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EXAMPLE 1 Solve a quadratic equation by finding square roots Solve x 2 – 8x + 16 = 25. x 2 – 8x + 16 = 25 Write original equation. (x – 4) 2 = 25 Write.

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Presentation on theme: "EXAMPLE 1 Solve a quadratic equation by finding square roots Solve x 2 – 8x + 16 = 25. x 2 – 8x + 16 = 25 Write original equation. (x – 4) 2 = 25 Write."— Presentation transcript:

1 EXAMPLE 1 Solve a quadratic equation by finding square roots Solve x 2 – 8x + 16 = 25. x 2 – 8x + 16 = 25 Write original equation. (x – 4) 2 = 25 Write left side as a binomial squared. x – 4 = +5 Take square roots of each side. x = 4 + 5 Solve for x. The solutions are 4 + 5 = 9 and 4 –5 = – 1. ANSWER

2 EXAMPLE 2 Make a perfect square trinomial Find the value of c that makes x 2 + 16x + c a perfect square trinomial. Then write the expression as the square of a binomial. SOLUTION STEP 1 Find half the coefficient of x. STEP 2 16 2 =8 Square the result of Step 1. 8 2 = 64 STEP 3 Replace c with the result of Step 2. x 2 + 16x + 64

3 EXAMPLE 2 Make a perfect square trinomial The trinomial x 2 + 16x + c is a perfect square when c = 64. Then x 2 + 16x + 64 = (x + 8)(x + 8) = (x + 8) 2. ANSWER

4 GUIDED PRACTICE for Examples 1 and 2 1. x 2 + 6x + 9 = 36. x 2 + 6x + 9 = 36 Write original equation. (x + 3) 2 = 36 Write left side as a binomial squared. x + 3 = + 6 Take square roots of each side. x = – 3 + 6 Solve for x. The solutions are – 3 + 6 = 3 and – 3 – 6 = – 9. ANSWER Solve the equation by finding square roots.

5 GUIDED PRACTICE for Examples 1 and 2 2. x 2 – 10x + 25 = 1. x 2 – 10x + 25 = 1 Write original equation. (x – 5) 2 = 1 Write left side as a binomial squared. x – 5 = + 1 Take square roots of each side. x = 5 + 1 Solve for x. The solutions are 5 – 1 = 4 and 5 + 1 = 6. ANSWER

6 GUIDED PRACTICE for Examples 1 and 2 3. x 2 – 24x + 144 = 100. x 2 – 24x + 144 = 100 Write original equation. (x – 12) 2 = 100 Write left side as a binomial squared. x – 12 = + 10 Take square roots of each side. x = 12 + 10 Solve for x. The solutions are 12 – 10 = 2 and 12 + 10 = 22. ANSWER

7 GUIDED PRACTICE for Examples 1 and 2 4. x 2 + 14x + c x2x2 7x7x 7x7x49 x x 7 7 SOLUTION STEP 1 Find half the coefficient of x. STEP 2 14 2 =7 Square the result of Step 1. 7 2 = 49 STEP 3 Replace c with the result of Step 2. x 2 + 14x + 49 Find the value of c that makes the expression a perfect square trinomial.Then write the expression as the square of a binomial.

8 GUIDED PRACTICE for Examples 1 and 2 The trinomial x 2 + 14x + c is a perfect square when c = 49. Then x 2 + 14x + 49 = (x + 7)(x + 7) = (x + 7) 2. ANSWER

9 GUIDED PRACTICE for Examples 1 and 2 5. x 2 + 22x + c x2x2 11x 121 x x 11 SOLUTION STEP 1 Find half the coefficient of x. STEP 2 22 2 = Square the result of Step 1. 11 2 = 121 STEP 3 Replace c with the result of Step 2. x 2 + 22x + 121

10 GUIDED PRACTICE for Examples 1 and 2 The trinomial x 2 + 22x + c is a perfect square when c = 144. Then x 2 + 22x + 144 = (x + 11)(x + 11) = (x + 11) 2. ANSWER

11 GUIDED PRACTICE for Examples 1 and 2 6. x 2 – 9x + c x2x2 x x SOLUTION STEP 1 Find half the coefficient of x. STEP 2 STEP 3 Replace c with the result of Step 2. 9 2 – x 9 2 – x 81 4 9 2 – 9 2 – Square the result of Step 1. x 2 – 9x + 81 4 9 2 9 2 = 9 2 ( ) = 81 4

12 GUIDED PRACTICE for Examples 1 and 2 ANSWER The trinomial x 2 – 9x + c is a perfect square when c =. Then x 2 – 9x + = (x – )(x – ) = (x – ) 2. 81 4 4 9 2 9 2 9 2


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