## Intermediate Algebra (12th Edition)

$\dfrac{4+3\sqrt{8}}{7}$
$\bf{\text{Solution Outline:}}$ To rationalize the given radical expression, $\dfrac{\sqrt{8}}{3-\sqrt{2}} ,$ multiply the numerator and the denominator by the conjugate of the denominator. Then use special products to multiply the result. $\bf{\text{Solution Details:}}$ Multiplying the numerator and the denominator of the given expression by the conjugate of the denominator, the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{\sqrt{8}}{3-\sqrt{2}} \cdot\dfrac{3+\sqrt{2}}{3+\sqrt{2}} \\\\= \dfrac{\sqrt{8}(3+\sqrt{2})}{(3-\sqrt{2})(3+\sqrt{2})} .\end{array} Using the product of the sum and difference of like terms which is given by $(a+b)(a-b)=a^2-b^2,$ the expression above is equivalent \begin{array}{l}\require{cancel} \dfrac{\sqrt{8}(3+\sqrt{2})}{(3)^2-(\sqrt{2})^2} \\\\= \dfrac{\sqrt{8}(3+\sqrt{2})}{9-2} \\\\= \dfrac{\sqrt{8}(3+\sqrt{2})}{7} .\end{array} Using the Distributive Property which is given by $a(b+c)=ab+ac,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{\sqrt{8}(3)+\sqrt{8}(\sqrt{2})}{7} \\\\= \dfrac{3\sqrt{8}+\sqrt{8}(\sqrt{2})}{7} .\end{array} Using the Product Rule of radicals which is given by $\sqrt[m]{x}\cdot\sqrt[m]{y}=\sqrt[m]{xy},$ the expression above is equivalent to\begin{array}{l}\require{cancel} \dfrac{3\sqrt{8}+\sqrt{8(2)}}{7} \\\\= \dfrac{3\sqrt{8}+\sqrt{16}}{7} \\\\= \dfrac{3\sqrt{8}+\sqrt{(4)^2}}{7} \\\\= \dfrac{3\sqrt{8}+4}{7} \\\\= \dfrac{4+3\sqrt{8}}{7} .\end{array}