Q50 / 160 · AP EAMCET 2024 (21 May Shift 1) · Mathematics

Roots of the equation $a(b-c) x^2+b(c-a) x+c(a-b)$ $=0$ are

  • A
    $\frac{a(b-c)}{c(a-b)}, 1$
  • B
    $\frac{b(c-a)}{c(a-b)}, 1$
  • Correct answer
    C
    $\frac{c(a-b)}{a(b-c)}, 1$
  • D
    $\frac{c(a-b)}{b(c-a)}, 1$

Solution

$a(b-c) x^2+b(c-a) x+c(a-b)=0$ $x=1$ satisfies his equation Product of roots $=\frac{c(a-b)}{a(b-c)}=1 \times \alpha$ $\alpha=\frac{c(a-b)}{a(b-c)}=\text { other root }$
Roots are $\frac{c(a-b)}{a(b-c)}$ and 1

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

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