Explanation Definition used: General quadratic equation : “Given that a , b , and c are constant ( a ≠ 0 ) , the equation a x 2 + b x 2 + c = 0 is called the general quadratic equation in x ”. Calculation: Consider the equation ( x 2 − 1 ) 2 + ( x 2 − 1 ) − 2 = 2 . In order to find the value of x , first rewrite the equation in quadratic form. Substitute y = ( x 2 − 1 ) − 2 in ( x 2 − 1 ) 2 + ( x 2 − 1 ) − 2 = 2 and simplify the equation as follows. 1 y + y − 2 = 0 y 2 − 2 y + 1 = 0 ( y − 1 ) 2 = 0 y − 1 = 0 y = 1 On further simplification, 1 ( x 2 − 1 ) 2 = 1 [ ∵ y = ( x 2 − 1 ) − 2 ] 1 = ( x 2 − 1 ) 2 x 4 − 2 x 2 + 1 = 1 x 4 − 2 x 2 = 0 On further simplification, x 2 ( x 2 − 2 ) = 0 x 2 = 0 or x 2 − 2 = 0 x = 0 or x 2 = 2 x = 0 or x = ± 2 Thus, the solutions of the equation are x = 0 , x = 2 and x = − 2 . Check the answer as follows. Substitute x = 0 in ( x 2 − 1 ) 2 + ( x 2 − 1 ) − 2 = 2

Student Solutions Manual For Basic Technical Mathematics And Basic Technical Mathematics With Calculus

11th Edition

ISBN: 9780134434636

Author: Allyn J. Washington, Richard Evans

Publisher: PEARSON

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Chapter 14.3, Problem 20E

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To solve: The equation (x2−1)2+(x2−1)−2=2 algebraically.

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Chapter 14.3, Problem 19E

Chapter 14.3, Problem 21E

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Chapter 14 Solutions

Student Solutions Manual For Basic Technical Mathematics And Basic Technical Mathematics With Calculus

Ch. 14.1 - In Example 7, determine how many times the rocket...Ch. 14.1 - Prob. 1E Ch. 14.1 - Prob. 2E Ch. 14.1 - Prob. 3E Ch. 14.1 - Prob. 4E Ch. 14.1 - Prob. 5E Ch. 14.1 - Prob. 6E Ch. 14.1 - Prob. 7E Ch. 14.1 - Prob. 8E Ch. 14.1 - Prob. 9E

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The equation ( x 2 − 1 ) 2 + ( x 2 − 1 ) − 2 = 2 algebraically.

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Chapter 14 Solutions