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Published byEmery Elliott Modified over 9 years ago

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

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Solve quadratic equations EXAMPLE 1 b. m 2 – 18 = – 18 Write original equation. m 2 = 0 Add 18 to each side. The square root of 0 is 0. m = 0m = 0 ANSWER The solution is 0.

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Solve quadratic equations EXAMPLE 1 c. b 2 + 12 = 5 Write original equation. b 2 = – 7 Subtract 12 from each side. ANSWER Negative real numbers do not have real square roots. So, there is no solution.

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EXAMPLE 2 Take square roots of a fraction Solve 4z 2 = 9. SOLUTION 4z 2 = 9 Write original equation. z 2 = 9 4 Divide each side by 4. Take square roots of each side. z = ± 9 4 3 2 Simplify.

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Take square roots of a fraction EXAMPLE 2 ANSWER The solutions are – and 3 2 3 2

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Approximate solutions of a quadratic equation EXAMPLE 3 Solve 3x 2 – 11 = 7. Round the solutions to the nearest hundredth. SOLUTION 3x 2 – 11 = 7 Write original equation. 3x 2 = 18 Add 11 to each side. x 2 = 6 Divide each side by 3. x = ± 6 Take square roots of each side.

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Approximate solutions of a quadratic equation EXAMPLE 3 x ± 2.45 Use a calculator. Round to the nearest hundredth. ANSWER The solutions are about – 2.45 and about 2.45.

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EXAMPLE 1 Solve quadratic equations Solve the equation. 1. c 2 – 25 = 0 GUIDED PRACTICE for Examples 1,2 and 3 ANSWER –5, 5. 2. 5w 2 + 12 = – 8 ANSWER no solution 3. 2x 2 + 11 = 11 ANSWER 0

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EXAMPLE 1 Solve quadratic equations Solve the equation. 4. 25x 2 = 16 GUIDED PRACTICE for Examples 1,2 and 3 ANSWER 4 5 4 5 –, 5. 9m 2 = 100 ANSWER 10 3 –, 3 6. 49b 2 + 64 = 0 ANSWER no solution

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EXAMPLE 1 Solve quadratic equations Solve the equation. Round the solutions to the nearest hundredth. 7. x 2 + 4 = 14 GUIDED PRACTICE ANSWER – 3.16, 3.16 for Examples 1,2 and 3 8. 3k 2 – 1 = 0 ANSWER – 0.58, 0.58 9. 2p 2 – 7 = 2 ANSWER – 2.12, 2.12

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