If $$x=8+2\sqrt{15}$$, then what is the value of $$\sqrt{x}+\frac{1}{\sqrt{x}}$$ ?

Given : $$x=8+2\sqrt{15}$$

=> $$x=5+3+2\sqrt{(5)(3)}$$

=> $$x=(\sqrt5)^2+(\sqrt3)^2+2(\sqrt5)(\sqrt3)$$

=> $$x=(\sqrt5+\sqrt3)^2$$

=> $$\sqrt{x}=\sqrt5+\sqrt3$$ ------------(i)

Now, $$\frac{1}{\sqrt{x}}=\frac{1}{\sqrt5+\sqrt3}$$

=> $$\frac{1}{\sqrt{x}}=\frac{1}{\sqrt5+\sqrt3}\times\frac{(\sqrt5-\sqrt3)}{(\sqrt5-\sqrt3)}$$

=> $$\frac{1}{\sqrt{x}}=\frac{\sqrt5-\sqrt3}{5-3}=\frac{\sqrt5-\sqrt3}{2}$$ -----------(ii)

Adding equations (i) and (ii), we get :

=> $$\sqrt{x}+\frac{1}{\sqrt{x}}=(\sqrt5+\sqrt3)+(\frac{\sqrt5-\sqrt3}{2})$$

= $$\frac{2\sqrt5+2\sqrt3+\sqrt5-\sqrt3}{2}=\frac{3\sqrt5+\sqrt3}{2}$$

=> Ans - (C)