Question

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

Answer 1

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

is the form of

$A{x}^2+Bx+C=0$

Where $A=a(b-c),B=b(c-a)$

and $C=c(a-b)$

$A+B+C=ab-ac+bc-ba+ca-cb$

$=0$

$\therefore$ one root is $1$

Let the other root be $\alpha$

$\Rightarrow$ product of roots $1\times \alpha =\frac{c(a-b)}{a(b-c)}$

$\therefore$ the roots are $1,\frac{c(a-b)}{a(b-c)}$