Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 10 - Section 10.5 - Rationalizing Numerators and Denominators of Radical Expressions - Exercise Set: 69

Answer

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

Work Step by Step

$\dfrac{\sqrt{x}+3}{\sqrt{x}}$ Multiply the numerator and the denominator of this expression by the conjugate of the numerator and simplify if possible: $\dfrac{\sqrt{x}+3}{\sqrt{x}}=\dfrac{\sqrt{x}+3}{\sqrt{x}}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}-3}=\dfrac{(\sqrt{x})^{2}-3^{2}}{\sqrt{x}(\sqrt{x}-3)}=...$ $...=\dfrac{x-9}{\sqrt{x^{2}}-3\sqrt{x}}=\dfrac{x-9}{x-3\sqrt{x}}$
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