Question

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

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

Answer 1

The correct option is A $1,\frac{c(a-b)}{a(b-c)}$

Given equation is
$a(b-c)x^2+b(c-a)x+c(a-b)=0$
Consider $a{x}^2+bx+c=0$ ; it has a root $1$ if
$a+b+c=0$
Here, $a(b-c)+b(c-a)+c(a-b)=0$
Hence, $1$ is a root.
Product of roots is $\frac{c(a-b)}{a(b-c)}$
$\therefore$ Other root is $\frac{c(a-b)}{a(b-c)}$.