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9-7

Solving Quadratic Equations by Using Square Roots. 9-7. Warm Up. Lesson Presentation. Lesson Quiz. Holt Algebra 1. 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 = 10. x = 80. x = 20.

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9-7

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  1. Solving Quadratic Equations by Using Square Roots 9-7 Warm Up Lesson Presentation Lesson Quiz Holt Algebra 1

  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 = 10 x= 80 x = 20

  3. Objective Solve quadratic equations by using square roots.

  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.

  5. Positive Square root of 9 Negative Square root of 9 When you take the square root of a positive number and the sign of the square root is not indicated, you must find both the positive and negative square root. This is indicated by ±√ Positive and negative Square roots of 9

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

  7. x2 = 169 Checkx2 = 169 (–13)2 169 (13)2 169   169 169 169 169 Example 1A: Using Square Roots to Solve x2 = a Solve using square roots. Check your answer. 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 and –13 into the original equation.

  8. Example 1B: Using Square Roots to Solve x2 = a Solve using square roots. x2 = –49 There is no real number whose square is negative. There is no real solution.

  9. x2 = 121 Check x2 = 121 (–11)2 121 (11)2 121   121 121 121 121 Check It Out! Example 1a Solve using square roots. Check your answer. x2 = 121 Solve for x by taking the square root of both sides. Use ± to show both square roots. x = ± 11 The solutions are 11 and –11. Substitute 11 and –11 into the original equation.

  10. Check It Out! Example 1c Solve using square roots. Check your answer. x2 = –16 There is no real number whose square is negative. There is no real solution.

  11. 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.

  12. x2 + 7 = 7 –7 –7 x2 = 0 Example 2A: Using Square Roots to Solve Quadratic Equations Solve using square roots. x2 + 7 = 7 Subtract 7 from both sides. Take the square root of both sides. The solution is 0.

  13. Example 2B: Using Square Roots to Solve Quadratic Equations Solve using square roots. 16x2 – 49 = 0 16x2 – 49 = 0 +49 +49 Add 49 to both sides. Divide by 16 on both sides. Take the square root of both sides. Use ± to show both square roots.

  14. . Example 2B Continued Solve using square roots. Check your answer. Check 16x2 – 49 = 0 16x2 – 49 = 0   49 – 49 0 49 – 49 0

  15. 100x2 + 49 = 0 –49 –49 100x2 =–49 Check It Out! Example 2a Solve by using square roots. Check your answer. 100x2 + 49 = 0 Subtract 49 from both sides. Divide by 100 on both sides. There is no real number whose square is negative. There is no real solution.

  16. . Check It Out! Example 2b Solve by using square roots. Check your answer. 36x2 = 1 Divide by 36 on both sides. Take the square root of both sides. Use ± to show both square roots.

  17. When solving quadratic equations by using square roots, you may need to find the square root of a number that is not a perfect square. In this case, the answer is an irrational number. You can approximate the solutions.

  18. Evaluate on a calculator. Example 3A: Approximating Solutions Solve. Round to the nearest hundredth. x2 = 15 Take the square root of both sides. x 3.87 The approximate solutions are 3.87 and –3.87.

  19. –3x2 + 90 = 0 –90 –90 Evaluate on a calculator. Example 3B: Approximating Solutions Solve. Round to the nearest hundredth. –3x2 + 90 = 0 Subtract 90 from both sides. Divide by – 3 on both sides. x2 = 30 Take the square root of both sides. x 5.48 The approximate solutions are 5.48 and –5.48.

  20. Example 3B Continued Solve. Round to the nearest hundredth. –3x2 + 90 = 0 The approximate solutions are 5.48 and –5.48. CheckUse a graphing calculator to support your answer. Use the zero function. The approximate solutions are 5.48 and – 5.48.

  21. 0 = 90 – x2 + x2 + x2 x2 = 90 Check It Out! Example 3a Solve. Round to the nearest hundredth. 0 = 90 – x2 Add x2 to both sides. Take the square root of both sides. The approximate solutions are 9.49 and –9.49.

  22. 2x2 – 64 = 0 + 64 + 64 Check It Out! Example 3b Solve. Round to the nearest hundredth. 2x2 – 64 = 0 Add 64 to both sides. Divide by 2 on both sides. x2 = 32 Take the square root of both sides. The approximate solutions are 5.66 and –5.66.

  23. – 45 – 45 x2 = –45 Check It Out! Example 3c Solve. Round to the nearest hundredth. x2 + 45 = 0 x2 + 45 = 0 Subtract 45 from both sides. There is no real number whose square is negative. There is no real solution.

  24. 2x x=578 ● Example 4: 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. lw = A Use the formula for area of a rectangle. l = 2w Length is twice the width. Substitute x for w, 2x for l, and 578 for A. 2x2 = 578

  25. Evaluate on a calculator. Example 4 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.

  26. 2x 2x x Check It Out! Example 4 A house is on a lot that is shaped like a trapezoid. The solid lines show the boundaries, where x represents the width of the front yard. Find the width of the front yard, given that the area is 6000 square feet. Round to the nearest foot. (Hint: Use ) Use the formula for area of a trapezoid.

  27. Evaluate on a calculator. Check It Out! Example 4 Substitute 2x for h and b1, x for b2 , and 6000 for A. Divide by 3 on both sides. Take the square root of both sides. Negative numbers are not reasonable for width, so x ≈ 45 is the only solution that makes sense. Therefore, the width of the front yard is about 45 feet.

  28. Lesson Quiz: Part 1 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 no real solutions ± 6.71

  29. (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

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