<s>[INST] Simplify $\frac{1}{\sqrt{4+2\sqrt{3}} - \sqrt{4-2\sqrt{3}}}$ Simplify $$\frac{1}{\sqrt{4+2\sqrt{3}} - \sqrt{4-2\sqrt{3}}}$$
I know there is another easier method except the one I answered. I cannot find it. Can you please help? Thanks in advance.
 [/INST] For a general approach related to lab's hint:  We want to (hopefully) simplify $\sqrt{4 \pm 2\sqrt{3}}$.  Let's look at $\sqrt{4 + 2\sqrt{3}}.$
$$
\sqrt{4 + 2\sqrt{3}} = a + b\sqrt{3}
$$
Let's find $a$ and $b$.  First square both sides.
$$
4 + 2\sqrt{3} = a^2 + 3b^2 + 2ab\sqrt{3}.
$$
This gives us two equations:
$$
  \left\{
    \begin{array}{rcl}
      a^2 + 3b^2 &= & 4\\
      2ab &= & 2
    \end{array}
  \right.
$$
From the second equation we can say $a = 1/b$.  Plugging this into the first equation gives us
$$
  \frac{1}{b^2} + 3b^2 = 4.
$$
This can be rewritten as $3b^4 - 4b^2 + 1 = 0$.  This is an equation of "quadratic type" (because it's quadratic in $b^2$) and when we solve it (by factoring or by using the quadratic formula) we get $b^2 = 1$ and $b^2 = 1/3$.  Therefore $b = \pm 1$ or $b = \pm 1/\sqrt 3$.  Since $a = 1/b$, this yields four possibilities for $a + b\sqrt{3}$:
$$
  \underbrace{1 + \sqrt{3}}_{b=1}, \qquad 
  \underbrace{-1 - \sqrt{3}}_{b=-1}, \qquad 
  \underbrace{\sqrt{3} + 1}_{b = 1/\sqrt{3}}, \qquad
  \underbrace{-\sqrt{3} - 1}_{b = -1/\sqrt{3}}
$$
Notice that this is actually only two distinct possibilities:  $1 + \sqrt{3}$ and $-1 - \sqrt{3}$.  Because, by convention, $\sqrt{4 + 2\sqrt{3}}$ must be positive, we can throw out the extraneous $-1 - \sqrt{3}$ solution and we end up with
$$ \sqrt{4 + 2\sqrt{3}} = 1 + \sqrt{3}.$$
A similar analysis will show that
$$ \sqrt{4 - 2 \sqrt{3}} = -1 + \sqrt{3}.$$
Therefore we have:
$$
  \frac{1}{\sqrt{4 + 2\sqrt{3}} - \sqrt{4 - 2\sqrt{3}}} = \frac{1}{(1+\sqrt{3}) - (-1 + \sqrt{3})} = \frac{1}{2}
$$
It's subjective whether or not this is really better/easier than what you did.
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