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.
It can be helpful to get rid of the square root symbols. If you let $m=x^2$ and $n=y^2$ with $x$ and $y$ understood to be positive, the equation becomes $x^2+4xy-2x-4y+4y^2=3$, which can be rewritten as
$$(x+2y)^2-2(x+2y)=3$$ or $$u^2-2u-3=0\qquad\text{with}\quad u=x+2y$$
The quadratic factors as $(u-3)(u+1)=0$. Recalling that $x=\sqrt m$ and $y=\sqrt n$ are supposed to be positive, this gives $u=x+2y=3$ as the only meaningful solution. We also see that the value we're after is
$${x+2y+2014\over4-(x+2y)}={3+2014\over4-3}=2017$$
The trivial answer is to realize that m = 1 and n = 1 meets the first criteria, so you can use those in the second equation.
But that's not much fun.
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)
$\implies \sqrt{m}+2\sqrt{n}=3$
$\implies \dfrac{\sqrt{m} +2\sqrt{n} +2014}{4-(\sqrt{m} +2\sqrt{n})}= \boxed{2017}$
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 $$ \frac{\sqrt{m}+2\sqrt{n}+2014}{4-\sqrt{m}-2\sqrt{n}}=\frac{3+2014}{4-3}=2017. $$
