Question

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

Answer 1

$$\begin{gathered} \mathrm{Given}: \\ (a-b)x^2+(b-c)x+(c-a)=0 \\ \Rightarrow x^2+\frac{b-c}{a-b}x+\frac{c-a}{a-b}=0 \\ \Rightarrow x^2-\frac{c-a}{a-b}x-x+\frac{c-a}{a-b}=0\left[\because \frac{b-c}{a-b}=\frac{-c+a-a+b}{a-b}=-\frac{c-a}{a-b}-1\right] \\ \Rightarrow x\left(x-\frac{c-a}{a-b}\right)-1\left(x+\frac{c-a}{a-b}\right)=0 \\ \Rightarrow \left(x-\frac{c-a}{a-b}\right)\left(x-1\right)=0 \\ \Rightarrow x-\frac{c-a}{a-b}=0\mathrm{or}x-1=0 \\ \Rightarrow x=\frac{c-a}{a-b}\mathrm{or}x=1 \\ \mathrm{Thus},\mathrm{roots}\mathrm{of}\mathrm{the}\mathrm{equation}\mathrm{are}\frac{c-a}{a-b}\mathrm{and}1. \end{gathered}$$

Now,
$$\begin{gathered} \mathrm{\alpha }+\mathrm{\beta }=-\frac{b-c}{a-b} \\ \Rightarrow 1+\mathrm{\beta }=-\frac{b-c}{a-b} \\ \Rightarrow \mathrm{\beta }=-\frac{b-c}{a-b}-1=\frac{c-a}{a-b} \end{gathered}$$