Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 11 - Quadratic Functions and Equations - 11.5 Equations Reducible to Quadratic - 11.5 Exercise Set - Page 732: 34


$w=64$ or $w=-8$

Work Step by Step

$ w^{\frac{2}{3}}-2w^{\frac{1}{3}}-8=0\qquad$...substitute $w^{\frac{1}{3}}$ for $u$ so that $u^{2}=w^{\frac{2}{3}}$ $ u^{2}-2u-8=0\qquad$... solve with the Quadratic formula. $u=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ $u=\displaystyle \frac{2\pm\sqrt{4+32}}{2}$ $u=\displaystyle \frac{2\pm\sqrt{36}}{2}$ $u=\displaystyle \frac{2\pm 6}{2}$ $u=\displaystyle \frac{2+6}{2}=\frac{8}{2}=4$ or $u=\displaystyle \frac{2-6}{2}=\frac{-4}{2}=-2$ Bring back $w^{\frac{1}{3}}=u$. $w^{\frac{1}{3}}=4$ or $ w^{\frac{1}{3}}=-2\qquad$...raise both sides of both expressions to the third power $(w^{\frac{1}{3}})^{3}=4^{3}$ or $(w^{\frac{1}{3}})^{3}=(-2)^{3} $ $w=64$ or $w=-8$
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