Difference between revisions of "2009 AMC 12A Problems/Problem 21"

Revision as of 22:09, 19 February 2012

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 $x^{12} + ax^8 + bx^4 + c$ ?

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

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 $4 + 2 + 2 = 8\ \mathbf{(C)}$ .

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

@@ Line 9: / Line 9: @@
 == Solutions ==
-=== Solution 1 ===
+=== Solution===
 From the three zeroes, we have <math>p(x) = (x - (2009 + 9002\pi i))(x - 2009)(x - 9002)</math>.