## Algebra 2 (1st Edition)

$\displaystyle \frac{2}{4+\sqrt{11}}=\frac{2\sqrt{11}-8}{7}$
$\displaystyle \frac{2}{4+\sqrt{11}}$ $\qquad$ ...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 $\displaystyle \frac{-1}{-1}$ $=\displaystyle \frac{2\sqrt{11}-8}{7}$