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4.7 Complete the Square

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**Solve a quadratic equation by finding square roots**

EXAMPLE 1 Solve a quadratic equation by finding square roots Solve x2 – 8x + 16 = 25. x2 – 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 = 9 and 4 –5 = – 1. ANSWER

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EXAMPLE 2 Make a perfect square trinomial Find the value of c that makes x2 + 16x + c a perfect square trinomial. Then write the expression as the square of a binomial. SOLUTION STEP 1 16 2 = 8 Find half the coefficient of x. STEP 2 Square the result of Step 1. 82 = 64 STEP 3 Replace c with the result of Step 2. x2 + 16x + 64

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EXAMPLE 2 Make a perfect square trinomial ANSWER The trinomial x2 + 16x + c is a perfect square when c = 64. Then x2 + 16x + 64 = (x + 8)(x + 8) = (x + 8)2.

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GUIDED PRACTICE for Examples 1 and 2 Solve the equation by finding square roots. x2 + 6x + 9 = 36. ANSWER 3 and –9. x2 – 10x + 25 = 1. ANSWER 4 and 6. x2 – 24x = 100. ANSWER 2 and 22.

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GUIDED PRACTICE for Examples 1 and 2 Find the value of c that makes the expression a perfect square trinomial.Then write the expression as the square of a binomial. 4. x2 + 14x + c ANSWER 49 ; (x + 7)2 5. x2 + 22x + c ANSWER 121 ; (x + 11)2 6. x2 – 9x + c 81 4 9 2 ANSWER ; (x – )2.

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**( ) EXAMPLE 3 Solve ax2 + bx + c = 0 when a = 1**

Solve x2 – 12x + 4 = 0 by completing the square. x2 – 12x + 4 = 0 Write original equation. x2 – 12x = –4 Write left side in the form x2 + bx. x2 – 12x + 36 = –4 + 36 Add –12 2 ( ) = (–6) 36 to each side. (x – 6)2 = 32 Write left side as a binomial squared. x – 6 = Take square roots of each side. x = Solve for x. x = Simplify: 32 = 16 2 4 The solutions are 6 + 4 and 6 – 4 2 ANSWER

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EXAMPLE 3 Solve ax2 + bx + c = 0 when a = 1 CHECK You can use algebra or a graph. Algebra Substitute each solution in the original equation to verify that it is correct. Graph Use a graphing calculator to graph y = x2 – 12x + 4. The x-intercepts are about – 4 2 and

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**( ) EXAMPLE 4 Solve ax2 + bx + c = 0 when a = 1**

Solve 2x2 + 8x + 14 = 0 by completing the square. 2x2 + 8x + 14 = 0 Write original equation. x2 + 4x + 7 = 0 Divide each side by the coefficient of x2. x2 + 4x = –7 Write left side in the form x2 + bx. Add 4 2 ( ) = to each side. x2 – 4x + 4 = –7 + 4 (x + 2)2 = –3 Write left side as a binomial squared. x + 2 = –3 Take square roots of each side. x = – –3 Solve for x. x = –2 + i 3 Write in terms of the imaginary unit i.

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EXAMPLE 4 Solve ax2 + bx + c = 0 when a = 1 The solutions are –2 + i 3 and –2 – i 3 . ANSWER

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EXAMPLE 5 Standardized Test Practice SOLUTION Use the formula for the area of a rectangle to write an equation.

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**( ) EXAMPLE 5 Standardized Test Practice 3x(x + 2) = 72 3x2 + 6x = 72**

Length Width = Area 3x2 + 6x = 72 Distributive property x2 + 2x = 24 Divide each side by the coefficient of x2. Add 2 ( ) = 1 to each side. x2 – 2x + 1 = (x + 1)2 = 25 Write left side as a binomial squared. x + 1 = + 5 Take square roots of each side. x = –1 + 5 Solve for x.

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EXAMPLE 5 Standardized Test Practice So, x = –1 + 5 = 4 or x = – 1 – 5 = –6. You can reject x = –6 because the side lengths would be –18 and –4, and side lengths cannot be negative. The value of x is 4. The correct answer is B. ANSWER

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GUIDED PRACTICE for Examples 3, 4 and 5 Solve the equation by completing the square. 7. x2 + 6x + 4 = 0 10. 3x2 + 12x – 18 = 0 –3+ 5 ANSWER –2 + 10 ANSWER 8. x2 – 10x + 8 = 0 11. 6x(x + 8) = 12 5 + 17 ANSWER –4 +3 2 ANSWER 9. 2n2 – 4n – 14 = 0 12. 4p(p – 2) = 100 1 + 2 2 ANSWER 1 + 26 ANSWER

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