Finite Math and Applied Calculus (6th Edition)

Published by Brooks Cole
ISBN 10: 1133607705
ISBN 13: 978-1-13360-770-0

Chapter 0 - Section 0.3 - Multiplying and Factoring Algebraic Expressions - Exercises - Page 21: 47


$x=\pm 2$ $or$ $x=\pm1$

Work Step by Step

$x^{4}-5x^{2}+4$ $Let$ $ x^{2}=y$ Then $x^{4}-5x^{2}+4=y^{2}-5y+4=y^{2}-4y-y+4=y(y-4)-1(y-4)=(y-4)(y-1)=(x^{2}-4)(x^{2}-1)$ $(x^{2}-4)(x^{2}-1)=0$ $x^{2}=4 $ $ or $ $x^{2}=1$ $x=\pm 2$ $or$ $x=\pm1$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.