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: 43

Answer

$\dfrac{2\sqrt{3}+\sqrt{6}}{4\sqrt{3}-\sqrt{6}}=\dfrac{3\sqrt{2}+5}{7}$

Work Step by Step

$\dfrac{2\sqrt{3}+\sqrt{6}}{4\sqrt{3}-\sqrt{6}}$ Multiply the numerator and the denominator of this expression by the conjugate of the denominator: $\dfrac{2\sqrt{3}+\sqrt{6}}{4\sqrt{3}-\sqrt{6}}=\dfrac{2\sqrt{3}+\sqrt{6}}{4\sqrt{3}-\sqrt{6}}\cdot\dfrac{4\sqrt{3}+\sqrt{6}}{4\sqrt{3}+\sqrt{6}}=...$ $...=\dfrac{(2\sqrt{3}+\sqrt{6})(4\sqrt{3}+\sqrt{6})}{(4\sqrt{3})^{2}-(\sqrt{6})^{2}}=...$ $...=\dfrac{8\sqrt{3^{2}}+2\sqrt{18}+4\sqrt{18}+\sqrt{6^{2}}}{16(3)-6}=...$ $...=\dfrac{8(3)+6\sqrt{18}+6}{48-6}=\dfrac{6(3)\sqrt{2}+24+6}{42}=\dfrac{18\sqrt{2}+30}{42}=...$ Take out common factor $6$ from the numerator and simplify: $...=\dfrac{6(3\sqrt{2}+5)}{42}=\dfrac{3\sqrt{2}+5}{7}$
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