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.

Yes you have applied Componendo and dividendo to get
$$\frac{\sqrt{3+x}}{\sqrt{3-x}}=\frac{\sqrt5+1}{\sqrt5-1}$$
Squaring we get  $$\frac{3+x}{3-x}=\frac{6+2\sqrt5}{6-2\sqrt5}$$
Again apply Componendo and dividendo to get $$\frac3x=\frac6{2\sqrt5}\implies x=?$$
