Question

If one root of the quadratic 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. $\frac{a(b-c)}{c(a-b)}$
• C. $\frac{a(b-c)}{b(c-a)}$
• D. $\frac{c(a-b)}{a(b-c)}$

Answer 1

The Correct Option is D

We are given the quadratic equation:

$a(b - c)x^2 + b(c - a)x + c(a - b) = 0$ 

One root is given as $x = 1$. Let the other root be $\alpha$.

By Vieta's formulas, the sum of the roots is given by:

$1 + \alpha = -\frac{\text{coefficient of } x}{\text{coefficient of } x^2}$

$1 + \alpha = -\frac{b(c-a)}{a(b-c)}$

Solving for $\alpha$:

$\alpha = -\frac{b(c-a)}{a(b-c)} - 1$

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

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

$\alpha = \frac{-bc + ab - ab + ac}{a(b-c)}$

$\alpha = \frac{ac - bc}{a(b-c)}$

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

Thus, the other root is: $\frac{c(a-b)}{a(b-c)}$.