#### Answer

$-\dfrac{7\sqrt{3}}{12}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To rationalize the given radical expression, $
\dfrac{-7}{\sqrt{48}}
,$ multiply both the numerator and the denominator by an expression that will make the denominator a perfect power of the index.
$\bf{\text{Solution Details:}}$
Expressing the radicand with a factor that is a perfect power of the index results to
\begin{array}{l}\require{cancel}
\dfrac{-7}{\sqrt{16\cdot3}}
\\\\
\dfrac{-7}{\sqrt{(4)^2\cdot3}}
.\end{array}
Multiplying both the numerator and the denominator by an expression that will make the denominator a perfect power of the index results to
\begin{array}{l}\require{cancel}
\dfrac{-7}{\sqrt{(4)^2\cdot3}}\cdot\dfrac{\sqrt{3}}{\sqrt{3}}
.\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{-7\sqrt{3}}{\sqrt{(4)^2\cdot3(3)}}
\\\\=
\dfrac{-7\sqrt{3}}{\sqrt{(4)^2\cdot(3)^2}}
\\\\=
\dfrac{-7\sqrt{3}}{4\cdot3}
\\\\=
\dfrac{-7\sqrt{3}}{12}
\\\\=
-\dfrac{7\sqrt{3}}{12}
.\end{array}