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 + 25 5. t2 + 14t + 49 6. n2 – 18n + 81 9-5

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) + 112 = 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 9-5

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 9-5

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 x2 + 5 + = 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 9-5

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. 9-5

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 = 16 + 25 Add , or 25, to each side of the equation. 10 2 (x + 5)2 = 41 Write x2 + 10x +25 as a square. 9-5

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 9-5

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 + 10 + 10 = x + 20 length = x + x + 10 + 10 = 2x + 20 Relate: length  width = area Write: (2x + 20)(x + 20) = 6000 2x2 + 60x + 400 = 6000 Step 1: Rewrite the equation in the form x2 + bx = c. 2x2 + 60x + 400 = 6000 2x2 + 60x = 5600 Subtract 400 from each side. x2 + 30x = 2800 Divide each side by 2. 9-5

Step 2: Complete the square. Completing the Square ALGEBRA 1 LESSON 9-5 (continued) Step 2: Complete the square. x2 + 30x + 255 = 2800 + 225 Add , or 225, to each side. 30 2 (x + 15)2 = 3025 Write x2 + 30x + 255 as a square. Step 3: Solve each equation. (x + 15) = ± 3025 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. 9-5

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 9-5