# Chapter 9 - Roots and Radicals - 9.4 - Products and Quotients Involving Radicals - Problem Set 9.4 - Page 418: 70

$\dfrac{6-3\sqrt{2}-2\sqrt{5}+\sqrt{10}}{4}$

#### Work Step by Step

Multiplying by the conjugate of the denominator and using $(a+b)(a-b)=a^2-b^2,$ the given expression, $\dfrac{3-\sqrt{5}}{4+\sqrt{8}} ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{3-\sqrt{5}}{4+\sqrt{8}}\cdot\dfrac{4-\sqrt{8}}{4-\sqrt{8}} \\\\= \dfrac{3(4)+3(-\sqrt{8})-\sqrt{5}(4)-\sqrt{5}(-\sqrt{8})}{(4)^2-(\sqrt{8})^2} \\\\= \dfrac{12-3\sqrt{8}-4\sqrt{5}+\sqrt{40}}{8} \\\\= \dfrac{12-3\sqrt{4\cdot2}-4\sqrt{5}+\sqrt{4\cdot10}}{8} \\\\= \dfrac{12-3\sqrt{(2)^2\cdot2}-4\sqrt{5}+\sqrt{(2)^2\cdot10}}{8} \\\\= \dfrac{12-3(2)\sqrt{2}-4\sqrt{5}+2\sqrt{10}}{8} \\\\= \dfrac{12-6\sqrt{2}-4\sqrt{5}+2\sqrt{10}}{8} \\\\= \dfrac{\cancel{2}(6-3\sqrt{2}-2\sqrt{5}+\sqrt{10})}{\cancel{2}(4)} \\\\= \dfrac{6-3\sqrt{2}-2\sqrt{5}+\sqrt{10}}{4} .\end{array}

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