## Intermediate Algebra for College Students (7th Edition)

$\displaystyle \frac{13\sqrt{11}+39}{2}$
We lose the square roots in the denominator by applying the difference of squares formula: $(a\sqrt{x}+b\sqrt{y})(a\sqrt{x}-b\sqrt{y})=(a\sqrt{x})^{2}-(b\sqrt{y})^{2}=a^{2}x-b^{2}y$ $\displaystyle \frac{13}{\sqrt{11}-3} \displaystyle \color{red}{ \cdot\frac{\sqrt{11}+3}{\sqrt{11}+3} }\qquad$ (rationalize) $=\displaystyle \frac{13(\sqrt{11}+3)}{(\sqrt{11})^{2}-3^{2}} =\frac{13(\sqrt{11}+31)}{11-9} =\frac{13(\sqrt{11}+3)}{2}$ = $\displaystyle \frac{13\sqrt{11}+39}{2}$