Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 7 - Review: 103



Work Step by Step

Multiplying both the numerator and the denominator by a factor that will make the denominator a perfect power of the radical, the rationalized-denominator form of the given expression, $ \dfrac{3}{\sqrt{7}} ,$ is \begin{array}{l}\require{cancel} \dfrac{3}{\sqrt{7}}\cdot\dfrac{\sqrt{7}}{\sqrt{7}} \\\\= \dfrac{3\sqrt{7}}{\sqrt{7(7)}} \\\\= \dfrac{3\sqrt{7}}{\sqrt{(7)^2}} \\\\= \dfrac{3\sqrt{7}}{7} .\end{array}
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