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. :)) |
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<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 13:25, 14 February 2009
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 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 ![$\pm \sqrt [4]{2009}$](//latex.artofproblemsolving.com/6/b/4/6b4295126a7f866b0b02c2a27927d6fbdb37db8c.png)
, and there are two complex roots.
: The real roots are ![$\pm \sqrt [4]{9002}$](//latex.artofproblemsolving.com/6/2/6/626b95f70ce28fb6a5281a84e3c3bf1a3469101f.png)
, and there are two complex roots.
: Clearly, all the fourth roots are going to be complex.
: The real roots are ![$\pm \sqrt [4]{2009}$](http://latex.artofproblemsolving.com/6/b/4/6b4295126a7f866b0b02c2a27927d6fbdb37db8c.png)
, and there are two complex roots.
: The real roots are ![$\pm \sqrt [4]{9002}$](http://latex.artofproblemsolving.com/6/2/6/626b95f70ce28fb6a5281a84e3c3bf1a3469101f.png)
, 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 | |
