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