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: 42

Answer

$\dfrac{2\sqrt{a}}{3ab}$

Work Step by Step

The variables are assumed to be non-zero as they are also in the denominator. RECALL: $\dfrac{\sqrt{a}}{\sqrt{b}} = \sqrt{\dfrac{a}{b}}, a\ge0, b\gt0.$ Use the rule above to have $\require{cancel} \sqrt{\dfrac{20ab}{45a^2b^3}} \\=\sqrt{\dfrac{\cancel{20}4a\cancel{b}}{\cancel{45}9a^2\cancel{b^3}b^2}} \\=\sqrt{\dfrac{4a}{9a^2b^2}} \\=\sqrt{\dfrac{2^2(a)}{(3ab)^2}} \\=\dfrac{2}{3ab} \cdot \sqrt{a} \\=\dfrac{2\sqrt{a}}{3ab}.$
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