 4i

15 + 20i + 6i + 8 i

2

________________ 9 - 19 i

2

15 + 26i + 8(-1)

_______________ 9 - 16(-1)

7 + 26i______

25

7___25

+ 26___25

i

94.8 - 4i_____1 + i

· 1 - i____1 - i

8 - 8i - 4i + 4 i

2

______________1 - i

2

8 - 12i + 4(-1)

______________ 1 - (-1)

4 - 12i______2

2 - 6i

95.-12 + 26i_________

2 + 4i· 2 - 4i_____

2 - 4i

-24 + 48i + 52i - 104 i2

____________________ 4 - 16 i

2

-24 + 100i - 104(-1)

___________________ 4 - 16(-1)

80 + 100i________

204 + 5i

CHAPTER TEST, PAGE 396

1. g is f translated 1 unit left and 2 units down.

2. h is f reflected across the x-axis, vertically


, and translated 2 units

up.

3. g(x) = 1__2

(x - 6 )2

4a. downward

b. x = -b___2a

= -4_____2(-1)

= 2

c. f(2) = -(2)2 + 4(2) + 1

= -4 + 8 + 1

= 5



5a. upward

b. x = -b___2a

= -(-2)______2(1)

= 1

c. g(1) = (1 )2 - 2(1) + 3

= 1 - 2 + 3

= 2



e. e.

6. x = -b___2a

= -16_____

2(-1)= 8

A(8) = -(8)2 + 16(8)

= -64 + 128 = 64

The maximum area is 64 c m2 .

7. x2 - 2x + 1 = 0

(x - 1 )2 = 0

x - 1 = 0

x = 1

8. x2 + 10x = -21

x2 + 10x + 21 = 0

(x + 3)(x + 7) = 0

x + 3 = 0 or x + 7 = 0

x = -3 or x = -7

9. x2 + 4x = 12

x 2 + 4x + 4 = 12 + 4

(x + 2 )2 = 16

x + 2 = ± √ �� 16

x = 2 or -6

10. x2 - 12x = 25

x 2 - 12x + 36 = 25 + 36

(x - 6 )2 = 61

x - 6 = ± √ �� 61

x = 6 ± √ �� 61


11. x2 + 25 = 0

x 2 = -25

x = ± √ �� -25 x = ±5i

12. x2 + 12x = -40

x 2 + 12x + 36 = -40 + 36

(x + 6 )2 = -4

x + 6 = ± √ �� -4 x = -6 ± 2i

13. f(x) = x2 - 4x + 9

= (x2 - 4x) + 9

= (x2 - 4x + 4) + 9 - 4

= (x - 2 )2 + 5


14. g(x) = x2 - 18x + 92

= (x2 - 18x) + 92

= (x2 - 18x + 81) + 92 - 81

= (x - 9 )2 + 11


15. (x - 1 )2 + 7 = 0

x 2 - 2x + 1 + 7 = 0

x 2 - 2x + 8 = 0

x = -b ± √ �� b

2 - 4ac

______________

= -(-2) ± √ �� (-2)

2 - 4(1)(8)

______________________ 2(1)

= 2 ± √ �� -28_________

2

= 2 ± 2i √ � 7________

2 = 1 ± i √ � 7

16. x = -b ± √ �� b

2 - 4ac

______________

= -(-1) ± √ �� (-1)

2 - 4(2)(5)

______________________ 2(2)

= 1 ± √ �� -39_________

4

= 1__4

± √ �� 39____

4i

17. -0.025t2 - 0.5t + 50 = 0

t = -b ± √ �� b

2 - 4ac

______________

= -(-0.5) ± √ �� (-0.5)

2 - 4(-0.025)(50)

_______________________________ 2(-0.025)

= 0.5 ± √ �� -5.25

____________-0.05

= 0.5 ± 0.5 √ �� 21

____________-0.05

= -10 ± 10 √ �� 21t ≈ 35.8 or t ≈ -55.8The ride will last for about 35.8 s.

18.

19. - x2 + 3x + 5 = 7

- x2 + 3x - 2 = 0

- ( x2 - 3x + 2) = 0

-(x - 1)(x - 2) = 0x - 1 = 0 or x - 2 = 0 x = 1 or x = 2Therefore 1 ≤ x ≤ 2.

20. x2 - 4x + 1 = 1

x 2 - 4x = 0

x(x - 4) = 0x = 0 or x - 4 = 0x = 0 or x = 4Therefore x < 0 or x > 4.

21. y ≈ 1.8 x2 + 52x - 662

22. y ≈ 1.8(42 )2 + 52(42) - 662 ≈ 4697

The cost is about $4697.

23. (12 - i) - (5 + 2i)(12 - 5) - (i + 2i)7 - 3i

24. (6 - 2i)(2 - 2i)12 - 12i - 4i + 4 i

2

12 - 16i + 4(-1)8 - 16i

25. -2i18

-2 (i2)9

-2(-1)9

-2(-1)2

26.1 - 8i_____

4i· -4i___

-4i

-4i + 32 i

2_________-16i

2

-4i + 32(-1)

____________-16(-1)

-32 - 4i________16

-2 - 1__4

i


chapter solutions key 5 quadratic functions

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