Intermediate Algebra (6th Edition)

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

Chapter 7 - Review: 114



Work Step by Step

Multiplying both the numerator and the denominator by a factor that will make the numerator a perfect power of the radical, the rationalized-numerator form of the given expression, $ \sqrt[]{\dfrac{24x^5}{3y^2}} ,$ is \begin{array}{l}\require{cancel} \sqrt[]{\dfrac{24x^5}{3y^2}} \\\\= \sqrt[]{\dfrac{\cancel{3}\cdot8x^5}{\cancel{3}y^2}} \\\\= \sqrt[]{\dfrac{8x^5}{y^2}} \\\\= \sqrt[]{\dfrac{8x^5}{y^2}\cdot\dfrac{2x}{2x}} \\\\= \sqrt[]{\dfrac{16x^6}{y^2}\cdot\dfrac{1}{2x}} \\\\= \sqrt[]{\left( \dfrac{4x^3}{y}\right)^2\cdot\dfrac{1}{2x}} \\\\= \dfrac{4x^3}{y}\sqrt[]{\dfrac{1}{2x}} \\\\= \dfrac{4x^3}{y\sqrt[]{2x}} .\end{array} Note that all variables represent positive real numbers.
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