<s>[INST] 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.
 [/INST] In terms of your method, I think the first thing I'd do is rewrite it as,
$$r + \sqrt{3} = \sqrt{4 + 2\sqrt{3}}.$$
If you square both sides,
$$r^2 + 3 + 2\sqrt{3}r = 4 + 2\sqrt{3}.$$
Isolating $\sqrt{3}$,
$$2\sqrt{3}(r - 1) = 1 - r^2.$$
Squaring again,
$$12(r-1)^2 = r^4 - 2r^2 + 1.$$
Expanding and rearranging,
$$(r-1)^2\left(r^2+2r-11\right)=r^4 - 14r^2 + 24r - 11 = 0.$$
Therefore, if $r$ is rational, then $r = \pm 1, \pm 11$, which narrows things down. The only rational root of this polynomial is $1$, which narrows it down again. So, this method tells us that we only need to verify that $r$ is or is not $1$ (it is $1$), then we'll definitely know whether or not it is irrational.
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