Answer
$-\dfrac{\sqrt { 6}}{4}$
Work Step by Step
Simplify the denominator by factoring the radicand such that one of the factors is a perfect square:
$\dfrac{-\sqrt 3}{\sqrt 8}\\
=\dfrac{-\sqrt 3}{\sqrt {4\cdot 2}}\\
=\dfrac{-\sqrt 3}{\sqrt {2^2\cdot 2}}\\
$
Simplify.
$=\dfrac{-\sqrt 3}{2\sqrt {2}}$
Rationalize the denominator by multiplying $\sqrt2$ to both the numerator and the denominator:
$=\dfrac{-\sqrt 3}{2\sqrt 2} \cdot \dfrac{\sqrt 2}{\sqrt 2}$
$=\dfrac{-\sqrt 3\cdot \sqrt 2}{2(\sqrt 2)^2}$
$=\dfrac{-\sqrt 3\cdot \sqrt 2}{2(2)}$
$=\dfrac{-\sqrt 3\cdot \sqrt 2}{4}$
Use the rule $\sqrt{a} \cdot \sqrt{b}=\sqrt{a\cdot b}$ then simplify:
$=\dfrac{-\sqrt {3\cdot 2}}{4}$
$=\dfrac{-\sqrt {6}}{4}$
Hence, the correct answer is $-\frac{\sqrt { 6}}{4}$.
