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.
 A: 
is there another approach to these type of problems

Since 
$$4=1+3,3=1\times 3$$
we can have
$$4+2\sqrt 3=1+3+2\sqrt{1\times 3}=(1+\sqrt 3)^2$$
A: HINT: use that $$4+2\sqrt{3}=(1+\sqrt{3})^2$$
A: 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 $r>0$.
A: 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 $N>0$,  
$\implies N=\boxed1$
A: 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.
