# Holt McDougal Algebra 1 8-7 Solving Quadratic Equations by Using Square Roots 8-7 Solving Quadratic Equations by Using Square Roots Holt Algebra 1 Warm.

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Holt McDougal Algebra 1 8-7 Solving Quadratic Equations by Using Square Roots 8-7 Solving Quadratic Equations by Using Square Roots Holt Algebra 1 Warm Up Warm Up Lesson Presentation Lesson Presentation Lesson Quiz Lesson Quiz Holt McDougal Algebra 1

8-7 Solving Quadratic Equations by Using Square Roots Solve quadratic equations by using square roots. Objective

Holt McDougal Algebra 1 8-7 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

Holt McDougal Algebra 1 8-7 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.

Holt McDougal Algebra 1 8-7 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

Holt McDougal Algebra 1 8-7 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.

Holt McDougal Algebra 1 8-7 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 both sides by 16. Take the square root of both sides. Use ± to show both square roots.

Holt McDougal Algebra 1 8-7 Solving Quadratic Equations by Using Square Roots Check It Out! Example 2b Solve by using square roots. Check your answer. (x – 5) 2 = 16 Take the square root of both sides. Use ± to show both square roots. (x – 5) 2 = 16 x – 5 = ±4 Write two equations, using both the positive and negative square roots, and solve each equation. x – 5 = 4 or x – 5 = –4 + 5 + 5 x = 9 or x = 1 The solutions are 9 and 1.

Holt McDougal Algebra 1 8-7 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

Holt McDougal Algebra 1 8-7 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

Holt McDougal Algebra 1 8-7 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.

Holt McDougal Algebra 1 8-7 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.

Holt McDougal Algebra 1 8-7 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.

Holt McDougal Algebra 1 8-7 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.

Holt McDougal Algebra 1 8-7 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.

Holt McDougal Algebra 1 8-7 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.

Holt McDougal Algebra 1 8-7 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.

Holt McDougal Algebra 1 8-7 Solving Quadratic Equations 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

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