College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 1, Equations and Graphs - Chapter 1 Review - Exercises - Page 170: 59


$x=\pm 3$

Work Step by Step

We solve by factoring: $x^{4}-8x^{2}-9=0$ $(x^{2}-9)(x^{2}+1)=0$ $(x-3)(x+3)(x^{2}+1)=0$ $x-3=0$ or $x+3=0$ or $x^2+1=0$ $x=3$ or $x=-3$ The last equation does not have a solution, since $x^2$ can not equal $-1$. Thus the solutions are $x=\pm 3$.
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