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.

Here is another simple way, exploiting the innate symmetry.
Let $\ \bar c = \sqrt{3\!+\!x}+\sqrt{3\!-\!x},\,\  c = \sqrt{3\!+\!x}-\sqrt{3\!-\!x}.\,$ Then $\,\color{#0a0}{\bar c c} = 3\!+\!x-(3\!-\!x) = \color{#0a0}{2x},\ $ so
$\,\displaystyle\sqrt{5} = \frac{\bar c}c\, \Rightarrow \color{#c00}{\frac{6}{\sqrt{5}}} = {\frac{1}{\sqrt{5}}\!+\!\sqrt{5}} \,=\, \frac{c}{\bar c}+\frac{\bar c}c \,=\,  \frac{(c+\bar c)^2}{\color{#0a0}{c\bar c}} - 2 \,=\,\frac{4(3\!+\!x)}{\color{#0a0}{2x}}-2 \,=\, \color{#c00}{\frac{6}x}\  $ so $\ \color{#c00}x = \ldots$
