If $$\frac{\sqrt{2+x}+\sqrt{2-x}}{\sqrt{2+x}-\sqrt{2-x}}=2$$, the value of x is 

Expression : $$\frac{\sqrt{2+x}+\sqrt{2-x}}{\sqrt{2+x}-\sqrt{2-x}}=2$$

=> $$\frac{\sqrt{2+x}+\sqrt{2-x}}{\sqrt{2+x}-\sqrt{2-x}} \times \frac{\sqrt{2+x}+\sqrt{2-x}}{\sqrt{2+x}+\sqrt{2-x}}=2$$

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

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

=> $$\frac{4+2\sqrt{4-x^2}}{2x}=2$$

=> $$2+\sqrt{4-x^2}=2x$$

=> $$\sqrt{4-x^2}=2x-2$$

Squaring both sides, we get : 

=> $$(\sqrt{4-x^2})^2=(2x-2)^2$$

=> $$4-x^2=4x^2+4-8x$$

=> $$5x^2-8x=0$$

=> $$x(5x-8)=0$$

=> $$x=\frac{8}{5}$$

=> Ans - (C)