Question

If one root of the equation $a(b-c)x^2+b(c-a)x+c(a-b)=0$ is $1$, then the other root is ______.

• A. $\frac{b(c-a)}{a(b-c)}$
• B. $\frac{a(b-c)}{b(c-a)}$
• C. $\frac{a(b-c)}{c(b-a)}$
• D. $\frac{c(a-b)}{a(b-c)}$

Answer 1

Answer: (D)

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

Given one root of the equation as $1$

We know that the product of the roots of a quadratic equation $a{x}^2+bx+c=0$ is $\frac{c}{a}$.

Here one root is $1$ and let the other be $x$.

Now, $x\times 1=\frac{c(a-b)}{a(b-c)}$

$\therefore$ $x=\frac{c(a-b)}{a(b-c)}$

$\therefore$ Option D is correct