1 / 21

9.6 Solving Quadratic Equations by Using Square Roots

9.6 Solving Quadratic Equations by Using Square Roots. Warm Up Find each square root. Solve each equation. 5 . – 6 x = – 60 6. 7. 2 x – 40 = 0 8. 5 x = 3. 1. 6. 11. 2. 4. –25. 3. x = 80. x = 10. x = 20. California Standards.

ivana
Download Presentation

9.6 Solving Quadratic Equations by Using Square Roots

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. 9.6 Solving Quadratic Equations by Using Square Roots

  2. Warm Up Find each square root. Solve each equation. 5. –6x = –60 6. 7. 2x – 40 = 0 8. 5x = 3 1. 6 11 2. 4. –25 3. x= 80 x = 10 x = 20

  3. California Standards 2.0 Students understand and use such operations as taking the opposite, finding the reciprocal, taking a root, and raising to a fractional power. They understand and use the rules for exponents. 23.0 Students apply quadratic equations to physical problems,such as the motion of an object under the force of gravity.

  4. Some quadratic equations cannot be easily solved by factoring. Square roots can be used to solve some of these quadratic equations. Recall from Lesson 1-5 that every positive real number has two square roots, one positive and one negative. (Remember also that the symbol indicates a nonnegative square root.)

  5. Positive square root of 9 Negative square root of 9 When you take the square root to solve an equation, you must find both the positive and negative square root. This is indicated by the symbol ±√ . Positive and negative square roots of 9 Review

  6. Reading Math The expression ±3 means “3 or –3” and is read “plus or minus three.”

  7. Checkx2 = 169 (13)2 169  169 169 Solve using square roots. x2 = 169 Solve for x by taking the square root of both sides. Use ± to show both square roots. x = ± 13 The solutions are 13 and –13. Substitute 13 into the original equation.

  8. Solve using square roots. x2 = –49 There is no real number whose square is negative. There is no real solution. The solution set is the empty set, ø.

  9. Now you try Solve using square roots • x2= 121 • x2 = -25 • x2 = 0 • x2 = -16 • x2 = 100 • x2 = 64 x = ±11 no real solution. ø x = 0 no real solution. ø x = ±10 x = ±8

  10. If a quadratic equation is not written in the form x2 = a, use inverse operations to isolate x2 before taking the square root of both sides.

  11. x2 + 7 = 32 –7 –7 x2 = 25 Solve using square roots. x2 + 7 = 32 Subtract 7 from both sides. Take the square root of both sides.

  12. Solve by using square roots. 16x2 – 49 = 0 Add 49 to both sides. 16x2 – 49 = 0 +49 +49 Divide by 16 on both sides. Take the square root of both sides. Use ± to show both square roots.

  13. Solve by using square roots. 36x2 = 1 Divide by 36 on both sides. Take the square root of both sides. Use ± to show both square roots.

  14. 100x2 + 49 = 0 –49 –49 100x2 =–49 Solve by using square roots. 100x2 + 49 = 0 Subtract 49 from both sides. Divide by 100 on both sides. There is no real number whose square is negative. ø

  15. 0 = 90 – x2 + x2 + x2 x2 = 90 Estimate The exact solutions are and The approximate solutions are 9.49 and –9.49. . Solve. Round to the nearest hundredth. 0 = 90 – x2 Add x2 to both sides. Take the square root of both sides.

  16. Now you try Solve using square roots • x2 -196 = 0 • 0 = 3x2 -48 • 24x2 +96 = 0 • 10x2 - 75= 15 • 0 = 4x2 +144 • 5x2 – 105 = 20 x = ±14 x = ±4 no real solution. ø x = ±3 no real solution. ø x = ±5

  17. 2x x=578 ● Application Ms. Pirzada is building a retaining wall along one of the long sides of her rectangular garden. The garden is twice as long as it is wide. It also has an area of 578 square feet. What will be the length of the retaining wall? Let x represent the width of the garden. Use the formula for area of a rectangle. lw= A Length is twice the width. l=2w Substitute x for w, 2x for l, and 578 for A. 2x2 = 578

  18. Continued 2x2 = 578 Divide both sides by 2. Take the square root of both sides. x = ±17 Negative numbers are not reasonable for width, so x = 17 is the only solution that makes sense. Therefore, the length is 2w or34 feet.

  19. Lesson Quiz: Part I Solve using square roots. Check your answers. 1. x2 – 195 = 1 2. 4x2 – 18 = –9 3. 2x2 – 10 = –12 4.Solve 0 = –5x2 + 225. Round to the nearest hundredth. ± 14 ø ± 6.71

  20. (Hint: Use ) Lesson Quiz: Part II 5. A community swimming pool is in the shape of a trapezoid. The height of the trapezoid is twice as long as the shorter base and the longer base is twice as long as the height. The area of the pool is 3675 square feet. What is the length of the longer base? Round to the nearest foot. 108 feet

More Related