## Elementary Algebra

Published by Cengage Learning

# Chapter 9 - Roots and Radicals - Chapter 9 Test - Page 431: 19

#### Answer

$\dfrac{3\sqrt{2}-\sqrt{3}}{5}$

#### Work Step by Step

Multiplying by the conjugate of the denominator and using $(a+b)(a-b)=a^2-b^2$, the simplified form of the given expression, $\dfrac{\sqrt{6}}{\sqrt{12}+\sqrt{2}} ,$ is \begin{array}{l}\require{cancel} \dfrac{\sqrt{6}}{\sqrt{12}+\sqrt{2}}\cdot\dfrac{\sqrt{12}-\sqrt{2}}{\sqrt{12}-\sqrt{2}} \\\\= \dfrac{\sqrt{6(12)}-\sqrt{6(2)}}{(\sqrt{12})^2-(\sqrt{2})^2} \\\\= \dfrac{\sqrt{6(6\cdot2)}-\sqrt{(2\cdot3)(2)}}{12-2} \\\\= \dfrac{\sqrt{(6)^2\cdot2}-\sqrt{(2)^2\cdot3}}{10} \\\\= \dfrac{6\sqrt{2}-2\sqrt{3}}{10} \\\\= \dfrac{2(3\sqrt{2}-\sqrt{3})}{10} \\\\= \dfrac{\cancel{2}(3\sqrt{2}-\sqrt{3})}{\cancel{2}(5)} \\\\= \dfrac{3\sqrt{2}-\sqrt{3}}{5} .\end{array}

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