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: 47


The solutions are $x=\dfrac{1}{5}$ and $x=-\dfrac{1}{4}$

Work Step by Step

$x^{-2}-x^{-1}-20=0$ Let $u$ be equal to $x^{-1}$ If $u=x^{-1}$, then $u^{2}=x^{-2}$ Rewrite the original equation using the new variable $u$: $u^{2}-u-20=0$ Solve by factoring: $(u-5)(u+4)=0$ Set both factors equal to $0$ and solve each individual equation for $u$: $u-5=0$ $u=5$ $u+4=0$ $u=-4$ Substitute $u$ back to $x^{-1}$ and solve for $x$: $x^{-1}=5$ $\dfrac{1}{x}=5$ $x=\dfrac{1}{5}$ $x^{-1}=-4$ $\dfrac{1}{x}=-4$ $x=-\dfrac{1}{4}$ The solutions are $x=\dfrac{1}{5}$ and $x=-\dfrac{1}{4}$
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