Algebra 2 (1st Edition)

$\displaystyle \frac{4}{8-\sqrt{3}}=\frac{32+4\sqrt{3}}{61}$
$\displaystyle \frac{4}{8-\sqrt{3}}$ $\qquad$ ...rationalize the denominator by multyplying both the numerator and the denominator with the conjugate of $8-\sqrt{3}$, which is $8+\sqrt{3}$. $=\displaystyle \frac{4}{8-\sqrt{3}}\cdot\frac{8+\sqrt{3}}{8+\sqrt{3}}\qquad$ ...apply $(a+b)(a-b)=a^{2}-b^{2}$ in the denominator ($a=8,\ b=\sqrt{3}$). $=\displaystyle \frac{32+4\sqrt{3}}{8^{2}-(\sqrt{3})^{2}}\qquad$ ...simplify ($8^{2}=64,\ (\sqrt{3})^{2}=3$). $=\displaystyle \frac{32+4\sqrt{3}}{64-3}\qquad$ ...add like terms. $=\displaystyle \frac{32+4\sqrt{3}}{61}$