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: 40

Answer

$\dfrac{-3}{\sqrt{6}-2}=-\dfrac{3(\sqrt{6}+2)}{2}$

Work Step by Step

$\dfrac{-3}{\sqrt{6}-2}$ Multiply the numerator and the denominator of this expression by the conjugate of the denominator and simplify if possible: $\dfrac{-3}{\sqrt{6}-2}=\dfrac{-3}{\sqrt{6}-2}\cdot\dfrac{\sqrt{6}+2}{\sqrt{6}+2}=\dfrac{-3(\sqrt{6}+2)}{(\sqrt{6})^{2}-2^{2}}=...$ $...=\dfrac{-3(\sqrt{6}+2)}{6-4}=-\dfrac{3(\sqrt{6}+2)}{2}$
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