Elementary Algebra

Published by Cengage Learning
ISBN 10: 1285194055
ISBN 13: 978-1-28519-405-9

Chapter 9 - Roots and Radicals - 9.3 - More on Simplifying Radicals - Problem Set 9.3: 36

Answer

$\dfrac{\sqrt{10}}{14}$

Work Step by Step

Recall, we are not allowed to have radicals in the denominator of a fraction. Using the properties of radicals, the given expression, $ \dfrac{2\sqrt{5}}{7\sqrt{8}} ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{2\sqrt{5}}{7\sqrt{8}}\cdot\dfrac{\sqrt{2}}{\sqrt{2}} \\\\= \dfrac{2\sqrt{10}}{7\sqrt{16}} \\\\= \dfrac{2\sqrt{10}}{7\sqrt{(4)^2}} \\\\= \dfrac{2\sqrt{10}}{7\cdot4} \\\\= \dfrac{\cancel{2}\sqrt{10}}{7\cdot\cancel{2}(2)} \\\\= \dfrac{\sqrt{10}}{14} .\end{array}
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