Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 7 - Review - Page 469: 108


$\dfrac{3\sqrt[4]{2x^2}}{2x^{3}} $

Work Step by Step

Multiplying both the numerator and the denominator by a factor that will make the denominator a perfect power of the radical, the rationalized-denominator form of the given expression, $ \sqrt[4]{\dfrac{81}{8x^{10}}} ,$ is \begin{array}{l}\require{cancel} \sqrt[4]{\dfrac{81}{8x^{10}}\cdot\dfrac{2x^{2}}{2x^{2}}} \\\\= \sqrt[4]{\dfrac{81}{16x^{12}}\cdot2x^2} \\\\= \sqrt[4]{\left( \dfrac{3}{2x^{3}} \right)^4\cdot2x^2} \\\\= \dfrac{3}{2x^{3}}\sqrt[4]{2x^2} \\\\= \dfrac{3\sqrt[4]{2x^2}}{2x^{3}} .\end{array} Note that all variables are assumed to have positive real numbers.
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