Standard 8 Solve a quadratic equation Solve 6(x – 4) 2 = 42. Round the solutions to the nearest hundredth. 6(x – 4) 2 = 42 Write original equation. (x – 4) 2 = 7 Divide each side by 6. x – 4 = ± 7  Take square roots of each side. 7  x = 4 ± Add 4 to each side. ANSWER The solutions are and 4 –  7 

Solve a quadratic equation Standard 8 CHECK To check the solutions, first write the equation so that 0 is on one side as follows: 6(x – 4) 2 – 42 = 0. Then graph the related function y = 6(x – 4) 2 – 42. The x -intercepts appear to be about 6.6 and about 1.3. So, each solution checks.

EXAMPLE 1 Solve quadratic equations Solve the equation. Round the solution to the nearest hundredth if necessary. 1. 2(x – 2) 2 = 18 GUIDED PRACTICE Solve a Quadratic Equation ANSWER –1, (q – 3) 2 = 28 ANSWER 0.35, (t + 5) 2 = 24 ANSWER –7.83, –2.17

EXAMPLE 1 Solve quadratic equations Solve the equation. a. 2x 2 = 8 SOLUTION a. 2x 2 = 8 Write original equation. x 2 = 4 Divide each side by 2. x = ± 4 = ± 2  Take square roots of each side. Simplify. ANSWER The solutions are –2 and 2.

Solve quadratic equations EXAMPLE 1 b. m 2 – 18 = – 18 Write original equation. m 2 = 0 Add 18 to each side. The square root of 0 is 0. m = 0m = 0 ANSWER The solution is 0.

Solve quadratic equations EXAMPLE 1 c. b = 5 Write original equation. b 2 = – 7 Subtract 12 from each side. ANSWER Negative real numbers do not have real square roots. So, there is no solution.

EXAMPLE 2 Take square roots of a fraction Solve 4z 2 = 9. SOLUTION 4z 2 = 9 Write original equation. z 2 = 9 4 Divide each side by 4. Take square roots of each side. z = ±  Simplify.

Take square roots of a fraction EXAMPLE 2 ANSWER The solutions are – and

Approximate solutions of a quadratic equation EXAMPLE 3 Solve 3x 2 – 11 = 7. Round the solutions to the nearest hundredth. SOLUTION 3x 2 – 11 = 7 Write original equation. 3x 2 = 18 Add 11 to each side. x 2 = 6 Divide each side by 3. x = ± 6  Take square roots of each side.

Approximate solutions of a quadratic equation EXAMPLE 3 x ± 2.45 Use a calculator. Round to the nearest hundredth. ANSWER The solutions are about – 2.45 and about 2.45.

EXAMPLE 1 Solve quadratic equations Solve the equation. 1. c 2 – 25 = 0 GUIDED PRACTICE ANSWER–5, w = – 8 ANSWER no solution 3. 2x = 11ANSWER0

EXAMPLE 1 Solve quadratic equations Solve the equation x 2 = 16 GUIDED PRACTICE ANSWER –, 5. 9m 2 = 100 ANSWER 10 3 –, b = 0 ANSWER no solution

EXAMPLE 1 Solve quadratic equations Solve the equation. Round the solutions to the nearest hundredth. 7. x = 14 GUIDED PRACTICE ANSWER – 3.16, k 2 – 1 = 0 ANSWER – 0.58, p 2 – 7 = 2 ANSWER – 2.12, 2.12

Standard 8 Complete the square Find the value of c that makes the expression x 2 + 5x + c a perfect square trinomial. Then write the expression as the square of a binomial. STEP 1 Find the value of c. For the expression to be a perfect square trinomial, c needs to be the square of half the coefficient of bx. Find the square of half the coefficient of bx. 2 2 = 25 4 c = 5

Standard 8 Complete the square STEP 2 Write the expression as a perfect square trinomial. Then write the expression as the square of a binomial. Substitute 25 for c. 4 x 2 + 5x + c = x 2 + 5x Square of a binomial x =

GUIDED PRACTICE Find the value of c that makes the expression a perfect square trinomial. Then write the expression as the square of a binomial. 1. x 2 + 8x + cANSWER16; (x + 4) 2 2. x 2  12x + c 3. x 2 + 3x + c ANSWER 36; (x  6) 2 ANSWER ; (x  )

EXAMPLE 2 Solve a quadratic equation Solve x 2 – 16x = –15 by completing the square. SOLUTION Write original equation. x 2 – 16x = –15 Add, or (– 8) 2, to each side. – x 2 – 16x + (– 8) 2 = –15 + (– 8) 2 Write left side as the square of a binomial. (x – 8) 2 = –15 + (– 8) 2 Simplify the right side. (x – 8) 2 = 49

EXAMPLE 2 Standardized Test Practice Take square roots of each side. x – 8 = ±7 Add 8 to each side. x = 8 ± 7 ANSWER The solutions of the equation are = 15 and 8 – 7 = 1.

EXAMPLE 2 Standardized Test Practice CHECK You can check the solutions in the original equation. If x = 15: (15) 2 – 16(15) –15 ? = –15 = –15 If x = 1: (1) 2 – 16(1) –15 ? = –15 = –15

EXAMPLE 3 Solve a quadratic equation in standard form Solve 2x x – 8 = 0 by completing the square. SOLUTION Write original equation. 2x x – 8 = 0 Add 8 to each side. 2x x = 8 Divide each side by 2. x x = 4 Add , or 5 2, to each side. x x = Write left side as the square of a binomial. (x + 5) 2 = 29

EXAMPLE 3 Solve a quadratic equation in standard form Take square roots of each side. x + 5 = ± 29 Subtract 5 from each side. x = –5 ± 29 ANSWER The solutions are – and – –10.39.

GUIDED PRACTICE 4. x 2 – 2x = 3 ANSWER  1, 3 5. m m = –8 ANSWER  9.12,  g 2 – 24g + 27 = 0 ANSWER1.35, 6.65