Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Review - Exercises: 54

Answer

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

Work Step by Step

$\dfrac{\sqrt{x}-2}{\sqrt{x}+2}$ Multiply the numerator and the denominator of this expression by the conjugate of the denominator and simplify: $\dfrac{\sqrt{x}-2}{\sqrt{x}+2}=\dfrac{\sqrt{x}-2}{\sqrt{x}+2}\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}-2}=...$ $...=\dfrac{(\sqrt{x}-2)^{2}}{(\sqrt{x})^{2}-2^{2}}=\dfrac{(\sqrt{x})^{2}-4\sqrt{x}+2^{2}}{x-4}=...$ $...=\dfrac{x-4\sqrt{x}+4}{x-4}$
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