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Question
Write roots of the equation \[(a - b) x^2 + (b - c)x + (c - a) = 0\] .
Solution
\[\text { Given: } \]
\[(a - b) x^2 + (b - c)x + (c - a) = 0\]
\[ \Rightarrow x^2 + \frac{b - c}{a - b}x + \frac{c - a}{a - b} = 0\]
\[ \Rightarrow x^2 - \frac{c - a}{a - b}x - x + \frac{c - a}{a - b} = 0 \left[ \because \frac{b - c}{a - b} = \frac{- c + a - a + b}{a - b} = - \frac{c - a}{a - b} - 1 \right]\]
\[ \Rightarrow x\left( x - \frac{c - a}{a - b} \right) - 1\left( x + \frac{c - a}{a - b} \right) = 0\]
\[ \Rightarrow \left( x - \frac{c - a}{a - b} \right)\left( x - 1 \right) = 0\]
\[ \Rightarrow x - \frac{c - a}{a - b} = 0 or x - 1 = 0\]
\[ \Rightarrow x = \frac{c - a}{a - b} or x = 1\]
\[\text { Thus, roots of the equation are } \frac{c - a}{a - b} \text { and }1 .\]
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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