Solving an irrational equation Solve for $x$ in:
$$\frac{\sqrt{3+x}+\sqrt{3-x}}{\sqrt{3+x}-\sqrt{3-x}}=\sqrt{5}$$
I used the property of proportions ($a=\sqrt{3+x}$, $b=\sqrt{3-x})$:
$$\frac{(a+b)+(a-b)}{(a+b)-(a-b)}=\frac{2a}{2b}=\frac{a}{b}$$
I'm not sure if that's correct.
Or maybe the notations $a^3=3+x$, $b^3=3-x$ ? I don't know how to continue. Thank you.

$$\frac{\sqrt{3+x}+\sqrt{3-x}}{\sqrt{3+x}-\sqrt{3-x}}\cdot 1=\frac{\sqrt{3+x}+\sqrt{3-x}}{\sqrt{3+x}-\sqrt{3-x}}\cdot\frac{\sqrt{3+x}+\sqrt{3-x}}{\sqrt{3+x}+\sqrt{3-x}} =\sqrt{5}$$
$$\frac{(\sqrt{3+x}+\sqrt{3-x})^2}{\sqrt{3+x}^2-\sqrt{3-x}^2} =\sqrt{5}$$
$$\frac{3+x+3-x+2\sqrt{9-x^2}}{{3+x}-3+x} =\sqrt{5}$$
$$\frac{2(3+\sqrt{9-x^2})}{2x} =\sqrt{5}$$
$$\frac{3+\sqrt{9-x^2}}{x} =\frac{\sqrt{5}}{1}$$
$$3+\sqrt{9-x^2}=x\sqrt{5}$$
$$\sqrt{9-x^2}=x\sqrt{5}-3/^2$$
$$9-x^2=5x^2-6x\sqrt 5+9 $$
$$6x^2-6x\sqrt 5=0$$
$$6x(x-\sqrt 5)=0$$
$$x_1=0, x_2=\sqrt 5$$
But $x=0$ is not a rots of the given equation, becouese: $\frac{\sqrt{3+x}+\sqrt{3-x}}{\sqrt{3+x}-\sqrt{3-x}}\neq\sqrt 5$ 
