Solve $\frac{\sqrt{5+2\sqrt{6}} + \sqrt{5-2\sqrt{6}}}{\sqrt{5+2\sqrt{6}} - \sqrt{5-2\sqrt{6}}} =\sqrt \frac{x}{2}$ The value of $x$ (considering only the positive root) satisfying the equation $$\frac{\sqrt{5+2\sqrt{6}} + \sqrt{5-2\sqrt{6}}}{\sqrt{5+2\sqrt{6}} - \sqrt{5-2\sqrt{6}}} = \sqrt{\frac{x}{2}}$$ is?
Please, can you guys help me out because I can't understand which formula to use, Do we have to use the Discriminant formula ($b^2-4ac$)?

$$\frac{\sqrt{5+2\sqrt{6}} + \sqrt{5-2\sqrt{6}}}{\sqrt{5+2\sqrt{6}} - \sqrt{5-2\sqrt{6}}} = \frac{x}{2}$$
Using componendo dividendo
$$\frac{\sqrt{5+2\sqrt{6}} + \sqrt{5-2\sqrt{6}}+({\sqrt{5+2\sqrt{6}} - \sqrt{5-2\sqrt{6}}})}{\sqrt{5+2\sqrt{6}} + \sqrt{5-2\sqrt{6}}-(\sqrt{5+2\sqrt{6}} - \sqrt{5-2\sqrt{6}})} = \frac{x+2}{x-2}$$
$$\frac{\sqrt{5+2\sqrt{6}} +({\sqrt{5+2\sqrt{6}} })}{\sqrt{5-2\sqrt{6}}+ \sqrt{5-2\sqrt{6}})} = \frac{x+2}{x-2}$$
$$\frac{\sqrt{5+2\sqrt{6}}}{\sqrt{5-2\sqrt{6}}}=\frac{x+2}{x-2}$$
$$\frac{5+2\sqrt{6}}{5-2\sqrt{6}}=(\frac{x+2}{x-2})^2$$
Using it again
$$\frac{5+2\sqrt{6}+(5-2\sqrt{6})}{5+2\sqrt{6}-(5-2\sqrt{6})}=(\frac{(x+2)^2+(x-2)^2}{(x+2)^2-(x-2)^2})$$
$$\frac{10}{4\sqrt6}=\frac{2x^2+8}{8x}$$
$$\frac{10}{4\sqrt6}=\frac{x^2+4}{4x}$$
$$\frac{10}{\sqrt6}=\frac{x^2+4}{x}$$
$$\sqrt6(x^2+4)-10x=0$$
Just apply quadratic formula.
Here you go
