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EXAMPLE 3

). (. –12. Add. to each side. 2. 36. (–6). 2. =. =. 2. x – 6 = + 32. x = 6 + 32. x = 6 + 4 2. 2. Simplify:. 32. 16. 2. 4. =. =. ANSWER. 2. 2. The solutions are 6 + 4. and 6 – 4. EXAMPLE 3. Solve ax 2 + bx + c = 0 when a = 1.

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EXAMPLE 3

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  1. ) ( –12 Add to each side. 2 36 (–6) 2 = = 2 x – 6 = + 32 x = 6 + 32 x = 6 + 4 2 2 Simplify: 32 16 2 4 = = ANSWER 2 2 The solutions are 6 + 4 and6 – 4 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 (x – 6)2 = 32 Write left side as a binomial squared. Take square roots of each side. Solve for x.

  2. 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 0.34 6 – 4 2 and 11.66 6 + 4 2 EXAMPLE 3 Solve ax2 + bx + c = 0 when a = 1 CHECK

  3. Solve ax2 + bx + c = 0 when a = 1 ) ( 4 Add to each side. 2 4 2 2 = = 2 x + 2 = + –3 x = –2 + –3 x = –2 +i 3 EXAMPLE 4 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. x2– 4x + 4 = –7 + 4 (x + 2)2 = –3 Write left side as a binomial squared. Take square roots of each side. Solve for x. Write in terms of the imaginary unit i.

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

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

  6. Length Width = Area ) ( 2 Add to each side. 2 1 1 2 = = 2 EXAMPLE 5 Standardized Test Practice 3x(x + 2) = 72 3x2 + 6x = 72 Distributive property x2 + 2x = 24 Divide each side by the coefficient of x2. x2– 2x + 1 = 24 + 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.

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

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

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