Explanation Given information: Given, equation is ( x x + 1 ) 2 − 2 x x + 1 − 8 = 0 . Calculation: The equation can be written as: ( x x + 1 ) 2 − 2 ( x x + 1 ) − 8 = 0 Substitute x x + 1 = y in the above equation. y 2 − 2 y − 8 = 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 = − 2 , c = − 8 . Plugging the values in the above equation, we get, x = − b ± b 2 − 4 a c 2 a y = − ( − 2 ) ± ( − 2 ) 2 − 4 ( 1 ) ( − 8 ) 2 ( 1 ) y = 2 ± 4 + 32 2 y = 2 ± 36 2 y = 2 ± 6 2 y = 2 + 6 2 or y = 2 − 6 2 y = 8 2 or y = − 4 2 y = 4 or y = − 2 Substituting back the value of y to obtain the value of x. y = 4 or y = − 2 x x + 1 = 4 or x x + 1 = − 2 x = 4 ( x + 1 ) or x = − 2 ( x + 1 ) x = 4 x + 4 or x = − 2 x − 2 3 x = − 4 or 3 x = − 2 x = − 4 3 or x = − 2 3 Verifying the solutions: When x = − 4 3 ( x x + 1 ) 2 − 2 x x + 1 − 8 = 0 ( − 4 3 − 4 3 +

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 45E

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To find the solution set of the given equation (xx+1)2−2xx+1−8=0.

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Chapter A.6, Problem 44E

Chapter A.6, Problem 46E

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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 x + 1 ) 2 − 2 x x + 1 − 8 = 0 .

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To find the solution set of the given equation (xx+1)2−2xx+1−8=0.

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Chapter A.6 Solutions