## Intermediate Algebra (12th Edition)

$-\dfrac{\sqrt{6}}{2}$
$\bf{\text{Solution Outline:}}$ To rationalize the given radical expression, $\dfrac{-3}{\sqrt{6}} ,$ multiply the numerator and the denominator by the denominator. $\bf{\text{Solution Details:}}$ Multiplying the numerator and the denominator by the denominator, the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{-3}{\sqrt{6}}\cdot\dfrac{\sqrt{6}}{\sqrt{6}} \\\\= \dfrac{-3\sqrt{6}}{(\sqrt{6})^2} \\\\= \dfrac{-3\sqrt{6}}{6} \\\\= \dfrac{\cancel{3}(-1)\sqrt{6}}{\cancel{3}(2)} \\\\= \dfrac{-1\sqrt{6}}{2} \\\\= -\dfrac{\sqrt{6}}{2} .\end{array}