# 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.

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The question has been edited because the formulae were unclear. Can you please confirm if this is your actual question? – barto Jun 3 '14 at 17:41
Yes this is my actual question. – snivysteel Jun 4 '14 at 5:01
Simply substitute $m=n=1$ and you will see that the equation holds. Hence the answer becomes 2017. If the IMO is for smaller grades, then I'm sure this is the method. – N.S.JOHN Apr 5 at 3:21

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

$${x+2y+2014\over4-(x+2y)}={3+2014\over4-3}=2017$$

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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.

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this is an excellent observation – Jonathan Jun 4 '14 at 3:09
Or $(m,n) = (9,0)$. – user21820 Jun 4 '14 at 10:55
You are assuming that $\frac{\sqrt{m}+2\sqrt{n}+2014}{4-\sqrt{m}-2\sqrt{n}}$ does not depend on the values of $m$ and/or $n$, that is, you are assuming it is equal to some number instead of a function of $m$ or/and $n$. You would not get all the points for this kind of a solution in a math competition. I'm just adding this so that everyone knows - this fact wasn't emphasized in the answer. – user26486 Jun 4 '14 at 14:19

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

$\sqrt{m}=x$

$\sqrt{n}=y$

$x^2+4xy-2x-4y+4y^2=3$

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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I think my answer was maybe four seconds ahead of yours! Great minds, and all that.... – Barry Cipra Jun 3 '14 at 18:05

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 $$\frac{\sqrt{m}+2\sqrt{n}+2014}{4-\sqrt{m}-2\sqrt{n}}=\frac{3+2014}{4-3}=2017.$$

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