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: 71

Answer

$\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{x-1}{x-2\sqrt{x}+1}$

Work Step by Step

$\dfrac{\sqrt{x}+1}{\sqrt{x}-1}$ Multiply the numerator and the denominator of this expression by the conjugate of the numerator and simplify if possible: $\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}-1}=\dfrac{(\sqrt{x})^{2}-1^{2}}{(\sqrt{x}-1)^{2}}=...$ $...=\dfrac{x-1}{(\sqrt{x})^{2}-2\sqrt{x}+1^{2}}=\dfrac{x-1}{x-2\sqrt{x}+1}$
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