AMC12 2009 A

AMC12 2009 A · Q21

AMC12 2009 A · Q21. It mainly tests Quadratic equations, Complex numbers (rare).

Let $p(x) = x^3 + ax^2 + bx + c$, where $a$, $b$, and $c$ are complex numbers. Suppose that \[p(2009 + 9002\pi i) = p(2009) = p(9002) = 0\] What is the number of nonreal zeros of $x^{12} + ax^8 + bx^4 + c$?

设 $p(x) = x^3 + ax^2 + bx + c$，其中 $a$、$b$ 和 $c$ 是复数。假设 \[p(2009 + 9002\pi i) = p(2009) = p(9002) = 0\] 那么 $x^{12} + ax^8 + bx^4 + c$ 的非实零点个数是多少？

(A) 4 4

(B) 6 6

(D) 10 10

(E) 12 12

Answer

Correct choice: (C)

正确答案：（C）

Solution

From the three zeroes, we have $p(x) = (x - (2009 + 9002\pi i))(x - 2009)(x - 9002)$. Then $p(x^4) = (x^4 - (2009 + 9002\pi i))(x^4 - 2009)(x^4 - 9002)=x^{12}+ax^8+bx^4+c$. Let's do each factor case by case: - $x^4 - (2009 + 9002\pi i) = 0$: Clearly, all the fourth roots are going to be complex. - $x^4 - 2009 = 0$: The real roots are $\pm \sqrt [4]{2009}$, and there are two complex roots. - $x^4 - 9002 = 0$: The real roots are $\pm \sqrt [4]{9002}$, and there are two complex roots. So the answer is $4 + 2 + 2 = 8\ \mathbf{(C)}$.

由三个零点可得 \[p(x) = (x - (2009 + 9002\pi i))(x - 2009)(x - 9002).\] 于是 \[p(x^4) = (x^4 - (2009 + 9002\pi i))(x^4 - 2009)(x^4 - 9002)=x^{12}+ax^8+bx^4+c.\] 分别讨论每个因子： - $x^4 - (2009 + 9002\pi i) = 0$：显然其四次方根全为复数（非实）。 - $x^4 - 2009 = 0$：实根为 $\pm \sqrt [4]{2009}$，另外还有两个复根。 - $x^4 - 9002 = 0$：实根为 $\pm \sqrt [4]{9002}$，另外还有两个复根。因此非实零点总数为 $4 + 2 + 2 = 8\ \mathbf{(C)}$。

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