I've been trying to solve this one without success... can anybody help me?
The result should be $x=\frac{17}{16}$ and it's correct, I've already checked.
This is the equation:
$$\frac{1}{\sqrt {x+2} - \sqrt x}+\frac{2}{\sqrt {x+4} + \sqrt x}=2$$
Thanks in advance!
Rationalize the denominators.
$$\frac{1}{\sqrt{x+2} - \sqrt{x}} = \frac{\sqrt{x+2} + \sqrt{x}}{2},$$ and $$\frac{2}{\sqrt{x+4} + \sqrt{x}} = \frac{2\sqrt{x+4} - 2\sqrt{x}}{4},$$ and so we find $$\sqrt{x+2} + \sqrt{x+4} = 4.$$
Thus, $$\sqrt{x+4} = 4 - \sqrt{x+2},$$ and so we find that
$$x+4 = x + 18 - 8\sqrt{x+2},$$ $$8\sqrt{x+2} = 14,$$ $$x+2 = \frac{196}{64},$$ and so $$x = \frac{68}{64} = \frac{17}{16}.$$