1 / 22

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.

trang
Download Presentation

Standard 8

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  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

More Related