Difficulty in "expressing radical as square" 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.
 A: Notice that $$\sqrt{8+4\sqrt{3}} = \sqrt{2+2\sqrt{12} + 6} = \sqrt{(\sqrt{2} + \sqrt{6})^2} = \sqrt{2} + \sqrt{6},$$
as desired.  The trick is to write the $4\sqrt{3}$ term as twice a radical.
A: Notice that
$$4+2\sqrt3=(1+\sqrt3)^2$$
and 
$$2-\sqrt3=\frac12(1-\sqrt3)^2$$
so the given expression is
$$\frac1{\sqrt2}(1+\sqrt3)^2(\sqrt3-1)=\sqrt2(\sqrt3+1)=\sqrt6+\sqrt2$$
A: The idea is that you can compute square roots of numbers of the form $\sqrt{a+b\sqrt{d}}$. Indeed,
$$(x+y\sqrt{d})^2 = x^2 + dy^2 + 2xy\sqrt{d}. $$
We want $a = x^2 + dy^2$ and $b = 2xy$, so using $y = b/(2x)$, we get $a = x^2 + db^2/(4x^2)$, or $4x^4 - 4ax^2 + db^2 = 0$, which is a quadratic equation in $x^2$ whose solution is
$$ x^2 = \frac{4a\pm\sqrt{16a^2-16db^2}}{8} = \frac{a\pm\sqrt{a^2-db^2}}{2}. $$
For example, suppose we want to calculate $\sqrt{2-\sqrt{3}}$, so $a=2$ and $b=-1$. The formula gives $x^2 = (2\pm\sqrt{4-3})/2 = (2\pm 1)/2=3/2,1/2$. This gives the solutions
$$ x + y\sqrt{3} = \pm \left(\sqrt{\frac{3}{2}} - \sqrt{\frac{1}{6}} \sqrt{3}\right), \pm \left(\sqrt{\frac{1}{2}} - \sqrt{\frac{1}{2}} \sqrt{3}\right).$$
Wolfram alpha suggests using the second solution:
$$
\begin{align*}
(4+2\sqrt{3})\sqrt{2-\sqrt{3}} &= (4+2\sqrt{3})\left(-\sqrt{\frac{1}{2}} + \sqrt{\frac{1}{2}} \sqrt{3}\right) \\ &= -4\sqrt{\frac{1}{2}} + 2\sqrt{\frac{1}{2}}\cdot 3 + \left(4\sqrt{\frac{1}{2}}-2\sqrt{\frac{1}{2}}\right)\sqrt{3} \\ &= 
2\sqrt{\frac{1}{2}} + 2\sqrt{\frac{1}{2}} \sqrt{3} \\ &= \sqrt{2} + \sqrt{6}.
\end{align*}
$$
Of course, this is pretty ad hoc. However, this method will find an integral square root $\sqrt{a+b\sqrt{d}} = x+y\sqrt{d}$ if one exists, which could be helpful in other situations.
