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.
 A: HINT:
$$(\sqrt3\pm1)^2=4\pm2\sqrt3$$ and 
$$\sqrt{4\pm2\sqrt3}=+(\sqrt3\pm1)$$
A: 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.
A: Here is another way, a few lines shorter,$$\sqrt\frac{1}{(\sqrt{4+2\sqrt3}-\sqrt{4-2\sqrt{3}})^2}=\frac{1}{\sqrt{4}}=\frac{1}{2}$$
A: My attempt
$$\frac{1}{\sqrt{4+2\sqrt{3}} - \sqrt{4-2\sqrt{3}}}$$
$$ =\frac{\sqrt{4+2\sqrt{3}} + \sqrt{4-2\sqrt{3}}}{4+2\sqrt{3} - (4-2\sqrt{3})} $$
$$ =\frac{\sqrt{4+2\sqrt{3}} + \sqrt{4-2\sqrt{3}}}{4\sqrt{3}} $$
$$ = \sqrt{\left(\frac{\sqrt{4+2\sqrt{3}} + \sqrt{4-2\sqrt{3}}}{4\sqrt{3}} \right)^2}$$
$$ = \sqrt{\frac{4 + 2\sqrt{3} - (4 - 2\sqrt{3}) - 2\sqrt{4^2 - 4\times3}}{48}}$$
$$ = \sqrt{\frac{4 + 2\sqrt{3} + (4 - 2\sqrt{3}) + 2\sqrt{4^2 - 4\times3}}{48}}$$
$$ = \sqrt{\frac{12}{48}} = \sqrt{\frac{1}{4}}$$
$$\mathbf{= \frac{1}{2}}$$
A: Let $\zeta_{\pm}=\sqrt{4\pm 2\sqrt{3}}$. We have:
$$ \frac{1}{\zeta_{+}-\zeta_{-}}=\frac{\zeta_{+}+\zeta_{-}}{\zeta_{+}^2-\zeta_{-}^2}=\sqrt{\frac{\zeta_{+}^2+\zeta_{-}^2+2\zeta_{+}\zeta_{-}}{(\zeta_{+}^2-\zeta_{-}^2)^2}} $$
but $\zeta_{+}\zeta_{-}=2$, $\zeta_{+}^2-\zeta_{-}^2=4\sqrt{3}$ and $\zeta_{+}^2+\zeta_{-}^2=8$, hence:
$$ \frac{1}{\zeta_{+}-\zeta_{-}}=\sqrt{\frac{8+4}{48}}=\color{red}{\frac{1}{2}}.$$
A: Hint:
Suppose the middle term of a quadratic perfect square was either $2x$ or $-2x$. What would the first and last terms be?
Now, suppose that $x=\sqrt 3$.
