The two roots of the equation a(b-c)x^2+b(c-a)x +c(a-b)&=0 are 1 and:

1954 AMC 12 Problem 25

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25.

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)}

Use the product of the roots rather than solving the quadratic

The product is the constant coefficient divided by the leading coefficient

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.