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2001 AIME I Problems/Problem 3: Difference between revisions

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== Problem ==
== Problem ==
Find the sum of the roots, real and non-real, of the equation <math>x^{2001}+\left(\frac 12-x\right)^{2001}=0</math>, given that there are no multiple roots.
Find the sum of the [[root]]s, real and non-real, of the equation <math>x^{2001}+\left(\frac 12-x\right)^{2001}=0</math>, given that there are no multiple roots.


== Solution ==
== Solution ==
From [[Vieta's formulas]], we just need to find the first two terms.
From [[Vieta's formulas]], in a [[polynomial]] of the form <math>a_nx^n + a_{n-1}x^{n-1} + \cdots + a_0 = 0</math>, then the sum of the roots is <math>\frac{-a_{n-1}}{a_n}</math>.  


From the [[Binomial Theorem]], the first term of <math>left(\frac 12-x\right)^{2001}</math> is <math>-x^{2001}</math>, but <math>x^{2001}+-x^{2001}=0</math>, so the first term has <math>x^{2000}</math> in it, not <math>x^{2001}</math>. So we find that term, and the term with <math>x^{1999}</math>.
From the [[Binomial Theorem]], the first term of <math>\left(\frac 12-x\right)^{2001}</math> is <math>-x^{2001}</math>, but <math>x^{2001}+-x^{2001}=0</math>, so the term with the largest degree is <math>x^{2000}</math>. So we need the coefficient of that term, as well as the coefficient of <math>x^{1999}</math>.


<math>x^{2000}*\binom{2001}{1}*\frac{1}{2}=\frac{2001x^{2000}}{2}</math>
<cmath>\begin{align*}\binom{2001}{1} \cdot (-x)^{2000} \cdot \left(\frac{1}{2}\right)^1&=\frac{2001x^{2000}}{2}\\
\binom{2001}{2} \cdot (-x)^{1999} \cdot \left(\frac{1}{2}\right)^2 &=\frac{-x^{1999}*2001*2000}{8}=-2001 \cdot 250x^{1999}
\end{align*}</cmath>


<math>-x^{1999}*\binom{2001}{2}*\frac{1}{4}=\frac{-x^{1999}*2001*2000}{8}=-x^{1999}2001*250</math>
Applying Vieta's formulas, we find that the sum of the roots is <math>-\frac{-2001 \cdot 250}{\frac{2001}{2}}=250 \cdot 2=\boxed{500}</math>.


Applying Vieta's Formulas, we get that the sum of the roots is
<math>-\frac{-2001*250}{\frac{2001}{2}}=250*2=\boxed{500}</math>
== See also ==
== See also ==
{{AIME box|year=2001|n=I|num-b=2|num-a=4}}
{{AIME box|year=2001|n=I|num-b=2|num-a=4}}
[[Category:Intermediate Algebra Problems]]

Revision as of 16:34, 11 June 2008

Problem

Find the sum of the roots, real and non-real, of the equation $x^{2001}+\left(\frac 12-x\right)^{2001}=0$, given that there are no multiple roots.

Solution

From Vieta's formulas, in a polynomial of the form $a_nx^n + a_{n-1}x^{n-1} + \cdots + a_0 = 0$, then the sum of the roots is $\frac{-a_{n-1}}{a_n}$.

From the Binomial Theorem, the first term of $\left(\frac 12-x\right)^{2001}$ is $-x^{2001}$, but $x^{2001}+-x^{2001}=0$, so the term with the largest degree is $x^{2000}$. So we need the coefficient of that term, as well as the coefficient of $x^{1999}$.

\begin{align*}\binom{2001}{1} \cdot (-x)^{2000} \cdot \left(\frac{1}{2}\right)^1&=\frac{2001x^{2000}}{2}\\ \binom{2001}{2} \cdot (-x)^{1999} \cdot \left(\frac{1}{2}\right)^2 &=\frac{-x^{1999}*2001*2000}{8}=-2001 \cdot 250x^{1999} \end{align*}

Applying Vieta's formulas, we find that the sum of the roots is $-\frac{-2001 \cdot 250}{\frac{2001}{2}}=250 \cdot 2=\boxed{500}$.

See also

2001 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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