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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.

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HINT:

$$(\sqrt3\pm1)^2=4\pm2\sqrt3$$ and

$$\sqrt{4\pm2\sqrt3}=+(\sqrt3\pm1)$$

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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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Here is another way, a few lines shorter,$$\sqrt\frac{1}{(\sqrt{4+2\sqrt3}-\sqrt{4-2\sqrt{3}})^2}=\frac{1}{\sqrt{4}}=\frac{1}{2}$$

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My attempt

$$\frac{1}{\sqrt{4+2\sqrt{3}} - \sqrt{4-2\sqrt{3}}}$$

$$ =\frac{\sqrt{4+2\sqrt{3}} + \sqrt{4-2\sqrt{3}}}{4+2\sqrt{3} - (4-2\sqrt{3})} $$

$$ =\frac{\sqrt{4+2\sqrt{3}} + \sqrt{4-2\sqrt{3}}}{4\sqrt{3}} $$

$$ = \sqrt{\left(\frac{\sqrt{4+2\sqrt{3}} + \sqrt{4-2\sqrt{3}}}{4\sqrt{3}} \right)^2}$$

$$ = \sqrt{\frac{4 + 2\sqrt{3} - (4 - 2\sqrt{3}) - 2\sqrt{4^2 - 4\times3}}{48}}$$

$$ = \sqrt{\frac{4 + 2\sqrt{3} + (4 - 2\sqrt{3}) + 2\sqrt{4^2 - 4\times3}}{48}}$$

$$ = \sqrt{\frac{12}{48}} = \sqrt{\frac{1}{4}}$$

$$\mathbf{= \frac{1}{2}}$$

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    $\begingroup$ The idea of computing the square (and then taking the square root) is good, but you could've done that immediately to find the same result a lot faster. $\endgroup$ – StackTD May 18 '16 at 12:57
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    $\begingroup$ Agreed. I blame the school curricula that, for reasons I'll never understand, teach students that denominators should always be rationalized. $\endgroup$ – tilper May 18 '16 at 13:19
  • $\begingroup$ Thanks for that (and blaming my school, i hate my school too) but it wasn't taught in school, they don't even do such questions. I learnt it when I saw someone doing something similar. $\endgroup$ – dark32 May 18 '16 at 14:04
  • $\begingroup$ Oh I just mean schools in general. It's how I was taught when I was in school, and then later I lived in various parts of the country and realized that it's a pervasive issue. This is just in the US though. Maybe other countries do it better. $\endgroup$ – tilper May 18 '16 at 19:07
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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: $$ \frac{1}{\zeta_{+}-\zeta_{-}}=\sqrt{\frac{8+4}{48}}=\color{red}{\frac{1}{2}}.$$

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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 $x=\sqrt 3$.

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