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: 33


$t=8$ or $t=-27$

Work Step by Step

$ t^{\frac{2}{3}}+t^{\frac{1}{3}}-6=0\qquad$...substitute $t^{\frac{1}{3}}$ for $u$ so that $u^{2}=t^{\frac{2}{3}}$ $ u^{2}+u-6=0\qquad$... solve with the Quadratic formula. $u=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ $u=\displaystyle \frac{-1\pm\sqrt{1+24}}{2}$ $u=\displaystyle \frac{-1\pm\sqrt{25}}{2}$ $u=\displaystyle \frac{-1\pm 5}{2}$ $u=\displaystyle \frac{-1+5}{2}=\frac{4}{2}=2$ or $u=\displaystyle \frac{-1-5}{2}=\frac{-6}{2}=-3$ Bring back $t^{\frac{1}{3}}=u$. $t^{\frac{1}{3}}=2$ or $ t^{\frac{1}{3}}=-3\qquad$...raise both sides of both expressions to the third power $(t^{\frac{1}{3}})^{3}=2^{3}$ or $(t^{\frac{1}{3}})^{3}=(-3)^{3} $ $t=8$ or $t=-27$
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