Difference between revisions of "2009 AMC 12A Problems/Problem 21"
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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 22:09, 19 February 2012
Contents
[hide]Problem
Let , where , , and are complex numbers. Suppose that
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 |