Difference between revisions of "2009 AMC 12A Problems/Problem 21"
m (→Solutions) |
CakeIsEaten (talk | contribs) (→Solution 1) |
||
Line 9: | Line 9: | ||
== Solutions == | == Solutions == | ||
− | === Solution | + | === Solution=== |
From the three zeroes, we have <math>p(x) = (x - (2009 + 9002\pi i))(x - 2009)(x - 9002)</math>. | From the three zeroes, we have <math>p(x) = (x - (2009 + 9002\pi i))(x - 2009)(x - 9002)</math>. | ||
Revision as of 23:09, 19 February 2012
Contents
Problem
Let , where
,
, and
are complex numbers. Suppose that
![\[p(2009 + 9002\pi i) = p(2009) = p(9002) = 0\]](http://latex.artofproblemsolving.com/1/7/2/1724d4fbd29ca94c593585be9881983817b0e269.png)
What is the number of nonreal zeros of ?
Solutions
Solution
From the three zeroes, we have .
Then .
Let's do each factor case by case:
: Clearly, all the fourth roots are going to be complex.
: The real roots are
, and there are two complex roots.
: The real roots are
, and there are two complex roots.
So the answer is .
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 |