Show that $\sqrt{4+2\sqrt{3}}-\sqrt{3}$ is rational using the rational zeros theorem What I've done so far: 
Let $$r = \sqrt{4+2\sqrt{3}}-\sqrt{3}.$$
Thus, $$r^2 = 2\sqrt{3}-2\sqrt{3}\sqrt{4+2\sqrt{3}}+7$$
and $$r^4=52\sqrt{3}-28\sqrt{3}\sqrt{4+2\sqrt{3}}-24\sqrt{4+2\sqrt{3}}+109.$$
I did this because in a similar example in class, we related $r^2$ and $r^4$ to find a polynomial such that $mr^4+nr^2 = 0$ for some integers $m,n$. However, I cannot find such relation here. Am I doing this right or is there another approach to these type of problems.

From $r = \sqrt{4+2\sqrt{3}}-\sqrt{3}$ we get $r+\sqrt{3}=\sqrt{4+2\sqrt{3}}$ and, squaring both sides,
$$
r^2+2r\sqrt{3}+3=4+2\sqrt{3}
$$
and so
$$
r^2-1=2(1-r)\sqrt{3}
$$
Square again:
$$
r^4-2r^2+1=12-24r+12r^2
$$
so
$$
r^4-14r^2+24r-11=0
$$
The rational root test only allows $1$, $-1$, $11$ and $-11$ as roots. Since $1$ is clearly a root we have
$$
(r-1)(r^3+r^2-13r+11)=0
$$
and $1$ is a root also of the second factor:
$$
(r-1)^2(r^2+2r-11)=0
$$
The roots of the second factor are
$$
-1+2\sqrt{3},\qquad -1-2\sqrt{3}
$$
Since $r>0$, we only have two possibilities: $r=1$ or $r=2\sqrt{3}-1$. The second possibility gives
$$
\sqrt{4+2\sqrt{3}}=3\sqrt{3}-1
$$
If we square this, we get
$$
4+2\sqrt{3}=28-6\sqrt{3}
$$
or
$$
24=8\sqrt{3}
$$
which is absurd. Thus we only remain with the possibility that $r=1$.

Easier: $4+2\sqrt{3}=3+2\sqrt{3}+1=(\sqrt{3}+1)^2$.
Alternatively, from $r^2-1=2(1-r)\sqrt{3}$ we deduce $r=1$ or
$$
r+1=-2\sqrt{3}
$$
that's absurd, because $r>0$.
