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.

Let $\displaystyle N=\sqrt{4+2\sqrt{3}}-\sqrt{3}$  
$=\sqrt{{(\sqrt{3})}^2+1^2+2\cdot\sqrt{3}\cdot1}-\sqrt{3}$   
$=(\sqrt{3}+1-\sqrt{3})$  
$=\boxed1$  
Aliter: If you want to use polynomials, you can see that  
$\displaystyle (N+\sqrt{3})^2=4+2\sqrt{3}$  
$\implies N^2+2\sqrt{3}\cdot N + 3=4+2\sqrt{3}$  
$\implies (N^2-1)=2\sqrt{3}\cdot(1-N)$   
$\implies N=-2\sqrt{3}-1$  
or 
$N=1$  
But since $N>0$,  
$\implies N=\boxed1$
