Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 8 - Section 8.4 - Equations Quadratic in Form - Exercise Set - Page 635: 2


{$\pm 2, \pm 3$}

Work Step by Step

Since, $u^2-13u+36=0$ Factorize the expression as follows: $(u-4)(u-9)=0$ or, $u=${$4,9$} Replace $u$ with $x^2$, we have $x^2=4$ and $x^2=9$ This implies that $x=\pm 2, \pm 3$ Hence, our solution set is $x=${$\pm 2, \pm 3$}
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