<s>[INST] Math Olympiad Algebraic Question Comprising Square Roots 
If $m$ and $n$ are positive real numbers satisfying the equation $$m+4\sqrt{mn}-2\sqrt{m}-4\sqrt{n}+4n=3$$ 
  find the value of $$\frac{\sqrt{m}+2\sqrt{n}+2014}{4-\sqrt{m}-2\sqrt{n}}$$

I came across this question in a Math Olympiad Competition and had no idea how to solve it. Can anyone help? Thanks.
 [/INST] We have
$$
\begin{align*}
m+4\sqrt{mn}-2\sqrt{m}-4\sqrt{n}+4n&=3\\
m+4\sqrt{mn}+4n-2\sqrt{m}-4\sqrt{n}-3&=0\\
(\sqrt{m}+2\sqrt{n})^2-2(\sqrt{m}+2\sqrt{n})-3&=0\\
(\sqrt{m}+2\sqrt{n}-3)(\sqrt{m}+2\sqrt{n}+1)&=0
\end{align*}
$$
which gives
$$
\sqrt{m}+2\sqrt{n}=3\,\,\text{or}\,\,\sqrt{m}+2\sqrt{n}=-1.
$$
We can disregard the second solution as $m$ and $n$ are real so
$$
\frac{\sqrt{m}+2\sqrt{n}+2014}{4-\sqrt{m}-2\sqrt{n}}=\frac{3+2014}{4-3}=2017.
$$
</s>