Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 8 - Radical Functions - Chapter Test - Page 671: 9

Answer

$\frac{2\sqrt {5ab}}{5a}$

Work Step by Step

First, we need to cancel out common terms in the numerator and denominator: $\sqrt {\frac{4b}{5a}}$ Separate the radical: $\frac{\sqrt {4b}}{\sqrt {5a}}$ We don't want to leave radicals in the denominator, so to get rid of radicals in the denominator, we multiply both the numerator and denominator by the denominator: $\frac{\sqrt {4b}}{\sqrt {5a}} • \frac{\sqrt {5a}}{\sqrt {5a}}$ Multiply to simplify: $\frac{\sqrt {4b • 5a}}{\sqrt {5a • 5a}}$ Perform the multiplication: $\frac{\sqrt {20ab}}{\sqrt {25a^2}}$ Rewrite radicands to separate out perfect squares: $\frac{\sqrt {4 • 5 • ab}}{\sqrt {25 • a^2}}$ Take the square root of any perfect squares: $\frac{2\sqrt {5ab}}{5a}$
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