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: 63

Answer

$\color{blue}{\dfrac{\sqrt{6x}}{3x}}$

Work Step by Step

Rationalize the denominator by multiplying $3x$ to both the numerator and the denominator of the radicand to obtain: $=\sqrt{\dfrac{2(3x)}{(3x)(3x)}} \\=\sqrt{\dfrac{6x}{(3x)^2}}$ Bring out the square root of the denominator to obtain: $=\color{blue}{\dfrac{\sqrt{6x}}{3x}}$
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