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

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A slightly different approach using symmetry:
 
A slightly different approach using symmetry:
−
let y = 10x -1/x
+
<br />
−
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
+
Let <math>y = 10x - 1/x</math>.
−
now it is easy to see that the equation reduces to 2(y^3+30y)+ (y^2+20) + 10 = 0
+
Notice that the equation can be rewritten (after dividing across by x^3) as  
−
OR 2y^3 +y^2 +60y+30 = 0
+
<cmath> 2( (10x)^3 - 1/x^3 ) + (10x)^2 + (1/x)^2 + 10 = 0 </cmath>
−
or y^2(2y+1) + 30(2y+1) = 0
+
Now it is easy to see that the equation reduces to  
−
which is (2y+1)(y^2+30)= 0
+
<cmath>
−
so for real solutions we have y= -1/2
+
\begin{align*}
−
and solve the quadratic in x to get final answer as \boxed{200}$.
+
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*}
 +
</cmath>
 +
so for real solutions we have <math>y = -1/2</math>. Solve the quadratic in <math>x</math> to get the final answer as <math>\boxed{200}</math>.
  
 
== Solution 2 (Complex Bash)==
 
== Solution 2 (Complex Bash)==

Revision as of 16:27, 24 August 2022

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 $x$.


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

Notice the original expression can be written as $2000x^6+100x^5-200x^4+200x^4+10x^3-20x^2+20x^2+x-2$.

Which equals to $(20x^2+x-2)(100x^4+10x^2+1)=0$

So our solution is to find what is the root for $20x^2+x-2=0$ since the determinant of $100t^2+10t+1<0$(Let $x^2=t$)

By solving the equation, we can get that $x = \frac{-1 \pm \sqrt{161}}{40}$ for a final answer of $-1 + 161 + 40 = \boxed{200}$

~bluesoul

Solution 4 (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

Video solution

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

See also

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