If the equation $(m - n) x^{2} + (n - 1) x + 1 - m = 0$ has equal roots, then $1$ , $m$ and $n$ satisfy

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Solution

From sri Dhara charya formula, roots of a quadratic equation is given by,
$x = \frac{- b \pm \sqrt{D}}{2 a}$ where $D = b^{2} - 4 a c$ (Discriminant)
Where a, b and c denote the coefficients when the quadratic equation is written in general form, $a x^{2} + b x + c = 0$
Here, the equation is, $(m - n) x^{2} + (n - D x + (1 -) = 0$
Therefore, $a = m - n, b = n - 1$ and c = 1-m$
Denoting the two roots as $x$ , and $x_{2}$ they are given by
$x_{1} = \frac{- b + \sqrt{D}}{2 a}$ and $x_{2} = \frac{- b - \sqrt{D}}{2 a}$
Given, $x_{1} = x_{2} \Rightarrow \frac{- b + \sqrt{D}}{2 a} = \frac{- b - \sqrt{D}}{2 a} \Rightarrow 2 \sqrt{D} = 0 \Rightarrow D = 0$
$\Rightarrow b^{2} - 4 a c = 0$
$\Rightarrow (n - 1)^{2} - 4 (m - n) (1 - m) = 0$
$\Rightarrow n^{2} - 2 n + 1 - 4 (m - m^{2} - n + m n) = 0$
$\Rightarrow n^{2} - 2 n + 1 - 4 m + 4 m^{2} + 4 n - 4 m n = 0$
$\Rightarrow n^{2} + 4 m^{2} + 2 n - 4 m - 4 m n + 1 = 0$

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