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.
 A: 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 $$\frac3x=\frac6{2\sqrt5}\implies x=?$$
A: Use
$$(\sqrt{3+x}-\sqrt{3-x})(\sqrt{3+x}+\sqrt{3-x})=2x$$
A: 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 $\ \color{#c00}x = \ldots$
A: $$\frac{\sqrt{3+x}+\sqrt{3-x}}{\sqrt{3+x}-\sqrt{3-x}}\cdot 1=\frac{\sqrt{3+x}+\sqrt{3-x}}{\sqrt{3+x}-\sqrt{3-x}}\cdot\frac{\sqrt{3+x}+\sqrt{3-x}}{\sqrt{3+x}+\sqrt{3-x}} =\sqrt{5}$$
$$\frac{(\sqrt{3+x}+\sqrt{3-x})^2}{\sqrt{3+x}^2-\sqrt{3-x}^2} =\sqrt{5}$$
$$\frac{3+x+3-x+2\sqrt{9-x^2}}{{3+x}-3+x} =\sqrt{5}$$
$$\frac{2(3+\sqrt{9-x^2})}{2x} =\sqrt{5}$$
$$\frac{3+\sqrt{9-x^2}}{x} =\frac{\sqrt{5}}{1}$$
$$3+\sqrt{9-x^2}=x\sqrt{5}$$
$$\sqrt{9-x^2}=x\sqrt{5}-3/^2$$
$$9-x^2=5x^2-6x\sqrt 5+9 $$
$$6x^2-6x\sqrt 5=0$$
$$6x(x-\sqrt 5)=0$$
$$x_1=0, x_2=\sqrt 5$$
But $x=0$ is not a rots of the given equation, becouese: $\frac{\sqrt{3+x}+\sqrt{3-x}}{\sqrt{3+x}-\sqrt{3-x}}\neq\sqrt 5$ 
