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: 37



Work Step by Step

$ t^{\frac{1}{3}}+2t^{\frac{1}{6}}=3\qquad$... add $-3$ to both sides. $ t^{\frac{1}{3}}+2t^{\frac{1}{6}}-3=0\qquad$...substitute $t^{\frac{1}{6}}$ for $u$ so that $u^{2}=t^{\frac{1}{3}}$ Note that $t^{\frac{1}{6}}$ is the sixth root of $t$, a positive number. $ u^{2}+2u-3=0\qquad$... solve with the Quadratic formula. $u=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ $u=\displaystyle \frac{-2\pm\sqrt{4+12}}{2}$ $u=\displaystyle \frac{-2\pm\sqrt{16}}{2}$ $u=\displaystyle \frac{-2\pm 4}{2}$ $u=\displaystyle \frac{-2+4}{2}=\frac{2}{2}=1$ or $u=\displaystyle \frac{-2-4}{2}=\frac{-6}{2}=-3$ Bring back $t^{\frac{1}{6}}=u$. $t^{\frac{1}{6}}=1$ or $t^{\frac{1}{3}}=-3$ We discard $-3$ because the sixth root of $t$ is nonnegative. $t^{\frac{1}{6}}=1 $ $(t^{\frac{1}{6}})^{6}=1^{6}$ $t=1$
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