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: 41

Answer

The solutions are $x=\pm1$ and $x=\pm2$

Work Step by Step

$x^{4}-5x^{2}+4=0$ Let $u$ be equal to $x^{2}$ If $u=x^{2}$, then $u^{2}=x^{4}$ Rewrite the original equation using the new variable $u$: $u^{2}-5u+4=0$ Solve by factoring: $(u-4)(u-1)=0$ Set both factors equal to $0$ and solve each individual equation for $u$: $u-4=0$ $u=4$ $u-1=0$ $u=1$ Substitute $u$ back to $x^{2}$ and solve for $x$: $x^{2}=4$ $\sqrt{x^{2}}=\pm\sqrt{4}$ $x=\pm2$ $x^{2}=1$ $\sqrt{x^{2}}=\pm\sqrt{1}$ $x=\pm1$ The solutions are $x=\pm1$ and $x=\pm2$
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