Answer
$\dfrac{4+3\sqrt{8}}{7}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
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}