College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 1, Equations and Graphs - Section 1.6 - Solving Other Types of Equations - 1.6 Exercises - Page 139: 59

Answer

The solutions are $x=\sqrt{8}$ and $x=\sqrt{27}$

Work Step by Step

$x^{4/3}-5x^{2/3}+6=0\hspace{0.7cm} \color{blue}{\text{Given equation}}$ $W^2-5W+6=0\hspace{3.3cm} \color{blue}{\text{Let } W=x^{2/3}}$ $(W-2)+(W-3)=0\hspace{2.3cm} \color{blue}{\text{Factor}}$ $W-2=0\hspace{0.5cm}or\hspace{0.5cm}W-3=0\hspace{1.3cm} \color{blue}{\text{Zero-Product Property}}$ $W=2\hspace{1.5cm}W=3\hspace{1.99cm} \color{blue}{\text{Solve}}$ Now change $W$ back into the correspondimg values of $x$ $x^{2/3}=2\hspace{1.5cm}x^{2/3}=3$ $x=2^{3/2}\hspace{1.5cm}x=3^{3/2}$ $x=\sqrt{8}\hspace{1.5cm}x=\sqrt{27}$ The solutions are $x=\sqrt{8}$ and $x=\sqrt{27}$
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