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 == | + | == 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>. | ||
− | 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>. | ||
+ | ==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
Contents
[hide]Problem
Let , where , , and are complex numbers. Suppose that
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 , and there are two complex roots.
- : The real roots are , 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 |
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