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4.7 Complete the Square

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

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4.7 Complete the Square

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  1. 4.7 Complete the Square

  2. ANSWER The solutions are 4 + 5 = 9 and 4 –5 = – 1. 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.

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

  4. EXAMPLE 2 Make a perfect square trinomial ANSWER The trinomialx2 + 16x + c is a perfect square whenc = 64. Thenx2 + 16x + 64 = (x + 8)(x + 8) = (x + 8)2.

  5. for Examples 1 and 2 GUIDED PRACTICE Solve the equation by finding square roots. 1. x2 + 6x + 9 = 36. ANSWER 3 and –9. 2. x2 – 10x + 25 = 1. ANSWER 4 and 6. 3. x2 – 24x + 144 = 100. ANSWER 2 and 22.

  6. 81 9 4 2 for Examples 1 and 2 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. x2 + 14x + c 4. ANSWER 49 ; (x + 7)2 x2 + 22x + c 5. ANSWER 121 ; (x + 11)2 x2 – 9x + c 6. ANSWER ; (x – )2.

  7. ) ( –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.

  8. 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

  9. 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.

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

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

  12. 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.

  13. 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.

  14. 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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