Algebra 2 Common Core

Published by Prentice Hall
ISBN 10: 0133186024
ISBN 13: 978-0-13318-602-4

Chapter 6 - Radical Functions and Rational Exponents - 6-2 Multiplying and Dividing Radical Expressions - Practice and Problem-Solving Exercises - Page 372: 64

Answer

$\dfrac{5\sqrt{14x}}{21x}$

Work Step by Step

Multiplying the numerator and the denominator by an expression equal to $1$ which will make the denominator a perfect power of the index, the given expression is equivalent to \begin{align*}\require{cancel} & =\dfrac{5\sqrt{2}}{3\sqrt{7x}}\cdot\dfrac{\sqrt{7x}}{\sqrt{7x}} \\\\&= \dfrac{5\sqrt{14x}}{3\sqrt{7^2x^2}} \\\\&= \dfrac{5\sqrt{14x}}{3\sqrt{(7x)^2}} \\\\&= \dfrac{5\sqrt{14x}}{3(7x)} \\\\&= \dfrac{5\sqrt{14x}}{21x} .\end{align*} Hence, the simplified form of the given expression is $ \dfrac{5\sqrt{14x}}{21x} $.
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