Finite Math and Applied Calculus (6th Edition)

Published by Brooks Cole
ISBN 10: 1133607705
ISBN 13: 978-1-13360-770-0

Chapter 0 - Section 0.6 - Solving Miscellaneous Equations - Exercises - Page 34: 17

Answer

$$x = \pm \root 4 \of 2 $$

Work Step by Step

$$\eqalign{ & \frac{{2\left( {{x^2} - 1} \right)\sqrt {{x^2} + 1} - \frac{{{x^4}}}{{\sqrt {{x^2} + 1} }}}}{{{x^2} + 1}} = 0 \cr & {\text{Multiply the numerator and denominator by }}\sqrt {{x^2} + 1} \cr & \frac{{2\left( {{x^2} - 1} \right){{\left( {\sqrt {{x^2} + 1} } \right)}^2} - \frac{{{x^4}\sqrt {{x^2} + 1} }}{{\sqrt {{x^2} + 1} }}}}{{\left( {{x^2} + 1} \right)\sqrt {{x^2} + 1} }} = 0 \cr & {\text{Simplify}} \cr & \frac{{2\left( {{x^2} - 1} \right)\left( {{x^2} + 1} \right) - {x^4}}}{{\left( {{x^2} + 1} \right)\sqrt {{x^2} + 1} }} = 0 \cr & \frac{{2\left( {{x^4} - 1} \right) - {x^4}}}{{\left( {{x^2} + 1} \right)\sqrt {{x^2} + 1} }} = 0 \cr & \frac{{2{x^4} - 2 - {x^4}}}{{\left( {{x^2} + 1} \right)\sqrt {{x^2} + 1} }} = 0 \cr & \frac{{{x^4} - 2}}{{\left( {{x^2} + 1} \right)\sqrt {{x^2} + 1} }} = 0 \cr & {\text{Cross product}} \cr & {x^4} - 2 = 0 \cr & x = \pm \root 4 \of 2 \cr} $$
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