1969 AMC 12 第 5 题

Question

若一个数

N

（

N\ne0

）减去其倒数的四倍等于给定实常数

R

，则对于该给定的

R

，所有可能的

N

值之和为：

If a number

N,

N\ne0,

diminished by four times its reciprocal, equals a given real constant

R,

then, for this given

R,

the sum of all such possible values of

N

is:

\dfrac1R

R

4

\dfrac14

-R

将

N-\frac{4}{N}=R

改写成关于

N

的二次方程

Rewrite

N-\frac{4}{N}=R

as a quadratic in

N

利用该二次方程的根之和

Use the sum of the roots of that quadratic

将

N-\frac{4}{N}=R

乘以

N

，得到

N^2-RN-4=0\text{。}

由韦达定理，两根之和为

R

。因为两根之积为

-4

，所以两根都非零。

因此，正确答案是 B。

Multiplying

N-\frac{4}{N}=R

by

N

gives

N^2-RN-4=0.

The sum of its two roots is

R

by Vieta’s formulas. Both roots are nonzero because their product is

-4.

Therefore, the correct answer is B.

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