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 $x$ . Note: After dividing the equation with $x^3$ , divide again with $10$ before substituting it with $t$ to get it right.

A slightly different approach using symmetry:
Let $y = 10x - 1/x$ . Notice that the equation can be rewritten (after dividing across by $x^3$ ) as $2( (10x)^3 - 1/x^3 ) + (10x)^2 + (1/x)^2 + 10 = 0$ Now it is easy to see that the equation reduces to $\begin{align*} 2(y^3+30y)+ (y^2+20) + 10 = 0 \\ 2y^3 + y^2 + 60y + 30 = 0 \\ y^2(2y+1) + 30(2y+1) = 0 \\ (2y+1)(y^2+30)= 0 \\ \end{align*}$ so for real solutions we have $y = -1/2$ . Solve the quadratic in $x$ to get the final answer as $\boxed{200}$ .

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

Solution 3 (Geometric Series)

Observe that the given equation may be rearranged as $2000x^6-2+(100x^5+10x^3+x)=0$ . The expression in parentheses is a geometric series with common factor $10x^2$ . Using the geometric sum formula, we rewrite as $2000x^6-2+\frac{1000x^7-x}{10x^2-1}=0, 10x^2-1\neq0$ . Factoring a bit, we get $2(1000x^6-1)+(1000x^6-1)\frac{x}{10x^2-1}=0, 10x^2-1\neq0 \implies$ $(1000x^6-1)(2+\frac{x}{10x^2-1})=0, 10x^2-1\neq0$ . Note that setting $1000x^6-1=0$ gives $10x^2-1=0$ , which is clearly extraneous. Hence, we set $2+\frac{x}{10x^2-1}=0$ and use the quadratic formula to get the desired root $x=\frac{-1+\sqrt{161}}{40} \implies -1+161+40=\boxed{200}$

~keeper1098

Solution 4

Let $y=10x^2$ . The equation then becomes $2y^3+xy^2+xy+x-2=0$ , or $x(y^2+y+1)=2(1-y^3)$ .

By the difference of cubes formula, $2(1-y^3)=2(1-y)(1+y+y^2)$ , so we have two cases: either $y^2+y+1=0$ , or $x=2(1-y)$ . We start with the second formula as it is simpler.

Solving with the quadratic formula after re-substitution, we see that $x=\frac{-1\pm\sqrt{161}}{40}$ , so the answer is $-1+161+40=\boxed{200}$ .

For the sake of completeness, if we check the other equation, we come to the conclusion that $y=\frac{-1\pm i\sqrt{3}}{2}$ , so no real solution exists for $x$ . Thus our solution is correct.

~eevee9406

Video solution

https://www.youtube.com/watch?v=mAXDdKX52TM

2000 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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