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: 53

Answer

$\dfrac{5+3\sqrt{2}}{7}$

Work Step by Step

Rationalizing the denominator of $ \dfrac{2\sqrt{3}+\sqrt{6}}{4\sqrt{3}-\sqrt{6}} $ results to \begin{array}{l} \dfrac{2\sqrt{3}+\sqrt{6}}{4\sqrt{3}-\sqrt{6}} \cdot \dfrac{4\sqrt{3}+\sqrt{6}}{4\sqrt{3}+\sqrt{6}} \\\\= \dfrac{(2\sqrt{3})(4\sqrt{3})+(2\sqrt{3})(\sqrt{6})+(\sqrt{6})(4\sqrt{3})+(\sqrt{6})(\sqrt{6})}{(4\sqrt{3})^2-(\sqrt{6})^2} \\\\= \dfrac{8\cdot3+2\sqrt{18}+4\sqrt{18}+6}{16\cdot3-6} \\\\= \dfrac{24+6\sqrt{18}+6}{42} \\\\= \dfrac{30+6\sqrt{9\cdot2}}{42} \\\\= \dfrac{30+18\sqrt{2}}{42} \\\\= \dfrac{5+3\sqrt{2}}{7} .\end{array}
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