# Difference between revisions of "2000 AIME II Problems/Problem 13"

## Problem

The equation $2000x^6+100x^5+10x^3+x-2=0$ has exactly two real roots, one of which is $\frac{m+\sqrt{n}}r$, where $m$, $n$ and $r$ are integers, $m$ and $r$ are relatively prime, and $r>0$. Find $m+n+r$.

## Solution

We may factor the equation as:[1]

\begin{align*} 2000x^6+100x^5+10x^3+x-2&=0\\ 2(1000x^6-1) + x(100x^4+10x^2+1)&=0\\ 2[(10x^2)^3-1]+x[(10x^2)^2+(10x^2)+1]&=0\\ 2(10x^2-1)[(10x^2)^2+(10x^2)+1]+x[(10x^2)^2+(10x^2)+1]&=0\\ (20x^2+x-2)(100x^4+10x^2+1)&=0\\ \end{align*}

Now $100x^4+10x^2+1\ge 1>0$ for real $x$. Thus the real roots must be the roots of the equation $20x^2+x-2=0$. By the quadratic formula the roots of this are:

$$x=\frac{-1\pm\sqrt{1^2-4(-2)(20)}}{40} = \frac{-1\pm\sqrt{1+160}}{40} = \frac{-1\pm\sqrt{161}}{40}.$$

Thus $r=\frac{-1+\sqrt{161}}{40}$, and so the final answer is $-1+161+40 = \boxed{200}$.

^ A well-known technique for dealing with symmetric (or in this case, nearly symmetric) polynomials is to divide through by a power of $x$ with half of the polynomial's degree (in this case, divide through by $x^3$), and then to use one of the substitutions $t = x \pm \frac{1}{x}$. In this case, the substitution $t = x\sqrt{10} - \frac{1}{x\sqrt{10}}$ gives $t^2 + 2 = 10x^2 + \frac 1{10x^2}$ and $2\sqrt{10}(t^3 + 3t) = 200x^3 - \frac{2}{10x^3}$, which reduces the polynomial to just $(t^2 + 3)\left(2\sqrt{10}t + 1\right) = 0$. Then one can backwards solve for $x$.

## Solution 2 (Complex Bash)

It would be really nice if the coefficients were symmetrical. What if we make the substitution, $x = -\frac{i}{\sqrt{10}}y$. The the polynomial becomes

$-2y^6 - (\frac{i}{\sqrt{10}})y^5 + (\frac{i}{\sqrt{10}})y^3 - (\frac{i}{\sqrt{10}})y - 2$

It's symmetric! Dividing by $y^3$ and rearranging, we get

$-2(y^3 + \frac{1}{y^3}) - (\frac{i}{\sqrt{10}})(y^2 + \frac{1}{y^2}) + (\frac{i}{\sqrt{10}})$

Now, if we let $z = y + \frac{1}{y}$, we can get the equations

$z = y + \frac{1}{y}$

$z^2 - 2 = y^2 + \frac{1}{y^2}$

and

$z^3 - 3z = y^3 + \frac{1}{y^3}$

(These come from squaring $z$ and subtracting $2$, then multiplying that result by $z$ and subtracting $z$) Plugging this into our polynomial, expanding, and rearranging, we get

$-2z^3 - (\frac{i}{\sqrt{10}})z^2 + 6z + (\frac{3i}{\sqrt{10}})$

Now, we see that the two $i$ terms must cancel in order to get this polynomial equal to $0$, so what squared equals 3? Plugging in $z = \sqrt{3}$ into the polynomial, we see that it works! Is there something else that equals 3 when squared? Trying $z = -\sqrt{3}$, we see that it also works! Great, we use long division on the polynomial by

$(z - \sqrt{3})(z + \sqrt{3}) = (z^2 - 3)$ and we get

$2z -(\frac{i}{\sqrt{10}}) = 0$.

We know that the other two solutions for z wouldn't result in real solutions for $x$ since we have to solve a quadratic with a negative discriminant, then multiply by $-(\frac{i}{\sqrt{10}})$. We get that $z = (\frac{i}{-2\sqrt{10}})$. Solving for $y$ (using $y + \frac{1}{y} = z$) we get that $y = \frac{-i \pm \sqrt{161}i}{4\sqrt{10}}$, and multiplying this by $-(\frac{i}{\sqrt{10}})$ (because $x = -(\frac{i}{\sqrt{10}})y$) we get that $x = \frac{-1 \pm \sqrt{161}}{40}$ for a final answer of $-1 + 161 + 40 = \boxed{200}$

-Grizzy