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Park Forest Math Team
Meet #5
Algebra
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,
x + (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.
2
2
2
! Check. If x = 5, 2x + 2 = 12 and 2x + 3 = 13. 5 + 12 = 13
2
Calculator Meet
Category 5
Algebra
Meet #5 - March, 2014
50th anniversary edition
1) What are the two values of N that solve
?
2) To fence a rectangular pasture, a farmer uses 110 meters of fencing
to enclose a 750 square meter area. The length of the pasture is
longer than the width. What is the length of the pasture ?
3) Hoses are attached to two outdoor faucets so that a backyard
swimming pool can be filled. If both faucets are used, it takes two
hours to fill the pool. If either faucet were used alone, then one
faucet 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 one
faucet is used?
ANSWERS
1) ____ and ____
2) _____ meters
3) _____ hours
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Solutions to Category 5
Algebra
Meet #5 - March, 2014
Answers
1) - 4 and 8
1)
2)
30
3)
3
1)
N 2 − 4 N = 32
N 2 − 4 N − 32 = 0
( N + 4)( N − 8) = 0
Either ( N + 4) = 0 or ( N − 8) = 0
So, N = - 4 or N = 8
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Meet #5
March 2012
Calculators allowed
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. _______________
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?
.
Meet #5
March 2012
Calculators allowed
Solutions to Category 5 - Algebra
1. If we call the width , then we know that
(
)
Answers
or
which we can solve:
√
1.
2.
and the positive solution is
Therefore the perimeter is
(
3.
)
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
√
their difference is
that √
√
. So we’re told that –
and we conclude that
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and
and also
or
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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].
𝑥 2 − 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. _______________
Solutions to Category 5 - Algebra
Meet #5, March 2010
1.
2.
3.
You may use a calculator today!
Answers
10
144
20
1. If we rearrange the equation 𝑥 2 − 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 𝑥 2 − 4 ∙ 𝑥 − 21 = 0 and then 𝑥 =
4± 4 2 +84
2
= 2 ± 5 to get the
same values.
The plot to the left shows
the graph of
𝑦 = 𝑥 2 − 4 ∙ 𝑥 − 21.
This shows visually that x
values of -3 and +7 result in
𝑦 = 0 and are therefore the
solutions.
2. If we call the original length d, then the original area is 𝑑 2 . The increased length is
3
3
9
2
2
4
𝑑 ∙ 150% = ∙ 𝑑 and the increased area is ( ∙ 𝑑)2 = ∙ 𝑑 2 so we know that the
5
180∙4
4
5
difference in areas is ∙ 𝑑 2 = 180 and we get 𝑑 2 =
A
B
O
d
= 144 square inches.
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.
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.
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?
3x - 4
2x - 6
2x + 1
Answers
1. _______________
2. _______________
3. _______________
Solutions to Category 5
Algebra
Meet #5, March 2008
Answers
1.
1
2. 81
3. 40
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 xintercepts (otherwise known as roots) is to put the equation into
34
3
3
the form , 0- / 1- / 2 and then the sum is 5
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.
Category 5
Algebra
Meet #5, March 2006
You may use a calculator today.
1. Find the negative value of x that makes the following equation true:
x(x + 1) = 210
9
. If the number is less
10
than one, what is this number? Express your answer as a common fraction in
simplest form.
2. A rational number and its reciprocal have a sum of 2
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.
8x – 7
x
7x + 7
Answers
1. _______________
2. _______________
3. _______________
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Solutions to Category 5
Algebra
Meet #5, March 2006
Answers
1. –15
2.
2
5
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.
3. 840
2. The English translates to the equation x +
1
9
= 2 , or
x
10
1 29
= . Multiplying both sides of the equation by 10x, we
x 10
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
−b ± b 2 − 4ac
formula, x =
, or we can try to factor this
2a
trinomial into the product of two binomials. Factoring gives us
the equation (5x − 2)(2x − 5) = 0 . If 5x − 2 = 0 , then the
2
5
solution is x = . If 2x − 5 = 0 , then the solution is x = . We
5
2
2
want the solution that is less than 1, which is .
5
x+
3. By the Pythagorean Theorem, we write the equation x 2 + (7x + 7) = (8x − 7) .
Expanding on both sides, we get x 2 + 49x 2 + 98x + 49 = 64 x 2 −112x + 49 .
Subtracting 49 from both sides and combining like terms, we get
50x 2 + 98x = 64 x 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 = 14 x 2 − 210x . Factoring out 14x from both terms, we rewrite the equation to
get 0 = 14 x (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 = 12 ×112 ×15 = 56 ×15 = 840 square units.
2
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2
Category 5
Algebra
Meet #5, April 2004
You may use a calculator
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.
y
(0,12)
Answers
1. _______________
2. _______________
3. _______________
x
(-4, 0)
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(-3, 0)
Solutions to Category 5
Algebra
Meet #5, April 2004
Answers
1. 46
2. -17
3. -4
Average team got 17.3 points, or 1.4 questions correct
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 x 2 + 9x square units. Since we know the
area is 112 square units, we can write the quadratic
equation x 2 + 9x = 112 . Setting this equal to zero, we
get x 2 + 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
x 2 + 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
n 2 −1 = 288 , which means that n2 must be 289. Both positive and negative 17
satisfy the equation n 2 = 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 = x 2 + (3 − a )x − 3a . If we
substitute the x- and y-values from the ordered pair (0, 12) into this equation, we
get 12 = 0 2 + (3 − a ) ⋅ 0 − 3a , which simplifies to 12 = −3a . Dividing both sides by
-3, we get a = -4.
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