If $$\frac{\sqrt{5+x}+\sqrt{5-x}}{\sqrt{5+x}-\sqrt{5-x}}=3$$, then what is the value of x?

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

Rationalizing the denominator,

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

=> $$\frac{[(\sqrt{5+x})+(\sqrt{5-x})]^2}{(\sqrt{5+x})^2-(\sqrt{5-x})^2}=3$$

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

=> $$\frac{10+2\sqrt{25-x^2}}{2x}=3$$

=> $$5+\sqrt{25-x^2}=3x$$

=> $$3x-5=\sqrt{25-x^2}$$

Squaring both sides, we get :

=> $$(3x-5)^2=(\sqrt{25-x^2})^2$$

=> $$9x^2+25-30x=25-x^2$$

=> $$10x^2-30x=0$$

=> $$10x(x-3)=0$$

=> $$x=0,3$$    [But $$x$$ can't be zero because the denominator can't be zero]

=> Ans - (D)