Intermediate Algebra (6th Edition)

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

Chapter 7 - Section 7.5 - Rationalizing Denominators and Numerators of Radical Expressions - Exercise Set - Page 445: 77

Answer

$\dfrac{x-9}{x-3\sqrt{x}}$

Work Step by Step

Multiplying by the conjugate of the numerator, then the rationalized-numerator form of the given expression, $ \dfrac{\sqrt{x}+3}{\sqrt{x}} ,$ is \begin{array}{l}\require{cancel} \dfrac{\sqrt{x}+3}{\sqrt{x}}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}-3} \\\\= \dfrac{(\sqrt{x})^2-(3)^2}{\sqrt{x}(\sqrt{x})+\sqrt{x}(-3)} \\\\= \dfrac{x-9}{x-3\sqrt{x}} .\end{array}
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