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Standard 8

. . . . 7. 7. 7. 7. x – 4 = ±. x = 4 ±. ANSWER. The solutions are 4 + 6.65 and 4 – 1.35. 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.

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Standard 8

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  1.    7 7 7 7 x – 4 = ± x= 4 ± ANSWER The solutions are 4 + 6.65 and 4 –1.35. 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. Take square roots of each side. Add 4 to each side.

  2. Standard 8 Solve a quadratic equation 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.

  3. ANSWER ANSWER –1, 5 0.35, 5.65 ANSWER –7.83, –2.17 EXAMPLE 1 Solve a Quadratic Equation Solve quadratic equations GUIDED PRACTICE Solve the equation. Round the solution to the nearest hundredth if necessary. 1. 2(x – 2)2 = 18 2. 4(q – 3)2 = 28 3. 3(t + 5)2 = 24

  4. x = ± 4=±2  EXAMPLE 1 Solve quadratic equations Solve the equation. a. 2x2 = 8 SOLUTION a.2x2 = 8 Write original equation. Divide each side by 2. x2 = 4 Take square roots of each side. Simplify. The solutions are–2and2. ANSWER

  5. ANSWER The solution is 0. EXAMPLE 1 Solve quadratic equations b. m2 – 18 = – 18 Write original equation. m2 =0 Add 18 to each side. m=0 The square root of 0 is 0.

  6. ANSWER Negative real numbers do not have real square roots. So, there is no solution. EXAMPLE 1 Solve quadratic equations c.b2+12=5 Write original equation. b2 = – 7 Subtract 12 from each side.

  7. 9 3 9 z2 = 4 2 4  z = ± z = ± EXAMPLE 2 Take square roots of a fraction Solve 4z2 = 9. SOLUTION 4z2 = 9 Write original equation. Divide each side by 4. Take square roots of each side. Simplify.

  8. The solutions are – and 3 3 2 2 EXAMPLE 2 Take square roots of a fraction ANSWER

  9. 6 x = ± EXAMPLE 3 Approximate solutions of a quadratic equation Solve 3x2 – 11 = 7. Round the solutions to the nearest hundredth. SOLUTION 3x2 – 11 = 7 Write original equation. 3x2 = 18 Add 11 to each side. x2 = 6 Divide each side by 3. Take square roots of each side.

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

  11. no solution ANSWER ANSWER ANSWER –5, 5. 0 EXAMPLE 1 Solve quadratic equations GUIDED PRACTICE Solve the equation. 1.c2 – 25 = 0 2. 5w2 + 12 = – 8 3. 2x2 + 11 = 11

  12. no solution ANSWER ANSWER – , 4 4 10 10 ANSWER – 5 5 , 3 3 EXAMPLE 1 Solve quadratic equations GUIDED PRACTICE Solve the equation. 4. 25x2 = 16 5. 9m2 = 100 6. 49b2 +64 = 0

  13. – 3.16, 3.16 – 0.58, 0.58 – 2.12, 2.12 ANSWER ANSWER ANSWER EXAMPLE 1 Solve quadratic equations GUIDED PRACTICE Solve the equation. Round the solutions to the nearest hundredth. 7. x2 +4 = 14 8. 3k2 –1 = 0 9. 2p2 –7 = 2

  14. 25 5 2 c = = 4 2 Standard 8 Complete the square Find the value of c that makes the expression x2 + 5x + c a perfect square trinomial. Then write the expression as the square of a binomial. STEP 1 Find the value ofc.For the expression to be a perfect squaretrinomial, c needs to be the square of half thecoefficient ofbx. Find the square of half the coefficient of bx.

  15. 5 2 x2 + 5x +c =x2+ 5x + Substitute 25 for c. 4 25 = 4 x 2 + 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. Square of a binomial

  16. 9 3 4 2 ANSWER ANSWER 16; (x + 4)2 36; (x 6)2 ANSWER ; (x )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. 1. x2 + 8x + c 2. x2 12x + c 3. x2 + 3x + c

  17. Add , or (– 8)2, to each side. –16 2 2 EXAMPLE 2 Solve a quadratic equation Solve x2 – 16x = –15 by completing the square. SOLUTION x2 – 16x = –15 Write original equation. x2 – 16x + (– 8)2=–15 + (– 8)2 (x – 8)2 = –15 + (– 8)2 Write left side as the square of a binomial. (x – 8)2 = 49 Simplify the right side.

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

  19. (15)2– 16(15) –15 (1)2– 16(1) –15 –15 = –15 –15 = –15 ? = ? = EXAMPLE 2 Standardized Test Practice CHECK You can check the solutions in the original equation. If x = 1: If x = 15:

  20. 10 2 , or52, to each side. Add 2 EXAMPLE 3 Solve a quadratic equation in standard form Solve 2x2 + 20x – 8 = 0 by completing the square. SOLUTION 2x2 + 20x – 8 = 0 Write original equation. 2x2 + 20x = 8 Add 8 to each side. x2 + 10x = 4 Divide each side by 2. x2+ 10x + 52= 4 + 52 (x + 5)2 = 29 Write left side as the square of a binomial.

  21. ± x + 5 = 29 ± 29 x = –5 ANSWER The solutions are – 5 + 29 0.39 and – 5 - 29 –10.39. EXAMPLE 3 Solve a quadratic equation in standard form Take square roots of each side. Subtract 5 from each side.

  22. ANSWER 1, 3 ANSWER ANSWER 9.12, 0.88 1.35, 6.65 GUIDED PRACTICE 4. x2 – 2x = 3 5. m2 + 10m = –8 6. 3g2 – 24g+ 27 = 0

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