Download section 1.9 solutions

Section 1.9: Applications that involve factoring
1) A number is 20 less than its square. Find all such numbers.
Let x = a number
then x2 = its square
I will replace “a number” with x
“is” with an equal sign
and “20 less than its square” with x2 - 20
Now I can create an equation.
𝑥 = 𝑥 2 − 20
−𝑥
−𝑥
________________
0 = x2 – x – 20
0 = (x + 4)(x – 5)
x+4=0
x–5=0
x = -4
x=5
Solution: The numbers are -4 and 5.
3) The square of a number is 6 more than the number. Find all such numbers.
A number = x
square of a number = x2
“Is” will turn into an equal sign
6 more than a number = x + 6
𝑥2 = 𝑥 + 6
−𝑥 − 6 − 𝑥 − 6
________________
x2 – x – 6 = 0
(x+2)(x – 3) = 0
x+2=0
x–3=0
x=-2
x=3
Solution: The numbers are -2, 3
5) The product of two consecutive numbers is 72. Find all such numbers.
I will call the first of the two numbers x,
First number x
Since they are consecutive numbers the second number will be called x + 1
Second number x + 1
I will replace the “product of the two numbers” with x(x+1)
I will replace the “is” with an equal sign
This is the equation I need to solve:
x(x+1) = 72 ( I will clear the parenthesis, set equal to 0, then solve by factoring)
𝑥 2 + 𝑥 = 72
−72 − 72
__________________
x2 + x – 72 = 0
(x + 9)(x – 8) = 0
x+9=0
x–8=0
x = -9
x = 8 First number
x+1 = -8
x + 1 = 9 second number
I need to give my answers in pairs. I will put each x with the x + 1 that corresponds to it.
Solution: There are two sets of answers. -9 and -8 is one set, and 8 and 9 is the other.
7) The product of two consecutive even numbers is 24. Find all such numbers.
I will call the first of the two numbers x
first number x
Since they are consecutive even numbers the second number will be called x + 2
second number x + 2
I will replace the “product of the two numbers” with x(x+2)
I will replace the “is” with an equal sign
This is the equation I need to solve:
x(x+2) = 24 ( I will clear the parenthesis, set equal to 0, then solve by factoring)
𝑥 2 + 2𝑥 = 24
−24 − 24
__________________
x2 + 2x – 24 = 0
(x + 6)(x – 4) = 0
x+6=0
x–4=0
x = -6
x = 4 First number
x+2 = -4
x + 2 = 6 second number
I need to give my answers in pairs. I will put each x with the x + 2 that corresponds to it.
Solution: There are two sets of answers. -6 and -4 is one set, and 4 and 6 is the other.
9) The product of two consecutive even numbers is 63. Find all such numbers.
I will call the first of the two numbers x
first number x
Since they are consecutive even numbers the second number will be called x + 2
second number x + 2
I will replace the “product of the two numbers” with x(x+2)
I will replace the “is” with an equal sign
This is the equation I need to solve:
x(x+2) = 63 ( I will clear the parenthesis, set equal to 0, then solve by factoring)
𝑥 2 + 2𝑥 = 63
−63 − 63
__________________
x2 + 2x – 63 = 0
(x + 9)(x – 7) = 0
x+9=0
x–7=0
x = -9
x = 7 First number
x+2 = -7
x + 2 = 9 Second number
I need to give my answers in pairs. I will put each x with the x + 2 that corresponds to it.
Solution: There are two sets of answers. -9 and -7 is one set, and 7 and 9 is the other.
11) The length of a rectangular bedroom is 2 feet longer than its width. The area of the
bedroom is 120 square feet. Find the dimensions of the room.
Width = W
Length: L = W + 2
The area is found by multiplying the length and the width
This is the equation I need to solve:
Area = 120
or LW = 120
(W + 2)(W) = 120
𝑊 2 + 2𝑊 = 120
−120
− 120
_________________
W2 + 2W – 120 = 0
(W + 12)(W – 10) = 0
W + 12 = 0
W – 10 = 0
W = -12
W = 10
The width can’t be (-12). The answer must be the width is 10 feet. Add 2 feet to get the length
of 12 feet.
Solution: Length 12 feet, Width 10 feet
13) A rectangular garden is 4 feet narrower than it is long. The garden has an area of 32 square
feet. Find the dimensions of the garden.
Length = L
Width W = L – 4
Area = LW
32 = L(L-4)
32 = 𝐿2 − 4𝐿
−32
− 32
______________
0 = L2 – 4L – 32
0 = (L + 4)(L – 8)
L+4=0
L–8=0
L = -4
L=8
The length can’t be (-4). The answer must be the length is 8 feet. Subtract 4 feet to get the
width of 4 feet.
Solution: Length 8 feet, Width 4 feet
15) The base of a triangle is 2 feet longer than its height. The area of the triangle is 7.5 square
feet. Find the height of the triangle.
Height = h
Base = height plus 2
B=h+2
1
Area = 2 ∗ 𝑏𝑎𝑠𝑒 ∗ ℎ𝑒𝑖𝑔ℎ𝑡
1
7.5 = 2 (ℎ + 2)(ℎ)
Multiply by 2 to clear the fraction.
1
2*7.5 = 2 ∗ 2 (ℎ + 2)(ℎ)
15 = (h + 2)(h)
15 = ℎ2 + 2ℎ
0 = h2 + 2h – 15
(h + 5)(h – 3) = 0
h+5=0
h–3=0
h = -5
h=3
The height can’t be (-5) so the height must be 3 feet. I am not asked for the base, so I won’t
include it in my answer. I could just add 2 to get the base of 5 feet.
Solution: height = 3 feet
.
Section 1.9: Applications that involve factoring
17) The height of a triangle is 2 feet shorter than its base. The area of the triangle is 17.5
square feet. Find the height of the triangle.
Base = B
Height H = B - 2
1
Area = 2 ∗ 𝑏𝑎𝑠𝑒 ∗ ℎ𝑒𝑖𝑔ℎ𝑡
1
17.5 = 2 𝐵(𝐵 − 2)
1
2*17.5 = 2 ∗ 2 𝐵(𝐵 − 2)
35 = B(B – 2)
35 = B2 – 2B
0 = B2 – 2B – 35
0 = (B+5)(B – 7)
B+5=0
B–7=0
B = -5
B=7
The base can’t be (-5) so the base must be 7 feet. Subtract 2 to find the height of 5 feet.
Solution: Height = 5 feet
19) The length of the hypotenuse of a right triangle is 8 inches more than the shortest leg. The
length of the longer leg is 7 inches more than the length of the shorter leg. Find the length of
each side of the triangle.
A = shortest leg
B = longer leg
C = hypotenuse
length of the hypotenuse of a right triangle is 8 inches more than the shortest leg gives
C=A+8
length of the longer leg is 7 inches more than the length of the shorter leg gives
B=A+7
I will use the A2 + B2 = C2 formula, and replace the B and C with the above values.
A2 + (A + 7)2 = (A + 8)2
A2 + (A + 7)(A+7) = (A+8)(A+8)
A2 + A2 + 7A + 7A + 49 = A2 + 8A + 8A + 64
2A2 + 14A + 49 = A2 + 16A + 64
-A2 - 16A - 64 -A2 -16A – 64
A2 – 2A – 15 = 0
(A + 3)(A – 5) = 0
A+3=0
A–5=0
A = -3
A=5
The length can’t be (-3), so A must be 5 inches.
B = 5 + 7 = 12 inches
C = 5 + 8 = 13 inches
Solution: short leg 5 inches, longer leg 12 inches, hypotenuse 13 inches.
21) The length of a hypotenuse of a right triangle is 1 foot more than the longer leg. The
length of the shorter leg is 1 foot less than the length of the longer leg. Find the length of each
side of the right triangle.
A = shortest leg
B = longer leg
C = hypotenuse
length of a hypotenuse of a right triangle is 1 foot more than the longer leg gives
C=B+1
length of the shorter leg is 1 foot less than the length of the longer leg gives
A=B-1
I will use the A2 + B2 = C2 formula, and replace the B and C with the above values.
(B – 1)2 + B2 = (B + 1)2
(B – 1)(B – 1) + B2 = (B+1)(B+1)
B2 – 1B – 1B + 1 + B2 = B2 + 1B + 1B + 1
2B2 - 2B + 1 = B2 + 2B + 1
-B2 - 2B - 1 -B2 – 2B – 1
B2 – 4B = 0
B(B – 4) = 0
B=0
B–4=0
B=4
B can’t be 0 feet, so B must be 4 feet.
A = 4 – 1 = 3 feet
C = 4 + 1 = 5 feet
Solution: short leg 3 feet, longer leg 4 feet, hypotenuse 5 feet
23) The length of the hypotenuse in a right triangle is 15 inches. The shortest leg is 3 inches
shorter than the length of the longest leg. Find the length of each of the legs.
A = shortest leg B - 3
B = longer leg
C = 15 (the hypotenuse)
(B – 3)2 + B2 = 152
(B – 3)(B – 3) + B2 = 225
B2 – 3B – 3B + 9 + B2 = 225
2B2 – 6B + 9= 225
2B2 – 6B – 216 = 0
This is a trick that will make the work easier. I can divide each number by 2 without changing
the solution.
B2 - 3B – 108 = 0
This is still hard to factor. The lasts will be -12 and 9.
(B– 12)(B+ 9) = 0
B – 12 = 0
B+9=0
B = 12
B = -9
The length can’t be (-9). B must equal 12 inches.
A = 12 – 3 = 9 inches
Solution: short leg 9 inches, long leg 12 inches, hypotenuse 15 inches.
25) The length of the short leg in a right triangle is 3 inches. The longest leg is 1 inch less than
the length of the hypotenuse. Find the length of each of the each unknown side.
A = 3 inches
B=C–1
C=C
32 + (C – 1)2 = C2
9 + (C – 1)(C - 1) = C2
9 + C2 – 1C – 1C + 1 = C2
C2 – 2C + 10 = C2
-C2
- C2
-2C + 10 = 0
+2C
+2C
10 = 2C
5 =C
I also need to find B: B = C – 1, so B = 4.
Solution: short side 3 inches, longer side 4 inches, hypotenuse is 5 inches.