Question

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

Answer 1

\[\text { Given: } \]

\[(a - b) x^2 + (b - c)x + (c - a) = 0\]

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

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

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

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

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

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

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

Now,

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

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

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