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

2. Holt Algebra 1 9-7 Solving Quadratic Equations by Using Square Roots Warm Up Find each square root. Solve each equation. 5. –6x = –606. 7. 2x – 40 = 08. 5x = 3 611 –25 1. 2. 3. 4. x = 10 x = 80 x = 20

3. Holt Algebra 1 9-7 Solving Quadratic Equations by Using Square Roots Solve quadratic equations by using square roots. Objective

4. Holt Algebra 1 9-7 Solving Quadratic Equations by Using Square Roots 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. Holt Algebra 1 9-7 Solving Quadratic Equations by Using Square Roots 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 Positive Square root of 9

6. Holt Algebra 1 9-7 Solving Quadratic Equations by Using Square Roots The expression ±3 is read “plus or minus three” Reading Math

7. Holt Algebra 1 9-7 Solving Quadratic Equations by Using Square Roots

8. Holt Algebra 1 9-7 Solving Quadratic Equations by Using Square Roots Example 1A: Using Square Roots to Solve x 2 = a Solve using square roots. Check your answer. x 2 = 169 x = ± 13 The solutions are 13 and –13. Solve for x by taking the square root of both sides. Use ± to show both square roots. Substitute 13 and –13 into the original equation. x 2 = 169 (–13) 2 169 169 169 Check x 2 = 169 (13) 2 169 169 169

9. Holt Algebra 1 9-7 Solving Quadratic Equations by Using Square Roots Example 1B: Using Square Roots to Solve x 2 = a Solve using square roots. x 2 = –49 There is no real number whose square is negative. There is no real solution.

10. Holt Algebra 1 9-7 Solving Quadratic Equations by Using Square Roots Check It Out! Example 1a Solve using square roots. Check your answer. x 2 = 121 x = ± 11 The solutions are 11 and –11. Solve for x by taking the square root of both sides. Use ± to show both square roots. Substitute 11 and –11 into the original equation. x 2 = 121 (–11) 2 121 121 121 Check x 2 = 121 (11) 2 121 121 121

11. Holt Algebra 1 9-7 Solving Quadratic Equations by Using Square Roots Check It Out! Example 1b Solve using square roots. Check your answer. x 2 = 0 x = 0 The solution is 0. Solve for x by taking the square root of both sides. Use ± to show both square roots. Substitute 0 into the original equation. Check x 2 = 0 (0) 2 0 0 0

12. Holt Algebra 1 9-7 Solving Quadratic Equations by Using Square Roots x 2 = –16 There is no real number whose square is negative. There is no real solution. Check It Out! Example 1c Solve using square roots. Check your answer.

13. Holt Algebra 1 9-7 Solving Quadratic Equations by Using Square Roots If a quadratic equation is not written in the form x 2 = a, use inverse operations to isolate x 2 before taking the square root of both sides.

14. Holt Algebra 1 9-7 Solving Quadratic Equations by Using Square Roots Example 2A: Using Square Roots to Solve Quadratic Equations Solve using square roots. x 2 + 7 = 7 –7 x 2 + 7 = 7 x 2 = 0 The solution is 0. Subtract 7 from both sides. Take the square root of both sides.

15. Holt Algebra 1 9-7 Solving Quadratic Equations by Using Square Roots Example 2B: Using Square Roots to Solve Quadratic Equations Solve using square roots. 16x 2 – 49 = 0 +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.

16. Holt Algebra 1 9-7 Solving Quadratic Equations by Using Square Roots Example 2B Continued Solve using square roots. Check your answer. Check 16x 2 – 49 = 0 49 – 49 0 16x 2 – 49 = 0 49 – 49 0.

17. Holt Algebra 1 9-7 Solving Quadratic Equations by Using Square Roots Check It Out! Example 2a Solve by using square roots. Check your answer. 100x 2 + 49 = 0 –49 100x 2 =–49 There is no real solution. There is no real number whose square is negative. Subtract 49 from both sides. Divide by 100 on both sides.

18. Holt Algebra 1 9-7 Solving Quadratic Equations by Using Square Roots Check It Out! Example 2b Solve by using square roots. Check your answer. 36x 2 = 1 Divide by 36 on both sides. Take the square root of both sides. Use ± to show both square roots..

19. Holt Algebra 1 9-7 Solving Quadratic Equations by Using Square Roots Check It Out! Example 2b Continued Solve by using square roots. Check your answer. Check 36x 2 = 136x 2 = 1 1 1

20. Holt Algebra 1 9-7 Solving Quadratic Equations by Using Square Roots 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.

21. Holt Algebra 1 9-7 Solving Quadratic Equations by Using Square Roots Example 3A: Approximating Solutions Solve. Round to the nearest hundredth. x 2 = 15 Take the square root of both sides. Evaluate on a calculator. The approximate solutions are 3.87 and –3.87. x  3.87

22. Holt Algebra 1 9-7 Solving Quadratic Equations by Using Square Roots Example 3B: Approximating Solutions Solve. Round to the nearest hundredth. –3x 2 + 90 = 0 –90 x 2 = 30 The approximate solutions are 5.48 and –5.48. Subtract 90 from both sides. Divide by – 3 on both sides. Take the square root of both sides. Evaluate on a calculator. x  5.48

23. Holt Algebra 1 9-7 Solving Quadratic Equations by Using Square Roots Example 3B Continued Solve. Round to the nearest hundredth. –3x 2 + 90 = 0 The approximate solutions are 5.48 and –5.48. Check Use a graphing calculator to support your answer. Use the zero function. The approximate solutions are 5.48 and – 5.48.

24. Holt Algebra 1 9-7 Solving Quadratic Equations by Using Square Roots Check It Out! Example 3a Solve. Round to the nearest hundredth. 0 = 90 – x 2 + x 2 0 = 90 – x 2 x 2 = 90 Add x 2 to both sides. Take the square root of both sides. The approximate solutions are 9.49 and –9.49.

25. Holt Algebra 1 9-7 Solving Quadratic Equations by Using Square Roots Check It Out! Example 3b Solve. Round to the nearest hundredth. 2x 2 – 64 = 0 + 64 x 2 = 32 The approximate solutions are 5.66 and –5.66. Add 64 to both sides. Divide by 2 on both sides. Take the square root of both sides.

26. Holt Algebra 1 9-7 Solving Quadratic Equations by Using Square Roots Check It Out! Example 3c Solve. Round to the nearest hundredth. x 2 + 45 = 0 – 45 x 2 = –45 There is no real number whose square is negative. There is no real solution. Subtract 45 from both sides.

27. Holt Algebra 1 9-7 Solving Quadratic Equations by Using Square Roots 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. Substitute x for w, 2x for l, and 578 for A. 2x x = 578 ● l = 2w 2x 2 = 578 Length is twice the width.

28. Holt Algebra 1 9-7 Solving Quadratic Equations by Using Square Roots Example 4 Continued 2x 2 = 578 x = ± 17 Take the square root of both sides. Evaluate on a calculator. Negative numbers are not reasonable for width, so x = 17 is the only solution that makes sense. Therefore, the length is 2w or 34 feet. Divide both sides by 2.

29. Holt Algebra 1 9-7 Solving Quadratic Equations by Using Square Roots 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 ) 2x2x 2x2x x Use the formula for area of a trapezoid.

30. Holt Algebra 1 9-7 Solving Quadratic Equations by Using Square Roots Check It Out! Example 4 Substitute 2x for h and b 1, x for b 2, and 6000 for A. Divide by 3 on both sides. Take the square root of both sides. Evaluate on a calculator. 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.