Precalculus (6th Edition) Blitzer

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

Chapter 10 - Cumulative Review Exercises - Page 1127: 13


The value of the expression ${{x}^{\frac{1}{2}}}-6{{x}^{\frac{1}{4}}}+8=0$ is $ x=16$ or $ x=256$.

Work Step by Step

Rewrite the provided equation as given below: $\begin{align} & {{x}^{\frac{1}{2}}}-6{{x}^{\frac{1}{4}}}+8=0 \\ & {{x}^{\frac{1}{2}}}+8=6{{x}^{\frac{1}{4}}} \end{align}$ Then, squaring both sides, and get: $\begin{align} & x+16{{x}^{\frac{1}{2}}}+64=36{{x}^{\frac{1}{2}}} \\ & x+64=20{{x}^{\frac{1}{2}}} \end{align}$ Again, square both sides: $\begin{align} & {{x}^{2}}+128x+{{64}^{2}}=400x \\ & {{x}^{2}}-272x+4096=0 \\ & {{x}^{2}}-256x-16x+4096=0 \\ & x\left( x-256 \right)-16\left( x-256 \right)=0 \end{align}$ Therefore, one has $\left( x-256 \right)\left( x-16 \right)=0$ Thus, the value of x is $ x=16$ or $ x=256$.
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