1952 AMC 12 第 23 题

Question

若方程

\dfrac{x^2-bx}{ax-c}=\dfrac{m-1}{m+1}

的两根绝对值相等、符号相反，则

m

的值必须为：

If

\dfrac{x^2-bx}{ax-c}=\dfrac{m-1}{m+1}

has roots which are numerically equal but of opposite signs, the value of

m

must be:

\dfrac{a-b}{a+b}

\dfrac{a+b}{a-b}

c

\dfrac1c

1

消去分母，并将方程整理成关于

x

的二次方程

Clear the denominator and collect the equation as a quadratic in

x

互为相反数的两根之和为零，所以

x

的系数必须为零

Opposite roots have sum zero, so the coefficient of

x

must vanish

交叉相乘，得到

\begin{aligned} (m+1)x^2 &-(m+1)bx\\ &-(m-1)ax\\ &\quad{}+(m-1)c=0 \end{aligned}\text{。}

绝对值相等、符号相反的两根之和为

0

，所以

x

的系数为

0

。因此

m(a+b)+(b-a)=0\text{，}

并且

m=\frac{a-b}{a+b}\text{。}

因此，正确答案是 A。

Cross-multiplication gives

\begin{aligned} (m+1)x^2 &-(m+1)bx\\ &-(m-1)ax\\ &\quad{}+(m-1)c=0. \end{aligned}

Roots that are equal in magnitude and opposite in sign have sum

0,

so the coefficient of

x

is

0.

Thus

m(a+b)+(b-a)=0,

and

m=\frac{a-b}{a+b}.

Thus, the correct answer is A.

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