Difference between revisions of "2009 AMC 12A Problems/Problem 21"
m (8 is choice C NOT choice D. Same mistake I made during the test. :)) |
(→Solution) |
||
Line 8: | Line 8: | ||
<math>\textbf{(A)}\ 4\qquad \textbf{(B)}\ 6\qquad \textbf{(C)}\ 8\qquad \textbf{(D)}\ 10\qquad \textbf{(E)}\ 12</math> | <math>\textbf{(A)}\ 4\qquad \textbf{(B)}\ 6\qquad \textbf{(C)}\ 8\qquad \textbf{(D)}\ 10\qquad \textbf{(E)}\ 12</math> | ||
− | == Solution == | + | == Solutions == |
+ | === Solution 1 === | ||
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>. | ||
− | Then <math>p(x^4) = (x^4 - (2009 + 9002\pi i))(x^4 - 2009)(x^4 - 9002)</math>. | + | Then <math>p(x^4) = (x^4 - (2009 + 9002\pi i))(x^4 - 2009)(x^4 - 9002)=x^{12}+ax^8+bx^4+c</math>. |
Let's do each factor case by case: | Let's do each factor case by case: | ||
*<math>x^4 - (2009 + 9002\pi i) = 0</math>: Clearly, all the fourth roots are going to be complex. | *<math>x^4 - (2009 + 9002\pi i) = 0</math>: Clearly, all the fourth roots are going to be complex. | ||
− | *<math>x^4 - 2009 = 0</math>: The real roots are <math>\pm \sqrt [4]{2009}</math>, there are two complex roots. | + | *<math>x^4 - 2009 = 0</math>: The real roots are <math>\pm \sqrt [4]{2009}</math>, and there are two complex roots. |
− | *<math>x^4 - 9002 = 0</math>: | + | *<math>x^4 - 9002 = 0</math>: The real roots are <math>\pm \sqrt [4]{9002}</math>, and there are two complex roots. |
So the answer is <math>4 + 2 + 2 = 8\ \mathbf{(C)}</math>. | So the answer is <math>4 + 2 + 2 = 8\ \mathbf{(C)}</math>. | ||
+ | === Solution 2 === | ||
== See also == | == See also == |
Revision as of 14:25, 14 February 2009
Problem
Let , where , , and are complex numbers. Suppose that
What is the number of nonreal zeros of ?
Solutions
Solution 1
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 .
Solution 2
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 |