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 "2009 AMC 12A Problems/Problem 21"

m (→‎Solutions)
 
Line 20: Line 20:
 
So the answer is <math>4 + 2 + 2 = 8\ \mathbf{(C)}</math>.
 
So the answer is <math>4 + 2 + 2 = 8\ \mathbf{(C)}</math>.
  
 +
==Alternative Thinking==
 +
Consider the graph of <math>x^4</math>. It is similar to a parabola, but with a wider "base". First examine <math>x^4-2009</math> and <math>x^4-9002</math>. Clearly they are just being translated down some large amount. This will result in the <math>x</math>-axis being crossed twice, indicating <math>2</math> real zeroes. From the [[Fundamental Theorem of Algebra]] we know that a polynomial must have exactly as many roots as its highest degree, so we are left with <math>4-2</math> or <math>2</math> nonreal roots for each of the graphs. For the graph of <math>x^4-(2009+9002\pi i)</math>, it's not even possible to graph it on the Cartesian plane, so all <math>4</math> roots will be nonreal. This is <math>2+2+4 = 8</math> total nonreal roots <math>\Rightarrow \boxed{\text{C}}</math>.
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2009|ab=A|num-b=20|num-a=22}}
 
{{AMC12 box|year=2009|ab=A|num-b=20|num-a=22}}

Latest revision as of 09:25, 29 July 2020

Problem

Let $p(x) = x^3 + ax^2 + bx + c$, where $a$, $b$, and $c$ are complex numbers. Suppose that

\[p(2009 + 9002\pi i) = p(2009) = p(9002) = 0\]

What is the number of nonreal zeros of $x^{12} + ax^8 + bx^4 + c$?

$\textbf{(A)}\ 4\qquad \textbf{(B)}\ 6\qquad \textbf{(C)}\ 8\qquad \textbf{(D)}\ 10\qquad \textbf{(E)}\ 12$

Solution

From the three zeroes, we have $p(x) = (x - (2009 + 9002\pi i))(x - 2009)(x - 9002)$.

Then $p(x^4) = (x^4 - (2009 + 9002\pi i))(x^4 - 2009)(x^4 - 9002)=x^{12}+ax^8+bx^4+c$.

Let's do each factor case by case:

  • $x^4 - (2009 + 9002\pi i) = 0$: Clearly, all the fourth roots are going to be complex.
  • $x^4 - 2009 = 0$: The real roots are $\pm \sqrt [4]{2009}$, and there are two complex roots.
  • $x^4 - 9002 = 0$: The real roots are $\pm \sqrt [4]{9002}$, and there are two complex roots.

So the answer is $4 + 2 + 2 = 8\ \mathbf{(C)}$.

Alternative Thinking

Consider the graph of $x^4$. It is similar to a parabola, but with a wider "base". First examine $x^4-2009$ and $x^4-9002$. Clearly they are just being translated down some large amount. This will result in the $x$-axis being crossed twice, indicating $2$ real zeroes. From the Fundamental Theorem of Algebra we know that a polynomial must have exactly as many roots as its highest degree, so we are left with $4-2$ or $2$ nonreal roots for each of the graphs. For the graph of $x^4-(2009+9002\pi i)$, it's not even possible to graph it on the Cartesian plane, so all $4$ roots will be nonreal. This is $2+2+4 = 8$ total nonreal roots $\Rightarrow \boxed{\text{C}}$.

See also

2009 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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=2009_AMC_12A_Problems/Problem_21&oldid=129723"