Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 7 - Sections 7.1-7.5 - Integrated Review - Radicals and Rational Exponents - Page 447: 37

Answer

$\dfrac{13-3\sqrt[]{21}}{5}$

Work Step by Step

Multiplying both the numerator and the denominator by $ 2\sqrt[]{3}+\sqrt{7} $, then the rationalized-denominator form of the given expression, $ \dfrac{\sqrt[]{3}-\sqrt{7}}{2\sqrt[]{3}+\sqrt{7}} ,$ is equivalent to \begin{array}{l}\require{cancel} \dfrac{\sqrt[]{3}-\sqrt{7}}{2\sqrt[]{3}+\sqrt{7}}\cdot\dfrac{2\sqrt[]{3}-\sqrt{7}}{2\sqrt[]{3}-\sqrt{7}} \\\\= \dfrac{\sqrt[]{3}(2\sqrt[]{3})+\sqrt[]{3}(-\sqrt{7})-\sqrt{7}(2\sqrt[]{3})-\sqrt{7}(-\sqrt{7})}{(2\sqrt[]{3})^2-(\sqrt{7})^2} \\\\= \dfrac{2\sqrt[]{3(3)}-\sqrt[]{3(7)}-2\sqrt[]{3(7)}+\sqrt{7(7)}}{4(3)-7} \\\\= \dfrac{2\sqrt[]{(3)^2}-\sqrt[]{21}-2\sqrt[]{21}+\sqrt{(7)^2}}{12-7} \\\\= \dfrac{2(3)-\sqrt[]{21}-2\sqrt[]{21}+7}{12-7} \\\\= \dfrac{6-\sqrt[]{21}-2\sqrt[]{21}+7}{5} \\\\= \dfrac{(6+7)+(-\sqrt[]{21}-2\sqrt[]{21})}{5} \\\\= \dfrac{13-3\sqrt[]{21}}{5} .\end{array}
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