<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] Given: $m$ and $n$ are positive real numbers satisfying the equation   
$$m+4\sqrt{mn}-2\sqrt{m}-4\sqrt{n}+4n=3$$
Just to get a better feeling, substitute   
$\sqrt{m}=x$   
$\sqrt{n}=y$   
Now your equation becomes  
$x^2+4xy-2x-4y+4y^2=3$   
Combining first, second and last term of L.H.S. , we get,    
$(x+2y)^2-2(x+2y)=3$  
Substitute: $(x+2y)=t$ to get,  
$t^2-2t-3=0$  
$\implies t=3$ or $t=-1$   
But since $t=x+2y=\sqrt{m}+2\sqrt{n} \implies t \ge 0$ (Since $\sqrt{}$ gives positive value in its domain)   
$\implies \sqrt{m}+2\sqrt{n}=3$
$\implies \dfrac{\sqrt{m} +2\sqrt{n} +2014}{4-(\sqrt{m} +2\sqrt{n})}= \boxed{2017}$
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