Answer
$ 4+\sqrt{15}$
Work Step by Step
We lose the square roots in the denominator by applying the difference of squares formula:
$(a\sqrt{x}+b\sqrt{y})(a\sqrt{x}-b\sqrt{y})=(a\sqrt{x})^{2}-(b\sqrt{y})^{2}=a^{2}x-b^{2}y$
$\displaystyle \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}color{red}{ \cdot\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}} }\qquad$ (rationalize)
$=\displaystyle \frac{(\sqrt{5}+\sqrt{3})^{2}}{(\sqrt{5})^{2}-(\sqrt{3})^{2}}$
... the numerator is a square of a sum, $(A+B)^{2}=A^{2}+2AB+B^{2}$
$=\displaystyle \frac{(\sqrt{5})^{2}+2\sqrt{5}\cdot\sqrt{3}+(\sqrt{3})^{2}}{5-3}$
$=\displaystyle \frac{5+2\sqrt{15}+3}{2}$
$=\displaystyle \frac{8+2\sqrt{15}}{2}$
$=\displaystyle \frac{2(4+\sqrt{15})}{2}$
$=4+\sqrt{15}$
