Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.5 - Equations - 1.5 Exercises - Page 57: 104

Answer

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

Work Step by Step

$x^{4}-5x^{2}+4=0$ To solve this equation, let $x^{2}$ be equal to $t$. $t=x^{2}$ so $t^{2}=x^{4}$ Substitute $x^{4}$ by $t^{2}$ and $x^{2}$ by $t$ in the original equation: $t^{2}-5t+4=0$ Solve this equation by factoring: $(t-1)(t-4)=0$ Set both factors equal to $0$ and solve each individual equation for $t$: $t-1=0$ $t=1$ $t-4=0$ $t=4$ Substitute $t$ back to $x^{2}$ and solve for $x$: $t=1$ $x^{2}=1$ $x=\sqrt{1}$ $x=\pm1$ $t=4$ $x^{2}=4$ $x=\sqrt{4}$ $x=\pm2$ The solutions are $x=\pm1$ and $x=\pm2$
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