Intermediate Algebra (6th Edition)

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

Chapter 7 - Section 7.5 - Rationalizing Denominators and Numerators of Radical Expressions - Exercise Set: 50

Answer

$\dfrac{-3\sqrt{6}-6}{2}$

Work Step by Step

Rationalizing the denominator of $ \dfrac{-3}{\sqrt{6}-2} $ results to \begin{array}{l} \dfrac{-3}{\sqrt{6}-2} \cdot \dfrac{\sqrt{6}+2}{\sqrt{6}+2} \\\\= \dfrac{-3\sqrt{6}-6}{(\sqrt{6})^2-(2)^2} \\\\= \dfrac{-3\sqrt{6}-6}{6-4} \\\\= \dfrac{-3\sqrt{6}-6}{2} .\end{array}
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