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: 54

Answer

$\dfrac{7+\sqrt{10}}{3}$

Work Step by Step

Rationalizing the denominator of $ \dfrac{4\sqrt{5}+\sqrt{2}}{2\sqrt{5}-\sqrt{2}} $ results to \begin{array}{l} \dfrac{4\sqrt{5}+\sqrt{2}}{2\sqrt{5}-\sqrt{2}} \cdot \dfrac{2\sqrt{5}+\sqrt{2}}{2\sqrt{5}+\sqrt{2}} \\\\= \dfrac{(4\sqrt{5})(2\sqrt{5})+(4\sqrt{5})(\sqrt{2})+(\sqrt{2})(2\sqrt{5})+(\sqrt{2})(\sqrt{2})}{(2\sqrt{5})^2-(\sqrt{2})^2} \\\\= \dfrac{8\cdot5+4\sqrt{10}+2\sqrt{10}+2}{20-2} \\\\= \dfrac{40+6\sqrt{10}+2}{18} \\\\= \dfrac{42+6\sqrt{10}}{18} \\\\= \dfrac{7+\sqrt{10}}{3} .\end{array}
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