1. Completing the Square Find each square.
ALGEBRA 1 LESSON 9-5
(For help, go to Lessons 9-4 and 9-7.)
Find each square.
1. (d – 4)2 2. (x + 11)2 3. (k – 8)2
Factor.
4. b2 + 10b t2 + 14t n2 – 18n + 81
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2. Completing the Square Solutions
ALGEBRA 1 LESSON 9-5
1. (d – 4)2 = d2 – 2d(4) + 42 = d2 – 8d + 16
2. (x + 11)2 = x2 + 2x(11) = x2 + 22x + 121
3. (k – 8)2 = k2 – 2k(8) + 82 = k2 – 16k + 64
4. b2 + 10b + 25 = b2 + 2b(5) + 52 = (b + 5)2
5. t2 + 14t + 49 = t2 + 2t(7) + 72 = (t + 7)2
6. n2 – 18n + 81 = n2 – 2n(9) + 92 = (n – 9)2
Solutions
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3. Find the value of c to complete the square for x2 – 16x + c.
Completing the Square
ALGEBRA 1 LESSON 9-5
Find the value of c to complete the square for x2 – 16x + c.
The value of b in the expression x2 – 16x + c is –16.
The term to add to x2 – 16x is or 64.
–16 2
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4. Solve the equation x2 + 5 = 50 by completing the square.
ALGEBRA 1 LESSON 9-5
Solve the equation x2 + 5 = 50 by completing the square.
Step 1: Write the left side of x2 + 5 = 50 as a perfect square.
x2 + 5 = 50
x = 50 + 5 2
Add , or , to each side of the equation. 25 4 5 2
x +
200 4 + 25 =
Write x2 + 5x as a square.
Rewrite 50 as a fraction with denominator 4. 5 2
x + =
225 4
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5. Step 2: Solve the equation.
Completing the Square
ALGEBRA 1 LESSON 9-5
(continued)
Step 2: Solve the equation.
Find the square root of each side. 5 2
x + =
225 4 15
Simplify. 5 2
x + = 15 or –
Write as two equations.
x = 5 or x = –10
Solve for x.
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6. Add 16 to each side of the equation.
Completing the Square
ALGEBRA 1 LESSON 9-5
Solve x2 + 10x – 16 = 0 by completing the square. Round to the nearest hundredth.
Step 1: Rewrite the equation in the form x2 + bx = c and complete the square.
x2 + 10x – 16 = 0
x2 + 10x = 16
Add 16 to each side of the equation.
x2 + 10x + 25 =
Add , or 25, to each side of the equation. 10 2
(x + 5)2 = 41
Write x2 + 10x +25 as a square.
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7. Step 2: Solve the equation.
Completing the Square
ALGEBRA 1 LESSON 9-5
(continued)
Step 2: Solve the equation.
x + 5 = ± 41
Find the square root of each side.
Use a calculator to find 41
x + 5
± 6.40
x + 5
6.40 or
–6.40
Write as two equations.
Subtract 5 from each side.
x 6.40 – 5 or x –6.40 – 5
x 1.40 or x –11.40
Simplify
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8. Relate: length  width = area
Completing the Square
ALGEBRA 1 LESSON 9-5
Suppose you wish to section off a soccer field as shown in the diagram to run a variety of practice drills. If the area of the field is 6000 yd2, what is the value of x?
Define: width = x = x + 20
length = x + x = 2x + 20
Relate: length  width = area
Write: (2x + 20)(x + 20) = 6000
2x2 + 60x = 6000
Step 1: Rewrite the equation in the form x2 + bx = c.
2x2 + 60x = 6000
2x2 + 60x = 5600
Subtract 400 from each side.
x2 + 30x = 2800
Divide each side by 2.
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9. Step 2: Complete the square.
Completing the Square
ALGEBRA 1 LESSON 9-5
(continued)
Step 2: Complete the square.
x2 + 30x =
Add , or 225, to each side. 30 2
(x + 15)2 = 3025
Write x2 + 30x as a square.
Step 3: Solve each equation.
(x + 15) = ±
Take the square root of each side.
x + 15 = ± 55
Use a calculator.
x + 15 = 55 or x + 15 = –55
x = 40 or x = –70
Use the positive answer for this problem.
The value of x is 40 yd.
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10. Completing the Square
ALGEBRA 1 LESSON 9-5
Solve each equation by completing the square. If necessary, round to the nearest hundredth.
1. x2 + 14x = –43
2. 3x2 + 6x – 24 = 0
3. 4x2 + 16x + 8 = 40
–9.45, –4.55
–4, 2
–5.46, 1.46
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