Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter R - Review of Basic Concepts - R.7 Radical Expressions - R.7 Exercises: 69

Answer

$\color{blue}{\dfrac{h\sqrt[4]{9g^3hr^2}}{3r^2}}$

Work Step by Step

Rationalize the denominator by multiplying $9r^2$ to both the numerator and denominator of the radicand to obtain: $=\sqrt[4]{\dfrac{g^3h^5(9r^2)}{9r^6(9r^2)}} \\=\sqrt[4]{\dfrac{9g^3h^5r^2}{81r^8}} \\=\sqrt[4]{\dfrac{9g^3h^5r^2}{3^4r^8}} \\=\sqrt[4]{\dfrac{9g^3h^5r^2}{(3r^2)^4}}$ Bring out the fourth root of the denominator to obtain: $\\=\dfrac{\sqrt[4]{9g^3h^5r^2}}{3r^2}$ Factor the radicand such that at least one factor is a perfect fourth power to obtain: $\\=\dfrac{\sqrt[4]{h^4(9g^3hr^2)}}{3r^2}$ Bring out the fourth root of the perfect fourth power factor/s of the numerator to obtain: $\\=\color{blue}{\dfrac{h\sqrt[4]{9g^3hr^2}}{3r^2}}$
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