Explanation Given information: Given, equation is ( x 2 − 4 x ) 2 + 7 ( x 2 − 4 x ) + 12 = 0 . Calculation: Substituting x 2 − 4 x = y y 2 + 7 y + 12 = 0 The solutions to the quadratic equation of the form a x 2 + b x + c = 0 is given by x = − b ± b 2 − 4 a c 2 a . Here x = y , a = 1 , b = 7 , c = 12 . Plugging the values in the above equation, we get, x = − b ± b 2 − 4 a c 2 a y = − ( 7 ) ± ( 7 ) 2 − 4 ( 1 ) ( 12 ) 2 ( 1 ) y = − 7 ± 49 − 48 2 y = − 7 ± 1 2 y = − 7 ± 1 2 y = − 7 + 1 2 or y = − 7 − 1 2 y = − 6 2 or y = − 8 2 y = − 3 or y = − 4 Substitute back the value of y to obtain the value of x. y = − 3 or y = − 4 x 2 − 4 x = − 3 or x 2 − 4 x = − 4 x 2 − 4 x + 3 = 0 or x 2 − 4 x + 4 = 0 When x 2 − 4 x + 3 = 0 Using the quadratic formula, we get, x = − b ± b 2 − 4 a c 2 a x = − ( − 4 ) ± ( − 4 ) 2 − 4 ( 1 ) ( 3 ) 2 ( 1 ) x = 4 ± 16 − 12 2 x = 4 ± 4 2 x = 4 ± 2 2 x = 4 + 2 2 or x = 4 − 2 2 x = 6 2 or x = 2 2 x = 3 or x = 1 When x 2 − 4 x + 4 = 0 Using the quadratic formula, we get, x = − b ± b 2 − 4 a c 2 a x = −

Precalculus: A Unit Circle Approach (3rd Edition)

3rd Edition

ISBN: 9780134433042

Author: J. S. Ratti, Marcus S. McWaters, Leslaw Skrzypek

Publisher: PEARSON

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Chapter A.6, Problem 44E

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To find the solution set of the given equation (x2−4x)2+7(x2−4x)+12=0.

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Chapter A.6, Problem 43E

Chapter A.6, Problem 45E

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Chapter A.6 Solutions

Precalculus: A Unit Circle Approach (3rd Edition)

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To find the solution set of the given equation ( x 2 − 4 x ) 2 + 7 ( x 2 − 4 x ) + 12 = 0 .

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To find the solution set of the given equation (x2−4x)2+7(x2−4x)+12=0.

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Chapter A.6 Solutions