1992 AMC 12 第 28 题

Question

令

i=\sqrt{-1}

。方程

z^2-z=5-5i

的两个根的实部之积为

Let

i=\sqrt{-1}.

The product of the real parts of the roots of

z^2-z=5-5i

is

-25

-6

-5

\frac14

25

使用求根公式，并令

\sqrt{21-20i}=a+bi

Apply the quadratic formula and write

\sqrt{21-20i}=a+bi

解

a^2-b^2=21

与

2ab=-20

，从而确定两个根的实部

Solve

a^2-b^2=21

and

2ab=-20

to determine the real parts of the two roots

由求根公式得

z=\frac{1\pm\sqrt{21-20i}}2\text{。}

因为

(5-2i)^2=21-20i

，所以两个根为

\begin{aligned} \frac{1+(5-2i)}2&=3-i,\\ \frac{1-(5-2i)}2&=-2+i \end{aligned}\text{。}

它们的实部之积为

3(-2)=-6

。

因此正确答案是 B。

The quadratic formula gives

z=\frac{1\pm\sqrt{21-20i}}2.

Since

(5-2i)^2=21-20i,

the roots are

\begin{aligned} \frac{1+(5-2i)}2&=3-i,\\ \frac{1-(5-2i)}2&=-2+i. \end{aligned}

Their real parts have product

3(-2)=-6.

Thus the correct answer is B.

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