Show that $\sqrt{6+4\sqrt{2}}-\sqrt{2}$ is rational using the rational zeros theorem Let $r=\sqrt{6+4\sqrt{2}}-\sqrt{2}$, then $r+\sqrt{2}=\sqrt{6+4\sqrt{2}}$.
Squaring both sides, we get $$r^2+2r\sqrt{2}+2=6+4\sqrt{2}$$ which is the same as $r^2-4=2\sqrt{2}$.
Squaring both sides again, we get $r^4-8r^2+16=8$ or $$r^4-8r^2+8=0.\tag{$\star$}$$ 
The rational zeros theorem tells us that the only possible rational solutions to ($\star$) are $±1$, $±2$, $±4$, $±8$.
I do not know where to go from here. Please help me complete this proof.   
 A: Check your second line, $r^2+2r\sqrt{2} + 2 = 6 + 4\sqrt{2} \ne r^2-4 -2\sqrt{2}$. 
It does, however, equal $ (r-2) (r+2 \sqrt{2}+2) = 0$ which would imply that $r = 2$ or $r= -2(1+\sqrt{2})$.
Thus $2$ is the only rational root of your polynomial and $\sqrt{6+4\sqrt{2}} -\sqrt{2}= 2 $ (since $r =\sqrt{6+4\sqrt{2}} -\sqrt{2}$ and $(r-2)=\sqrt{6+4\sqrt{2}} -\sqrt{2} -2 = 0$). 
A: \begin{align*}r=\sqrt{6+4\sqrt 2}-\sqrt 2&\Rightarrow r^2+2r\sqrt 2+2=6+4\sqrt2\iff r^2-4=2\sqrt 2(2-r)\\
&\Rightarrow(r^2-4)^2=8(r-2)^2\iff(r-2)^2((r+2)^2-8)=0\\
\end{align*}
The only rational root of this polynomial is $r=2$. We have to explain why $r$ can't be one of the irrational roots $\;-2(1\pm\sqrt2)$.
Anyway it can't be the negative one: $r>0$ since $6+4\sqrt 2>2$. The other root is ${}<1$ since $(1+2)^2-8>0$. However, it's easy to see that $r>1$ since $\sqrt{6+4\sqrt 2}-\sqrt 2>\sqrt 9-\sqrt 2>3-2$.
Consequently $r$ has to be equal to $2$.
Note: It's a standard exercise in high school to observe that 
$$\sqrt{6+4\sqrt 2}=\sqrt{(2+\sqrt2)^2}=2+\sqrt 2,\enspace\text{whence}\quad r=2+\sqrt 2-\sqrt 2.$$
