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: 13



Work Step by Step

Since, $u^2-4u-21=0$ Factorize the expression as follows: $(u-7)(u+3)=0$ or, $u=${$-3,7$} Replace $u$ with $x-5$, we have $x-5=-3 \implies x=2$ and $x-5=7$ or, $x=12$ Hence, our solution set is $x=${$2,12$}
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