Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 7 - Section 7.6 - Radical Equations - Exercise Set - Page 562: 84

Answer

$x= 129$

Work Step by Step

The solution is straightforward: \begin{equation} \begin{aligned} (x-4)^{\frac{2}{3}}&=25 \\ \left[ (x-4)^{\frac{2}{3}} \right]^{\frac{3}{2}}&= 25^{\frac{3}{2}}\\ x-4&= \sqrt{25^3}\\ x&= 4+\sqrt{25^3}\\ x&= 4+125\\ x&= 129\\ \end{aligned} \end{equation} Check: \begin{equation} \begin{aligned} (129-4)^{\frac{2}{3}}&\stackrel{?}{=}25\\ (125)^{\frac{2}{3}}&\stackrel{?}{=}25\\ \sqrt[3]{125^2}&\stackrel{?}{=}25\\ 25&= 25\checkmark\\ \end{aligned} \end{equation} The solution is $x= 129$.
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