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 - Page 445: 41

Answer

$-4-2\sqrt{7}$

Work Step by Step

Rationalizing the denominator of $ \dfrac{6}{2-\sqrt{7}} $ results to \begin{array}{l} \dfrac{6}{2-\sqrt{7}} \cdot \dfrac{2+\sqrt{7}}{2+\sqrt{7}} \\\\= \dfrac{12+6\sqrt{7}}{(2)^2-(\sqrt{7})^2} \\\\= \dfrac{12+6\sqrt{7}}{4-7} \\\\= \dfrac{12+6\sqrt{7}}{-3} \\\\= -4-2\sqrt{7} .\end{array}
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