Precalculus (6th Edition) Blitzer

$\displaystyle \frac{11(\sqrt{7}+\sqrt{3})}{4}$
Use $(a+b)(a-b)=a^{2}-b^{2}\quad$ when rationalizing. $[(a+\sqrt{b})(a-\sqrt{b})=a^{2}-b]$ $\displaystyle \frac{11}{\sqrt{7}-\sqrt{3}}\cdot\frac{\sqrt{7}+\sqrt{3}}{\sqrt{7}+\sqrt{3}}=\frac{11(\sqrt{7}+\sqrt{3})}{7-3}$ $=\displaystyle \frac{11(\sqrt{7}+\sqrt{3})}{4}$