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: 44

Answer

$\dfrac{4\sqrt{5}+\sqrt{2}}{2\sqrt{5}-\sqrt{2}}=\dfrac{7+\sqrt{10}}{3}$

Work Step by Step

$\dfrac{4\sqrt{5}+\sqrt{2}}{2\sqrt{5}-\sqrt{2}}$ Multiply the numerator and the denominator of this expression by the conjugate of the denominator: $\dfrac{4\sqrt{5}+\sqrt{2}}{2\sqrt{5}-\sqrt{2}}=\dfrac{4\sqrt{5}+\sqrt{2}}{2\sqrt{5}-\sqrt{2}}\cdot\dfrac{2\sqrt{5}+\sqrt{2}}{2\sqrt{5}+\sqrt{2}}=...$ $...=\dfrac{(4\sqrt{5}+\sqrt{2})(2\sqrt{5}+\sqrt{2})}{(2\sqrt{5})^{2}-(\sqrt{2})^{2}}=...$ $...=\dfrac{8\sqrt{5^{2}}+4\sqrt{10}+2\sqrt{10}+\sqrt{2^{2}}}{4(5)-2}=...$ $...=\dfrac{8(5)+6\sqrt{10}+2}{20-2}=\dfrac{40+2+6\sqrt{10}}{18}=\dfrac{42+6\sqrt{10}}{18}=...$ Take out common factor $6$ from the numerator and simplify: $...=\dfrac{6(7+\sqrt{10})}{18}=\dfrac{7+\sqrt{10}}{3}$
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