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: 10



Work Step by Step

Since, $u^2-6u+8=0$ Factorize the expression as follows: $(u-4)(u-2)=0$ or, $u=${$2,4$} Replace $u$ with $\sqrt x$, we have $\sqrt x=2$ or, $x=4$ or, $\sqrt x=4 \implies x=4^2$ and $x=16$ Hence, our solution set is $x=${$4,16$}
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