# Write roots of the equation

Question:

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

Solution:

Given:

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

$\Rightarrow x^{2}+\frac{b-c}{a-b} x+\frac{c-a}{a-b}=0$

$\Rightarrow x^{2}-\frac{c-a}{a-b} x-x+\frac{c-a}{a-b}=0 \quad\left[\because \frac{b-c}{a-b}=\frac{-c+a-a+b}{a-b}=-\frac{c-a}{a-b}-1\right]$

$\Rightarrow x\left(x-\frac{c-a}{a-b}\right)-1\left(x+\frac{c-a}{a-b}\right)=0$

$\Rightarrow\left(x-\frac{c-a}{a-b}\right)(x-1)=0$

$\Rightarrow x-\frac{c-a}{a-b}=0$ or $x-1=0$

$\Rightarrow x=\frac{c-a}{a-b}$ or $x=1$

Thus, roots of the equation are $\frac{c-a}{a-b}$ and 1 .

Now,

$\alpha+\beta=-\frac{b-c}{a-b}$

$\Rightarrow 1+\beta=-\frac{b-c}{a-b}$

$\Rightarrow \beta=-\frac{b-c}{a-b}-1=\frac{c-a}{a-b}$

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