Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 7 - Cumulative Review Exercises - Page 880: 19


The solution of the equation is $ x=625$.

Work Step by Step

Let us consider the provided equation: ${{x}^{\frac{1}{2}}}-2{{x}^{\frac{1}{4}}}-15=0$. Assume ${{x}^{\frac{1}{4}}}=t $ and put it into the equation: $\begin{align} & {{x}^{\frac{1}{2}}}-2{{x}^{\frac{1}{4}}}-15=0 \\ & {{\left( {{x}^{\frac{1}{4}}} \right)}^{2}}-2{{x}^{\frac{1}{4}}}-15=0 \\ & {{t}^{2}}-2t-15=0 \\ & \left( t-5 \right)\left( t+3 \right)=0 \end{align}$ Further solve the equation as given below: $\begin{align} & t=5,-3 \\ & {{x}^{\frac{1}{4}}}=5,-3 \end{align}$ When $ t=5$, the value of $ x $ is $\begin{align} & {{x}^{\frac{1}{4}}}=5 \\ & x=625 \end{align}$ When $ t=-3$, the value of $ x $ is $\begin{align} & {{x}^{\frac{1}{4}}}=-3 \\ & x=81 \end{align}$ But $ x=81$, does not satisfy the expression ${{x}^{\frac{1}{2}}}-2{{x}^{\frac{1}{4}}}-15=0$. Thus, the solution of the equation is $ x=625$.
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