Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2001 AMC 12 Problems/Problem 23"

(→‎See Also)
(→‎Solution)
Line 26: Line 26:
  
 
(As an example, the polynomial <math>x^4 - 2x^3 + 4x^2 - 3x</math> has zeroes <math>0</math>, <math>1</math>, and <math>\frac {1 \pm i \sqrt {11}}{2}</math>.)
 
(As an example, the polynomial <math>x^4 - 2x^3 + 4x^2 - 3x</math> has zeroes <math>0</math>, <math>1</math>, and <math>\frac {1 \pm i \sqrt {11}}{2}</math>.)
 +
 +
== Solution 2 ==
 +
 +
By Vieta, we know that the product of all four zeros of the polynomial equals the constant at the end of the polynomial.
 +
We also know that the two imaginary roots are a conjugate pair (I.E if one is a+bi, the other is a-bi).
 +
So the two imaginary roots must multiply to give you an integer.
 +
Taking the 5 answers into hand, we find that <math>\boxed{\frac {1 + i \sqrt {11}}{2}}</math> is our only integer giving solution.
  
 
== See Also ==
 
== See Also ==

Revision as of 00:08, 20 July 2013

Problem

A polynomial of degree four with leading coefficient 1 and integer coefficients has two zeros, both of which are integers. Which of the following can also be a zero of the polynomial?

$\text{(A) }\frac {1 + i \sqrt {11}}{2} \qquad \text{(B) }\frac {1 + i}{2} \qquad \text{(C) }\frac {1}{2} + i \qquad \text{(D) }1 + \frac {i}{2} \qquad \text{(E) }\frac {1 + i \sqrt {13}}{2}$

Solution

Let the polynomial be $P$ and let the two integer zeros be $z_1$ and $z_2$. We can then write $P(x)=(x-z_1)(x-z_2)(x^2+ax+b)$ for some integers $a$ and $b$.

If a complex number $p+qi$ with $q\not=0$ is a root of $P$, it must be the root of $x^2+ax+b$, and the other root of $x^2+ax+b$ must be $p-qi$.

We can then write $x^2+ax+b = (x-p-qi)(x-p+qi) = (x-p)^2 - (qi)^2 = x^2 - 2px + p^2 + q^2$.

We can now examine each of the five given complex numbers, and find the one for which the values $-2p$ and $p^2+q^2$ are integers. This is $\boxed{\frac {1 + i \sqrt {11}}{2}}$, for which we have $-2p = -2\cdot\frac 12 = -1$ and $p^2+q^2 = \left( \frac 12 \right)^2 + \left( \frac {\sqrt{11}}2 \right)^2 = \frac 14 + \frac {11}4 = \frac {12}4 = 3$.

(As an example, the polynomial $x^4 - 2x^3 + 4x^2 - 3x$ has zeroes $0$, $1$, and $\frac {1 \pm i \sqrt {11}}{2}$.)

Solution 2

By Vieta, we know that the product of all four zeros of the polynomial equals the constant at the end of the polynomial. We also know that the two imaginary roots are a conjugate pair (I.E if one is a+bi, the other is a-bi). So the two imaginary roots must multiply to give you an integer. Taking the 5 answers into hand, we find that $\boxed{\frac {1 + i \sqrt {11}}{2}}$ is our only integer giving solution.

See Also

2001 AMC 12 (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2001_AMC_12_Problems/Problem_23&oldid=56669"