Answer
$\displaystyle \frac{2}{4+\sqrt{11}}=\frac{2\sqrt{11}-8}{7}$
Work Step by Step
$\displaystyle \frac{2}{4+\sqrt{11}}$
$\qquad$ ...rationalize the denominator by multiplying both the numerator and the denominator
with the conjugate of $4+\sqrt{11}$, which is $4-\sqrt{11}$.
$=\displaystyle \frac{2(4-\sqrt{11})}{(4+\sqrt{11})(4-\sqrt{11})}\qquad$ ...apply $(a+b)(a-b)=a^{2}-b^{2}$ in the denominator ($a=4,\ b=\sqrt{11}$).
$=\displaystyle \frac{2(4-\sqrt{11})}{4^{2}-(\sqrt{11})^{2}}\qquad$ ...simplify the numerator.
$=\displaystyle \frac{8-2\sqrt{11}}{4^{2}-(\sqrt{11})^{2}}\qquad$ ...simplify $(\sqrt{11})^{2}=11$
$=\displaystyle \frac{8-2\sqrt{11}}{4-11}\qquad$ ...simplify
$=\displaystyle \frac{8-2\sqrt{11}}{-7}\qquad$ ...multiply with $\displaystyle \frac{-1}{-1}$
$=\displaystyle \frac{2\sqrt{11}-8}{7}$
