Q250 / 270 · NDA 2011 (Phase 2) · Mathematics

If one of the roots of the equation $a(b-c) x^2+b(c-a) x+c(a-b)=0$ is 1 , what is the second root?

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

Solution

Let other root is $\beta$ of the equation $a(b-c) x^2+b(c-a) x+c(a-b)=0$ product of roots $=\frac{c}{a}$ $1 \times \beta=\frac{c(a-b)}{a(b-c)}$ $\therefore \quad$ other root $(\beta)=\frac{c(a-b)}{a(b-c)}$.

Hence, the correct answer is option: 'C'.

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