Difference between revisions of "2009 AMC 12A Problems/Problem 21"
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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== | |
| − | |||
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>. | ||
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So the answer is <math>4 + 2 + 2 = 8\ \mathbf{(C)}</math>. | So the answer is <math>4 + 2 + 2 = 8\ \mathbf{(C)}</math>. | ||
| + | ==Alternative Thinking== | ||
| + | Consider the graph of <math>x^4</math>. It is similar to a parabola, but with a wider "base". First examine <math>x^4-2009</math> and <math>x^4-9002</math>. Clearly they are just being translated down some large amount. This will result in the <math>x</math>-axis being crossed twice, indicating <math>2</math> real zeroes. From the [[Fundamental Theorem of Algebra]] we know that a polynomial must have exactly as many roots as its highest degree, so we are left with <math>4-2</math> or <math>2</math> nonreal roots for each of the graphs. For the graph of <math>x^4-(2009+9002\pi i)</math>, it's not even possible to graph it on the Cartesian plane, so all <math>4</math> roots will be nonreal. This is <math>2+2+4 = 8</math> total nonreal roots <math>\Rightarrow \boxed{\text{C}}</math>. | ||
== See also == | == See also == | ||
{{AMC12 box|year=2009|ab=A|num-b=20|num-a=22}} | {{AMC12 box|year=2009|ab=A|num-b=20|num-a=22}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 09:25, 29 July 2020
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 
?
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 ![$\pm \sqrt [4]{2009}$](//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 
.
Alternative Thinking
Consider the graph of 
. It is similar to a parabola, but with a wider "base". First examine 
and 
. Clearly they are just being translated down some large amount. This will result in the 
-axis being crossed twice, indicating 
real zeroes. From the Fundamental Theorem of Algebra we know that a polynomial must have exactly as many roots as its highest degree, so we are left with 
or nonreal roots for each of the graphs. For the graph of 
, it's not even possible to graph it on the Cartesian plane, so all 
roots will be nonreal. This is 
total nonreal roots 
.
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
