1954 AMC 12 第 25 题

Question

方程

\begin{aligned} a(b-c)x^2+b(c-a)x\\ {}+c(a-b)&=0 \end{aligned}

的两个根为

1

和：

The two roots of the equation

\begin{aligned} a(b-c)x^2+b(c-a)x\\ {}+c(a-b)&=0 \end{aligned}

are

1

and:

\dfrac{b(c-a)}{a(b-c)}

\dfrac{a(b-c)}{c(a-b)}

\dfrac{a(b-c)}{b(c-a)}

\dfrac{c(a-b)}{a(b-c)}

\dfrac{c(a-b)}{b(c-a)}

使用两根之积，不必直接解二次方程

Use the product of the roots rather than solving the quadratic

两根之积等于常数项除以首项系数

The product is the constant coefficient divided by the leading coefficient

由韦达定理，两根之积为

\frac{c(a-b)}{a(b-c)}\text{。}

因为其中一个根是

1

，所以另一个根等于这个乘积。

因此，正确答案是 D。

By Vieta’s formulas, the product of the two roots is

\frac{c(a-b)}{a(b-c)}.

Since one root is

1,

the other root equals this product.

Thus, the correct answer is D.

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