If $$2\sqrt{x} = \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}} + \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}$$

it is given that

$$2\sqrt{x} = \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}} - \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}$$

here , $$\frac{\sqrt5+\sqrt3}{\sqrt5-\sqrt3}$$ = $$\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}$$ x $$\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}$$ = $$\frac{(\sqrt5 + \sqrt3)^2}{2}$$

similarly , $$\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}$$ = $$\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}$$ x $$\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}}$$ = $$\frac{(\sqrt5 - \sqrt3)^2}{2}$$

$$\frac{(\sqrt5 + \sqrt3)^2}{2}$$ + $$\frac{(\sqrt5 - \sqrt3)^2}{2}$$ = 2$$\sqrt(x)$$

8 = 2$$\sqrt(x)$$

x = 16