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: 31

Answer

$\dfrac{6}{2-\sqrt{7}}=-4-2\sqrt{7}$

Work Step by Step

$\dfrac{6}{2-\sqrt{7}}$ Multiply the numerator and the denominator by the conjugate of the denominator and simplify : $\dfrac{6}{2-\sqrt{7}}=\dfrac{6}{2-\sqrt{7}}\cdot\dfrac{2+\sqrt{7}}{2+\sqrt{7}}=\dfrac{6(2+\sqrt{7})}{2^{2}-(\sqrt{7})^{2}}=...$ $...=\dfrac{6(2+\sqrt{7})}{4-7}=\dfrac{6(2+\sqrt{7})}{-3}=-2(2+\sqrt{7})=-4-2\sqrt{7}$
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