Show that $\sqrt{4 + 2\sqrt{3}} - \sqrt{3}$ is rational. 
Show that $\sqrt{4 + 2\sqrt{3}} - \sqrt{3}$ is rational.

I've tried to attempt algebra on this problem. I noticed that there is some kind of nesting effect when trying to solve this. Please help me to understand how to attempt to denest this number.
Any help would be greatly appreciated.
 A: Such square roots can  be computed by a very Simple Denesting Rule:
Here $\, 4+2\sqrt 3\ $ has norm $= 4.\:$ $\rm\ \color{blue}{Subtracting\ out}\,\ \sqrt{norm}\ = 2\,\ $ yields $\,\  2+2\sqrt 3\:$
which has $\, {\rm\ \sqrt{trace}}\, =\, \sqrt{4}\, =\, 2.\,\ \  \rm \color{brown}{Dividing\ it\ out}\,\ $ of the above  yields $\ \ 1+\sqrt 3$ 
Remark $\ $ Many more worked examples are in prior posts on this denesting rule.
A: Let $x=\sqrt{4+2\sqrt{3}}-\sqrt{3}$. Then:
$$(x+\sqrt3)^2=4+2\sqrt{3}\implies x^2-1=2(1-x)\sqrt3\implies(x-1)(x+1+2\sqrt3)=0$$
So, certainly $x=1$ or $x=-1-2\sqrt3$. But a moment's thought (e.g. considering that $x>0$) convinces us that the first of these must be the case - i.e., $x=1$ (a known rational number).
A: Personally I like this method which is the same thing as the other answers.$\sqrt{4+2\sqrt{3}}=\sqrt{a}+\sqrt b$ squaring both sides $4+2\sqrt 3=a+2\sqrt{ab}+b$ for this to be true we must equate the parts with and without the radicals so $a+b=4$ and $2\sqrt{ab}=2\sqrt{3}$ so $ab=3$ now we have a system of equations that can be solved through substitution $a(4-a)=3\to a^2-4a+3=0$ which has roots  3, 1. So if a=1 then b=3 and vice versa if a=3. So our answer is $\sqrt{4+2\sqrt{3}}-\sqrt 3=1+\sqrt{3}-\sqrt3$
A: Note that $4+2\sqrt{3}=(1+\sqrt{3})^2$
A: Method 1: Consider this denesting algorithm:

Denested square roots: Given a radical of the form $\sqrt{X\pm {Y}}$ with $X,Y\in\mathbb{R}$ and $X>Y$, we have a possible simplification as$$\sqrt{X\pm Y}=\sqrt{\dfrac {X+\sqrt{X^2-Y^2}}2}\pm\sqrt{\dfrac {X-\sqrt{X^2-Y^2}}2}\tag1$$

Using $(1)$ on $\sqrt{4+2\sqrt3}$, we have$$\sqrt{4+2\sqrt3}=\sqrt{\dfrac {4+2}2}+\sqrt{\dfrac {4-2}2}=\sqrt3+1\tag2$$
So the original expression becomes$$\color{brown}{\sqrt{4+2\sqrt3}}-\sqrt3=\color{brown}{\sqrt3+1}-\sqrt3=1$$
Which is rational.

Method 2: If you don't like $(1)$ and think it's too complicated, I present you an alternative method. Simply set $\sqrt{4+2\sqrt{3}}-\sqrt3$ equal to a variable and simplify!
Here, we have$$\begin{align*} & \sqrt{4+2\sqrt3}-\sqrt3=\alpha\\ & \sqrt{4+2\sqrt3}=\alpha+\sqrt3\\ & 4+2\sqrt3=(\alpha+\sqrt3)^2\\ & 4+2\sqrt3=\alpha^2+3+2\alpha\sqrt3\end{align*}$$
To solve for $\alpha$, we have $2\alpha\sqrt3=2\sqrt3\implies\alpha=1$. Checking with the other half, $\alpha^2+3=1+3=4$ holds. Hence,$$\sqrt{4+2\sqrt3}-\sqrt3=1$$
A: Many questions with
sum or difference of square roots
can be solved with
conjugating.
So,
if $s = \sqrt{4 + 2\sqrt{3}} - \sqrt{3}$,
and $t = \sqrt{4 + 2\sqrt{3}} + \sqrt{3}$,
$\begin{array}\\
st
&=(\sqrt{4 + 2\sqrt{3}} - \sqrt{3})(\sqrt{4 + 2\sqrt{3}} + \sqrt{3})\\
&=4 + 2\sqrt{3}-3\\
&=1 + 2\sqrt{3}\\
\end{array}
$
Since
$t-s = 2\sqrt{3}$,
$st = 1+t-s$
or
$s(t+1) = 1+t$,
and, by magic,
we get
$s = 1$
(unless $t+1 = 0$
which it does not since
$t > 0$).
I am surprised that this worked so well.
I will suppress my need to generalize
and submit this as is.
