AMC12 2019 A

AMC12 2019 A · Q14

AMC12 2019 A · Q14. It mainly tests Complex numbers (rare), Coordinate geometry.

For a certain complex number $c$, the polynomial $P(x) = (x^2 - 2x + 2)(x^2 - cx + 4)(x^2 - 4x + 8)$ has exactly 4 distinct roots. What is $|c|$?

对于某个复数$c$，多项式$P(x) = (x^2 - 2x + 2)(x^2 - cx + 4)(x^2 - 4x + 8)$恰有4个不同根。求$|c|$？

(A) 2 2

(B) $\sqrt{6}$ $\sqrt{6}$

(D) 3 3

(E) $\sqrt{10}$ $\sqrt{10}$

Answer

Correct choice: (E)

正确答案：（E）

Solution

Answer (E): The roots of $x^2-2x+2$ are $1\pm i$, and the roots of $x^2-4x+8$ are $2\pm 2i=2(1\pm i)$. The two roots of $x^2-cx+4$ must be chosen from among those four numbers, and their product must be $4$. The only two pairs of possible roots with a product of $4$ are $\{1+i,\,2-2i\}$ and $\{1-i,\,2+2i\}$, and in each case the value of $c$ is the sum of the roots. Thus the only possible values of $c$ are $3\pm i$, and in both cases $\lvert c\rvert=\sqrt{10}$.

答案（E）：$x^2-2x+2$ 的根是 $1\pm i$，$x^2-4x+8$ 的根是 $2\pm 2i=2(1\pm i)$。$x^2-cx+4$ 的两个根必须从这四个数中选取，并且它们的乘积必须为 $4$。乘积为 $4$ 的可能根对只有两组：$\{1+i,\,2-2i\}$ 和 $\{1-i,\,2+2i\}$；在每种情况下，$c$ 的值等于两根之和。因此 $c$ 的唯一可能取值为 $3\pm i$，并且两种情况下都有 $\lvert c\rvert=\sqrt{10}$。

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