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 ($b^2-4ac$)?
 A: Let
$ a = \sqrt{5+ 2\sqrt{6}}, \ \ b=  \sqrt{5- 2\sqrt{6}}.$
$$ \frac{ a +b}{a-b} = \sqrt{\frac{x}{2}}$$
$$ \frac{(a+b)(a+b)}{(a-b)(a+b)} = \frac{(a+b)^2}{a^2 - b^2} = \sqrt{\frac{x}{2}}$$
$$ \frac{\left(\sqrt{5+ 2\sqrt{6}} + \sqrt{5-2\sqrt{6}}\right)^2}{\left(\sqrt{5+2\sqrt{6}}\right)^2-\left(\sqrt{5-2\sqrt{6}})\right)^2} = \sqrt{\frac{x}{2}}$$
$$ \frac{5+ 2\sqrt{6}+ 2\sqrt{(5+2\sqrt{6})(5 - 2\sqrt{6})}+5 -2\sqrt{6}}{5 +2\sqrt{6} -5 +2\sqrt{6}} = \sqrt{\frac{x}{2}}$$
$$ \frac{10 +2\sqrt{1}}{4\sqrt{6}} = \sqrt{\frac{x}{2}}$$
$$ \frac{12}{4\sqrt{6}}=\sqrt{\frac{x}{2}}$$
$$ \frac{3}{\sqrt{6}} = \sqrt{\frac{x}{2}}$$
$$ x = 3.$$
A: Hint:  Multiply the numerator and denominator of 
$$\frac{\sqrt{5 + 2\sqrt{6}} + \sqrt{5 - 2\sqrt{6}}}{\sqrt{5 + 2\sqrt{6}} - \sqrt{5 - 2\sqrt{6}}}$$
by the conjugate of the denominator to obtain 
$$\frac{\sqrt{5 + 2\sqrt{6}} + \sqrt{5 - 2\sqrt{6}}}{\sqrt{5 + 2\sqrt{6}} - \sqrt{5 - 2\sqrt{6}}} \cdot \frac{\sqrt{5 + 2\sqrt{6}} + \sqrt{5 - 2\sqrt{6}}}{\sqrt{5 + 2\sqrt{6}} + \sqrt{5 - 2\sqrt{6}}}$$
then simplify.
A: $$\frac{\sqrt{5+2\sqrt{6}} + \sqrt{5-2\sqrt{6}}}{\sqrt{5+2\sqrt{6}} - \sqrt{5-2\sqrt{6}}} = \frac{x}{2}$$
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
A: Hint: Calculate
$$(\sqrt2+\sqrt3)^2.$$
Then do the same with a sign change.
