Park Forest Math Team

Meet #5

AlgebraAlgebra

Self-study Packet

Problem Categories for this Meet: 1. Mystery: Problem solving

2. Geometry: Angle measures in plane figures including supplements and complements

3. Number Theory: Divisibility rules, factors, primes, composites

4. Arithmetic: Order of operations; mean, median, mode; rounding; statistics

5. Algebra: Simplifying and evaluating expressions; solving equations with 1 unknown including identities

Important things you need to know about ALGEBRA: Solving quadratics with rational solutions, including word problems

• If xy = 0, then x = 0 or y = 0. This is called the Zero Product Property • If (x – 3) (x + 2) = 0, then x – 3 = 0 or x + 2 = 0. The solutions to this problem are x =3

and x = -2 • When a graph crosses the x-axis, y = 0.

• To multiply binomials, such as (x – 4) (x + 6), we can use the distributive property. A

mnemonic is FOIL. Foil means multiply the First, Outside, Inside, and Last Terms.

(x – 4) (x + 6) = x2 + 6x – 4x – 24 = x2 + 2x – 24

• You should notice that in the above example, the -4 and 6 add to equal 2 and multiply to equal -24. Use this knowledge to work backward to factor a trinomial.

• Factor x2 – 7x + 12 (Think: What are two numbers that multiply to equal 12 and add to

equal -7? -3 and -4) So, x2 – 7x + 12 = (x – 3) (x – 4) • If x2 – 7x + 12 = 0, then (x – 3) (x – 4) = 0, so x = 3 or x = 4 • The quadratic formula can also be used to solve quadratic equations.

If Ax2 + Bx + C = 0, then

Example: How many units are in the shortest length of the right triangle below?

2x + 3 x

2x + 2 By Pythagorean Theorem, x2 + (2x + 2)2 = (2x + 3)2 So, x2 + (2x + 2)(2x + 2) = (2x + 3)(2x + 3) Using FOIL, x2 + 4x2 + 4x + 4x + 4 = 4x2 + 6x = 6x + 9 Combining like terms, 5x2 + 8x + 4 = 4x2 + 12x + 9 Subtracting the right side to get everything on the left, we get x2 – 4x – 5 = 0 So (x – 5)(x+1) = 0, Therefore, x = 5 or x = -1 A side length cannot be negative, so x must be 5.

! Check. If x = 5, 2x + 2 = 12 and 2x + 3 = 13. 52 + 122 = 132

Category 5 Calculator Meet Algebra

Meet #5 - March, 2014 50th anniversary edition50th anniversary edition50th anniversary edition50th anniversary edition

1) What are the two values of N that solve ?

2) To fence a rectangular pasture, a farmer uses 110 meters of fencingto enclose a 750 square meter area. The length of the pasture islonger than the width. What is the length of the pasture ?

3) Hoses are attached to two outdoor faucets so that a backyardswimming pool can be filled. If both faucets are used, it takes twohours to fill the pool. If either faucet were used alone, then onefaucet would take three hours less than the other to fill the pool.What is the least amount of time required to fill the pool if only onefaucet is used?

ANSWERS

1) ____ and ____

2) _____ meters

3) _____ hours

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1)

Solutions to Category 5 Algebra Meet #5 - March, 2014

Answers 1) 2 4 32N N− =2 4 32 0N N− − =

1) - 4 and 8 ( 4)( 8) 0N N+ − =Either ( 4) 0N + = or ( 8) 0N − =

2) 30 So, N = - 4 or N = 8

3) 3

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Meet #5 March 2012 Calculators allowed

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Category 5 – Algebra

1. A rectangle is inch longer than it is wide. Its area is square inches.

How many inches are there in its perimeter?

2. The price of a square solar panel is calculated in the following way:

cents for each square inch in its area, plus cents for each inch in its

perimeter.

How many inches are there in the side of a solar panel that costs ?

3. There are two solutions to the quadratic equation .

Their sum is and the positive difference between them is .

What is the value of the parameter ?

Answers

1. ___________inches

2. ___________inches

3. _______________

Meet #5 March 2012 Calculators allowed

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Solutions to Category 5 - Algebra

1. If we call the width , then we know that ( ) or

which we can solve:

√

and the positive solution is

Therefore the perimeter is ( )

2. If we call the length of a side , then we know that:

(note we have to translate the dollars to cents).

Solving this we get: √

and the positive solution is inches.

3. Given that the solutions of any quadratic equation are given by

√

( in our case), we get that their sum is and

their difference is √ . So we’re told that – and also

that √ √ and we conclude that or

Answers

1.

2.

3.

You may use a calculator today!

Category 5 - Algebra

Meet #5, March 2010

1. What is the positive difference between the two roots (solutions for x) of the equation

below? [Hint: both are integers].

𝑥2 − 4 ∙ 𝑥 = 21

2. When each side of a square was increased in length by 50%, its area increased by 180

square inches.

How many square inches are in the original square?

3. The diagram below shows a circle inscribed inside a square.

The shaded rectangle measures 4 × 8 inches and touches the circle with one corner.

How many inches are in the radius of the circle?

Answers

1. _______________

2. _______________

3. _______________

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Solutions to Category 5 - Algebra

Meet #5, March 2010

1. If we rearrange the equation 𝑥2 − 4 ∙ 𝑥 = 21 we can write it in the form:

𝑥 + 3 ∙ 𝑥 − 7 = 0 which makes clear that the solutions (roots) are

𝑥 = −3 𝑎𝑛𝑑 𝑥 = 7 and the difference is 10. If you could not factor this way, you

should have written 𝑥2 − 4 ∙ 𝑥 − 21 = 0 and then 𝑥 =4± 42+84

2= 2 ± 5 to get the

same values.

2. If we call the original length d, then the original area is 𝑑2. The increased length is

𝑑 ∙ 150% =3

2∙ 𝑑 and the increased area is (

3

2∙ 𝑑)2 =

9

4∙ 𝑑2 so we know that the

difference in areas is 5

4∙ 𝑑2 = 180 and we get 𝑑2 =

180∙4

5= 144 square inches.

3. In triangle 𝑂𝐴𝐵 we have 𝑂𝐵 = 𝑅 (𝑇ℎ𝑒 𝑐𝑖𝑟𝑐𝑙𝑒′𝑠 𝑟𝑎𝑑𝑖𝑢𝑠), 𝑂𝐴 = 𝑅 − 4, 𝐴𝐵 =

𝑅 − 8 and therefore: (𝑅 − 4)2 + (𝑅 − 8)2 = 𝑅2, the solutions of which are 𝑅 = 4

(invalid in our case) and 𝑅 = 20.

Answers

1. 10

2. 144

3. 20

O

A

B

The plot to the left shows

the graph of

𝑦 = 𝑥2 − 4 ∙ 𝑥 − 21.

This shows visually that x

values of -3 and +7 result in

𝑦 = 0 and are therefore the

solutions.

If the shaded area measures

180 square inches, then

each little square is 36

square inches and so the

original square is 4 times

that area, or 144 square

inches. d

Category 5

Algebra

Meet #5, March 2008

1. What is the sum of the x-coordinates of the x-intercepts of the equation below?

(the x-intercepts are known as the roots and are the values of x when y = 0.)

, � 2-2 .2-. 12

2. A square is changed into a new rectangle by adding 3 cm to the length and

subtracting 5 cm from the width. The area of the new rectangle is 48 cm. How

many square centimeters are in the area of the original square?

3. The triangle below is a right triangle with hypotenuse of length ( 3x – 4 ).

What is the value of the perimeter of this triangle?

Answers

1. _______________

2. _______________

3. _______________

2x - 6

2x + 1

3x - 4

Solutions to Category 5

Algebra

Meet #5, March 2008

1. , � 2-� . 2- . 12

, � 2- / 4�- . 3� 0 � 2- / 4�- . 3� 2- / 4 � 0 or - . 3 � 0

- � .2 or - � 3

The sum is .2 / 3 � �

Another way to find the sum of the x-coordinates of the x-

intercepts (otherwise known as roots) is to put the equation into

the form , � 0-� / 1- / 2 and then the sum is 345 � 33��

� � �

2. If the square’s original side length is s then the new rectangle

has length s +3 and width s – 5 and the area would be :

6 / 3�6 . 5� � 48

6� . 26 . 15 � 48

6� . 26 . 63 � 0 6 . 9�6 / 7� � 0

6 . 9 � 0 or 6 / 7 � 0

6 � 9 or 6 � .7 but since s is a side of a square, it can’t be

negative. So s = 9 and the area of the square was 92 =81

3. Using the Pythagorean Theorem, we know that :

2- . 6�� / 2- / 1�� � 3- . 4��

4-� . 24- / 36 / 4-� / 4- / 1 � 9-� . 24- / 16

8-� . 20- / 37 � 9-� . 24- / 16

0 � -� . 4- . 21

0 � - . 7�- / 3� 0 � - . 7 or 0 � - / 3

- � 7 or - � .3 but if - � .3 then all the sides would be negative which we can’t

have, so - � 7. That makes the three sides of the triangle 8, 15, and 17 and the

perimeter is 40.

Answers

1. 1

2. 81

3. 40

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Category 5 Algebra Meet #5, March 2006 1. Find the negative value of x that makes the following equation true:

x x + 1( )= 210

2. A rational number and its reciprocal have a sum of 29

10. If the number is less

than one, what is this number? Express your answer as a common fraction in simplest form. 3. The triangle below is a right triangle with side lengths given in terms of x. How many square units are in the area of the triangle? Express your answer to the nearest whole number of square units.

Answers 1. _______________ 2. _______________ 3. _______________

x

7x + 7

8x – 7

You may use a calculator today.

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Solutions to Category 5 Algebra Meet #5, March 2006

1. Rather than set the quadratic equation equal to zero, consider what the equation says in its current form. The product of two numbers is equal to 210. Since the two numbers are one apart, the square root of 210 will put us close to the correct values. If x were positive, it would be 14 times 15 equals 210. Since x is negative, it’s –15 times –14. So x = –15 is it.

2. The English translates to the equation x + 1x

= 29

10, or

x + 1x

= 2910

. Multiplying both sides of the equation by 10x, we

get 10x 2 + 10 = 29x . We now subtract 29x from both sides to get 10x 2 − 29x + 10 = 0. Now we can either use the quadratic

formula, x = −b ± b2 − 4ac2a

, or we can try to factor this

trinomial into the product of two binomials. Factoring gives us the equation 5x − 2( ) 2x − 5( )= 0 . If 5x − 2 = 0, then the

solution is x = 25

. If 2x − 5 = 0, then the solution is x = 52

. We

want the solution that is less than 1, which is 25

.

3. By the Pythagorean Theorem, we write the equation x 2 + 7x + 7( )2 = 8x − 7( )2. Expanding on both sides, we get x 2 + 49x 2 + 98x + 49 = 64x 2 −112x + 49. Subtracting 49 from both sides and combining like terms, we get 50x 2 + 98x = 64x 2 −112x. We can now set this equation equal to zero by subtracting 50x2 and 98x from both sides. This gives us the equation 0 = 14x 2 − 210x . Factoring out 14x from both terms, we rewrite the equation to get 0 = 14x x −15( ). The solutions are x = 0, which is not useful to us, or x = 15, which is useful to us. The height of the triangle is 15 units and the base is 7 ×15 + 7 = 105 + 7 = 112 units. The area of the triangle is thus A = 1

2 ×112 ×15 = 56 ×15 = 840 square units.

Answers 1. –15

2. 25

3. 840

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Category 5 Algebra Meet #5, April 2004 1. The length of a rectangle is nine units longer than the width of the rectangle and the area of the rectangle is 112 square units. Find the number of units in the perimeter of the rectangle. 2. The product of three consecutive negative integers is equal to 288 times the middle integer. What is the value of the middle integer? 3. The graph of the equation y = x + 3( ) x − a( ) is shown below. The ordered pairs of the x- and y-intercepts have also been given. Find the value of a.

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Answers 1. _______________ 2. _______________ 3. _______________

(0,12)

(-4, 0) (-3, 0)

x

y

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Solutions to Category 5 Average team got 17.3 points, or 1.4 questions correct Algebra Meet #5, April 2004

1. Let the width of the rectangle be x units. The length is then x + 9 units, and the area would be calculated as x x + 9( ) or x2 + 9x square units. Since we know the area is 112 square units, we can write the quadratic equation x2 + 9x = 112 . Setting this equal to zero, we get x2 + 9x −112 = 0 . Now we would like to factor this trinomial into the product of two binomials in the form x + ( ) x − ( ). To fill these blanks, we must find a pair

of numbers whose product is 112 and whose difference is 9. We could list factors of 112 in pairs as follows: 1 × 112, 2 × 56, 4 × 28, 7 × 16, and 8 × 14. The desired pair is 7 and 16 since 16 – 7 = 9. We now have x2 + 9x −112 = x + 16( ) x − 7( )= 0 . For this equation to equal zero, x has to equal either -16 or 7. The negative value doesn’t make sense for the width of our rectangle, so the width must be 7. This means the length is 7 + 9 = 16, and the perimeter is 2 7 + 16( )= 2 ⋅ 23 = 46 units.

2. If we let n be the middle number, the product of the three consecutive numbers is n −1( )⋅ n ⋅ n + 1( ). This product is said to be equal to 288 times the middle number, so we can write the equation n −1( )⋅ n ⋅ n + 1( )= 288n . Dividing both sides of the equation by n, we avoid a cubic and get the quadratic equation n −1( ) n + 1( )= 288 . Multiplying out the left side of the equation, we get

n2 −1 = 288, which means that n2 must be 289. Both positive and negative 17 satisfy the equation n2 = 289 , but we know that the three consecutive integers are negative. The value of the middle integer is -17. 3. Expanding the right side of the equation, we get y = x2 + 3 − a( )x − 3a . If we substitute the x- and y-values from the ordered pair (0, 12) into this equation, we get 12 = 02 + 3 − a( )⋅ 0 − 3a , which simplifies to 12 = −3a . Dividing both sides by -3, we get a = -4.

Answers 1. 46 2. -17 3. -4