How to show $\sqrt{4+2\sqrt{3}}-\sqrt{3} = 1$ I start with $x=\sqrt{4+2\sqrt{3}}-\sqrt{3}$, then
$\begin{align*}
x +\sqrt{3} &= \sqrt{4+2\sqrt{3}}\\
(x +\sqrt{3})^2 &= (\sqrt{4+2\sqrt{3}})^2\\
x^2 + (2\sqrt{3})x + 3 &= 4+ 2\sqrt{3}\\
x^2 + (2\sqrt{3})x - 1 - 2\sqrt{3} &= 0
\end{align*}$
So I have shown that there is some polynomial whose solution is $x=\sqrt{4+2\sqrt{3}}-\sqrt{3}$, but I have not shown it to be 1.
 A: It is useful, when approaching a problem like this, to eliminate the more complex square roots. So given:
$$\sqrt{4 + 2 \sqrt{3}} - \sqrt{3} = 1$$
Rearrange to (so as to isolate the more complex square root on the LHS):
$$\sqrt{4 + 2 \sqrt{3}} = \sqrt{3} + 1$$
Square both sides:
$$4 + 2 \sqrt{3} = (\sqrt{3} + 1)(\sqrt{3} + 1)$$
Multiply out the RHS:
$$4 + 2 \sqrt{3} = 3 + 2\sqrt{3} + 1$$
So
$$4 + 2 \sqrt{3} = 4 + 2\sqrt{3}$$
This does not rely on you seeing the factorization, you just methodically work on simplifying the expression you were given. 
A: Hint: $$4+2\sqrt{3} = 1+2\sqrt{3}+\sqrt{3}^2 = (1+\sqrt{3})^2.$$
A: Here’s an approach that doesn’t require you to spot a not-terribly-obvious factorization. Multiplying by the conjugate to get rid of some of the square roots is a fairly natural thing to do, and
$$\left(\sqrt{4+2\sqrt3}-\sqrt3\right)\left(\sqrt{4+2\sqrt3}+\sqrt3\right)=1+2\sqrt3\;.$$
Thus, the desired result holds if and only if 
$$1+2\sqrt3=\sqrt{4+2\sqrt3}+\sqrt3\;,$$
or
$$1+\sqrt3=\sqrt{4+2\sqrt3}\;,$$
which is easily verified.
A: We have $4+2 \sqrt 3= 3+ 2 \sqrt 3+1= \left( \sqrt 3+1\right)^2$. Therefore $\sqrt{4+2 \sqrt 3}= \sqrt 3+1$.
Thus, $\sqrt{4+2 \sqrt 3}- \sqrt 3=1$.
