I have to get from this expression: $(4+2\sqrt3)(\sqrt{2-\sqrt3})$
To this expression: $\sqrt2+\sqrt6$
I tried to square $(4+2\sqrt3)$ and put it inside the radical, so: $\sqrt{(16+12+16\sqrt3)(2-\sqrt3)}$
but eventually I get to: $\sqrt{52+32\sqrt3-28\sqrt3-48}$ and therefore $\sqrt{8+4\sqrt3}$ which will give me $2\sqrt{2+\sqrt3}$
From there I stopped because it didn't seem that I was getting anywhere.
I checked on wolfram alpha and using the step-by-step solution the first thing that it does is: "Express $2-\sqrt3$ as a square using $2-\sqrt3 = \frac{1-2\sqrt3+(\sqrt3)^2}{2}$ which is clearly correct and from there it is relatively easy to proceed to the solution.
So my questions are: (1) what is that does not work in my method of putting the squared expression inside the radical (2) where can I fond more information on how to express a radical as a square, is there a procedure (something similar to completing the square for example)?
Thanks in advance.