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Square Root Equations and California Standards Lesson

This lesson presentation focuses on solving square root equations and aligning with California standards. Students will learn how to find square roots and solve equations involving square roots. The lesson also covers quadratic equations in physical problems and the concept of positive and negative square roots.

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Square Root Equations and California Standards Lesson

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  1. Preview Warm Up California Standards Lesson Presentation

  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

  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 Additional 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 into the original equation.

  8. x2 = 169 (–13)2 169  169 169 Additional Example 1A Continued Solve using square roots. Check your answer. Check Substitute –13 into the original equation.

  9. Additional 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. The solution set is the empty set, ø.

  10. Check x2 = 121 (11)2 121  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 into the original equation.

  11. x2 = 121 (–11)2 121  121 121 Check It Out! Example 1a Continued Solve using square roots. Check your answer. Check Substitute –11 into the original equation.

  12. Checkx2 = 0 (0)2 0  0 0 Check It Out! Example 1b Solve using square roots. Check your answer. x2 = 0 Solve for x by taking the square root of both sides. Use ± to show both square roots. x = 0 The solution is 0. Substitute 0 into the original equation.

  13. 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. The solution set is the empty set, ø.

  14. 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.

  15. x2 + 7 = 7 –7 –7 x2 = 0 Additional 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.

  16. Additional 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.

  17. Additional Example 2B Continued Solve using square roots. Check 16x2 – 49 = 0 16x2 – 49 = 0   0 0 0 0

  18. 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. ø

  19. 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.

  20.  1 1 1 1 Check It Out! Example 2b Continued Solve by using square roots. Check your answer. Check36x2 = 1 36x2 = 1

  21. When solving quadratic equations by using square roots, the solutions may be irrational. In this case, you can give the exact solutions by leaving the square root in your answer, or you can approximate the solutions.

  22. Estimate . The exact solutions are and . The approximate solutions are 3.87 and –3.87. Additional Example 3A: Approximating Solutions Solve. Round to the nearest hundredth. x2 = 15 Take the square root of both sides. x 3.87

  23. –3x2 + 90 = 0 –90 –90 Estimate . The exact solutions are and . The approximate solutions are 5.48 and –5.48. Additional 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

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

  25. Remember! To review estimating square roots, see Lesson 1-5.

  26. 0 = 90 – x2 + x2 + x2 x2 = 90 Estimate The exact solutions are and The approximate solutions are 9.49 and –9.49. . 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.

  27. Check It Out! Example 3a Continued Solve. Round to the nearest hundredth. 0 = 90 – x2 CheckUse a graphing calculator to support your answer. Use the zero function. The approximate solutions are 9.49 and –9.49.

  28. 2x2 – 64 = 0 + 64 + 64 Estimate The exact solutions are and . The approximate solutions are 5.66 and –5.66. 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.

  29. Check It Out! Example 3b Continued Solve. Round to the nearest hundredth. 2x2 – 64 = 0 CheckUse a graphing calculator to support your answer. Use the zero function. The approximate solutions are 5.66 and –5.66.

  30. – 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. ø

  31. 2x x=578 ● Additional 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

  32. Additional 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.

  33. 2x 2x x Check It Out! Example 4 A lot is shaped like a trapezoid with bases x and 2x. Its area is 6000 ft2. Find x. Round to the nearest foot. (Hint: Use ) Use the formula for area of a trapezoid. Substitute 2x for h and b1, x for b2 , and 6000 for A.

  34. Estimate . Check It Out! Example 4 Continued 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, x is approximately 45 feet.

  35. 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

  36. (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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