Then/Now You solved quadratic equations by using the square root property. Complete the square to write perfect square trinomials. Solve quadratic equations by completing the square.

Concept

Example 1 Complete the Square Find the value of c that makes x 2 – 12x + c a perfect square trinomial. Complete the square. Answer: Thus, c = 36. Notice that x 2 – 12x + 36 = (x – 6) 2. Step 1 Step 2Square the result(–6) 2 = 36 of Step 1. Step 3Add the result ofx 2 –12x + 36 Step 2 to x 2 – 12x.

Example 1 A.7 B.14 C.156 D.49 Find the value of c that makes x x + c a perfect square.

Example 2 Solve an Equation by Completing the Square Solve x 2 + 6x + 5 = 12 by completing the square. Isolate the x 2 - and x-terms. Then complete the square and solve. x 2 + 6x + 5 = 12Original equation x 2 + 6x – 5 – 5= 12 – 5 Subtract 5 from each side. x 2 + 6x = 7Simplify. x 2 + 6x + 9 =7 + 9

Example 2 Solve an Equation by Completing the Square (x + 3) 2 =16Factor x 2 + 6x + 9. = –7 = 1 Simplify. Answer: The solutions are –7 and 1. x + 3 =±4 Take the square root of each side. x + 3 – 3=±4 – 3Subtract 3 from each side. x=±4 – 3 Simplify. x = –4 – 3 or x = 4 – 3Separate the solutions.

Example 2 A.{–2, 10} B.{2, –10} C.{2, 10} D.Ø Solve x 2 – 8x + 10 = 30.

Example 3 Equation with a ≠ 1 Solve –2x x – 10 = 24 by completing the square. –2x x – 10 = 24Original equation Isolate the x 2 - and x-terms. Then complete the square and solve. x 2 – 18x + 5= –12Simplify. x 2 – 18x + 5 – 5= –12 – 5 Subtract 5 from each side. x 2 – 18x = –17Simplify. Divide each side by –2.

Example 3 Equation with a ≠ 1 (x – 9) 2 =64Factor x 2 – 18x = 17 = 1Simplify. x – 9 =±8 Take the square root of each side. x – =±8 + 9Add 9 to each side. x =9 ± 8 Simplify. x = or x = 9 – 8Separate the solutions. x 2 – 18x + 81 =–

Example 3 Equation with a ≠ 1 Answer: The solutions are 1 and 17.

Example 3 A.{–1} B.{–1, –7} C.{–1, 7} D.Ø Solve x 2 + 8x + 10 = 3 by completing the square.