Intermediate Algebra (6th Edition)

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

Chapter 7 - Sections 7.1-7.5 - Integrated Review - Radicals and Rational Exponents - Page 447: 40

Answer

$\dfrac{x-4}{x+2\sqrt{x}}$

Work Step by Step

Multiplying both the numerator and the denominator by the conjugate of the numerator, then the rationalized-numerator form of the given expression, $ \dfrac{\sqrt{x}-2}{\sqrt{x}} ,$ is \begin{array}{l}\require{cancel} \dfrac{\sqrt{x}-2}{\sqrt{x}}\cdot\dfrac{\sqrt{x}+2}{\sqrt{x}+2} \\\\= \dfrac{(\sqrt{x})^2-(2)^2}{\sqrt{x}(\sqrt{x})+\sqrt{x}(2)} \\\\= \dfrac{x-4}{\sqrt{x(x)}+2\sqrt{x}} \\\\= \dfrac{x-4}{\sqrt{(x)^2}+2\sqrt{x}} \\\\= \dfrac{x-4}{x+2\sqrt{x}} .\end{array}
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