1977 AMC 12 第 21 题

Question

对多少个系数

a

的取值，方程

\begin{aligned} x^2+ax+1&=0,\\ x^2-x-a&=0 \end{aligned}

有公共实数解？

For how many values of the coefficient

a

do the equations

\begin{aligned} x^2+ax+1&=0,\\ x^2-x-a&=0 \end{aligned}

have a common real solution?

0

1

2

3

无穷多个

infinitely many

将两个方程相减，得到一个可因式分解的条件

Subtract the two equations to obtain a factored condition

分别检验

a=-1

和

x=-1

两种情形

Check separately the cases

a=-1

and

x=-1

用第一个方程减去第二个方程，得到

\begin{gathered} (a+1)x+(a+1)\\ =(a+1)(x+1)=0 \end{gathered}\text{。}

若

a=-1

，两个方程均化为

x^2-x+1=0

，它没有实根。否则

x=-1

，代入

x^2-x-a=0

得

a=2

。这个值确实满足条件，所以恰有一个

a

的取值符合要求。

因此，正确答案是 B。

Subtracting the second equation from the first gives

\begin{gathered} (a+1)x+(a+1)\\ =(a+1)(x+1)=0. \end{gathered}

If

a=-1,

the common equation is

x^2-x+1=0,

which has no real roots. Otherwise

x=-1,

and substitution into

x^2-x-a=0

gives

a=2.

This value works, so exactly one value of

a

qualifies.

Therefore, the correct answer is B.

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