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**2.13 Warm Up x² - 2x + 15 = 0; 3 x² + 3x – 4 = 0; 1**

Determine whether the given value is a solution of the equation. x² - 2x + 15 = 0; 3 x² + 3x – 4 = 0; 1 x² - 4x – 12 = 0; 2

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**2.13 Using Square Roots to Solve Quadratic Equations**

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**All positive real numbers have 2 square roots**

Positive square root (principle square root) Negative square root Radicand: number or expression inside the radical symbol Perfect square: squared integer If there is no b, you can isolate the x2 and then solve using square roots.

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**Solve quadratic equations**

EXAMPLE 1 Solve quadratic equations Solve the equation. a. 2x2 = 8 SOLUTION a. 2x2 = 8 Write original equation. x2 = 4 Divide each side by 2. x = ± 4 = ± 2 Take square roots of each side. Simplify. The solutions are –2 and 2. ANSWER

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**Solve quadratic equations**

EXAMPLE 1 Solve quadratic equations b. m2 – 18 = – 18 Write original equation. m2 = 0 Add 18 to each side. m = 0 The square root of 0 is 0. ANSWER The solution is 0.

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**Solve quadratic equations**

EXAMPLE 1 Solve quadratic equations c. b = 5 Write original equation. b2 = – 7 Subtract 12 from each side. ANSWER Negative real numbers do not have real square roots. So, there is no real solution.

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** EXAMPLE 2 Take square roots of a fraction Solve 4z2 = 9. SOLUTION**

Write original equation. z2 = 9 4 Divide each side by 4. z = ± 9 4 Take square roots of each side. z = ± 3 2 Simplify. The solutions are – and 3 2

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**Approximate solutions of a quadratic equation**

EXAMPLE 3 Approximate solutions of a quadratic equation Solve 3x2 – 11 = 7. Round the solutions to the nearest hundredth. SOLUTION 3x2 – 11 = 7 Write original equation. 3x2 = 18 Add 11 to each side. x2 = 6 Divide each side by 3. x = ± 6 Take square roots of each side. The solutions are about – 2.45 and about 2.45.

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EXAMPLE 1 GUIDED PRACTICE Solve quadratic equations for Examples 1,2 and 3 Solve the equation. 1. c2 – 25 = 0 ANSWER –5, 5. w = – 8 ANSWER no solution x = 11 ANSWER

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EXAMPLE 1 GUIDED PRACTICE Solve quadratic equations for Examples 1,2 and 3 Solve the equation. ANSWER 4 5 – , x2 = 16 ANSWER 10 3 – , m2 = 100 b = 0 ANSWER no solution Solve the equation. Round the solutions to the nearest hundredth. ANSWER – 3.16, 3.16 7. x2 + 4 = 14 k2 – 1 = 0 ANSWER – 0.58, 0.58 p2 – 7 = 2 ANSWER – 2.12, 2.12

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**Solve a quadratic equation**

EXAMPLE 4 Solve a quadratic equation Solve 6(x – 4)2 = 42. Round the solutions to the nearest hundredth. 6(x – 4)2 = 42 Write original equation. (x – 4)2 = 7 Divide each side by 6. x – 4 = ± 7 Take square roots of each side. 7 x = 4 ± Add 4 to each side. ANSWER The solutions are and 4 – 7

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EXAMPLE 1 GUIDED PRACTICE Solve quadratic equations for Examples 4 and 5 Solve the equation. Round the solution to the nearest hundredth if necessary. (x – 2)2 = 18 ANSWER –1, 5 (q – 3)2 = 28 ANSWER 0.35, 5.65 (t + 5)2 = 24 ANSWER –7.83, –2.17

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