College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 1 - Equations and Inequalities - Exercise Set 1.6 - Page 179: 52


The solutions are $x=32$ and $x=-243$

Work Step by Step

$x^{2/5}+x^{1/5}-6=0$ Let $u$ be equal to $x^{1/5}$ If $u=x^{1/5}$, then $u^{2}=x^{2/5}$ Rewrite the original equation using the new variable $u$: $u^{2}+u-6=0$ Solve by factoring: $(u+3)(u-2)=0$ Set both factors equal to $0$ and solve each individual equation for $u$: $u+3=0$ $u=-3$ $u-2=0$ $u=2$ Substitute $u$ back to $x^{1/5}$ and solve for $x$: $x^{1/5}=-3$ $(x^{1/5})^{5}=(-3)^{5}$ $x=-243$ $x^{1/5}=2$ $(x^{1/5})^{5}=2^{5}$ $x=32$ The solutions are $x=32$ and $x=-243$
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