Solving Quadratic Equations 9 7 by Using Square

Solving Quadratic Equations 9 -7 by Using Square Roots Warm Up Lesson Presentation Lesson Quiz Holt Algebra 11

Solving Quadratic Equations 9 -7 by Using Square Roots Warm Up Find each square root. 1. 3. 6 2. – 25 11 4. Solve each equation. x = 10 5. – 6 x = – 60 6. 7. 2 x – 40 = 0 8. 5 x = 3 x = 20 Holt Algebra 1 x = 80

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

Solving Quadratic Equations 9 -7 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. Holt Algebra 1

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

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

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

Solving Quadratic Equations 9 -7 by Using Square Roots Example 1 A: Using Square Roots to Solve x 2 = a Solve using square roots. Check your answer. x 2 = 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. x 2 = 169 Substitute 13 and – 13 (– 13)2 169 into the original 169 169 equation. Check x 2 = 169 (13)2 169 Holt Algebra 1

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

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

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

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

Solving Quadratic Equations 9 -7 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. Holt Algebra 1

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

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

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

Solving Quadratic Equations 9 -7 by Using Square Roots Check It Out! Example 2 a Solve by using square roots. Check your answer. 100 x 2 + 49 = 0 – 49 100 x 2 =– 49 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. Holt Algebra 1

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

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

Solving Quadratic Equations 9 -7 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. Holt Algebra 1

Solving Quadratic Equations 9 -7 by Using Square Roots Example 3 A: Approximating Solutions Solve. Round to the nearest hundredth. x 2 = 15 Take the square root of both sides. x 3. 87 Evaluate on a calculator. The approximate solutions are 3. 87 and – 3. 87. Holt Algebra 1

Solving Quadratic Equations 9 -7 by Using Square Roots Example 3 B: Approximating Solutions Solve. Round to the nearest hundredth. – 3 x 2 + 90 = 0 – 90 Subtract 90 from both sides. Divide by – 3 on both sides. x 2 = 30 Take the square root of both sides. x 5. 48 Evaluate on a calculator. The approximate solutions are 5. 48 and – 5. 48. Holt Algebra 1

Solving Quadratic Equations 9 -7 by Using Square Roots Example 3 B Continued Solve. Round to the nearest hundredth. – 3 x 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. Holt Algebra 1

Solving Quadratic Equations 9 -7 by Using Square Roots Check It Out! Example 3 a Solve. Round to the nearest hundredth. 0 = 90 – x 2 Add x 2 to both sides. 0 = 90 – x 2 + x 2 = 90 Take the square root of both sides. The approximate solutions are 9. 49 and – 9. 49. Holt Algebra 1

Solving Quadratic Equations 9 -7 by Using Square Roots Check It Out! Example 3 b Solve. Round to the nearest hundredth. 2 x 2 – 64 = 0 + 64 Add 64 to both sides. Divide by 2 on both sides. x 2 = 32 Take the square root of both sides. The approximate solutions are 5. 66 and – 5. 66. Holt Algebra 1

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

Solving Quadratic Equations 9 -7 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 l = 2 w 2 x ● x = 578 2 x 2 = 578 Holt Algebra 1 Use the formula for area of a rectangle. Length is twice the width. Substitute x for w, 2 x for l, and 578 for A.

Solving Quadratic Equations 9 -7 by Using Square Roots Example 4 Continued 2 x 2 = 578 Divide both sides by 2. Take the square root of both sides. x = ± 17 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 2 w or 34 feet. Holt Algebra 1

Solving Quadratic Equations 9 -7 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 ) 2 x 2 x x Use the formula for area of a trapezoid. Holt Algebra 1

Solving Quadratic Equations 9 -7 by Using Square Roots Check It Out! Example 4 Substitute 2 x 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. Holt Algebra 1

Solving Quadratic Equations 9 -7 by Using Square Roots Lesson Quiz: Part 1 Solve using square roots. Check your answers. 1. x 2 – 195 = 1 ± 14 2. 4 x 2 – 18 = – 9 3. 2 x 2 – 10 = – 12 no real solutions 4. Solve 0 = – 5 x 2 + 225. Round to the nearest hundredth. ± 6. 71 Holt Algebra 1

Solving Quadratic Equations 9 -7 by Using Square Roots 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. (Hint: Use 108 feet Holt Algebra 1 )