## Intermediate Algebra (12th Edition)

$\dfrac{\sqrt{2}}{8}$
$\bf{\text{Solution Outline:}}$ To simplify the given radical expression, $\dfrac{2\sqrt{25}}{8\sqrt{50}} ,$ use the laws of radicals. Then rationalize the denominator. $\bf{\text{Solution Details:}}$ Using the Quotient Rule of radicals which is given by $\sqrt[n]{\dfrac{x}{y}}=\dfrac{\sqrt[n]{x}}{\sqrt[n]{y}}{},$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{2}{8}\sqrt{\dfrac{25}{50}} \\\\= \dfrac{\cancel2}{\cancel24}\sqrt{\dfrac{\cancel{25}}{\cancel{25}\cdot2}} \\\\= \dfrac{1}{4}\sqrt{\dfrac{1}{2}} .\end{array} Rationalizing the numerator by multiplying the radicand by an expression equal to $1$ which will make the denominator a perfect power of the index, the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{1}{4}\sqrt{\dfrac{1}{2}\cdot\dfrac{2}{2}} \\\\= \dfrac{1}{4}\sqrt{\dfrac{2}{2^2}} .\end{array} Using the Quotient Rule of radicals which is given by $\sqrt[n]{\dfrac{x}{y}}=\dfrac{\sqrt[n]{x}}{\sqrt[n]{y}}{},$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{1}{4}\cdot\dfrac{\sqrt{2}}{\sqrt{2^2}} \\\\= \dfrac{1}{4}\cdot\dfrac{\sqrt{2}}{2} \\\\= \dfrac{\sqrt{2}}{8} .\end{array}