If $m, n$ are the roots of the quadratic equation $a x^{2} + b x + c = 0,$ then the roots of the equation $(4 a + 2 b + c) x^{2} + (4 a - b - 2 c) x + (a - b + c) = 0$ can be expressed as $\frac{m + e}{m + f}, \frac{n + g}{n + h}, m \neq 2, n \neq 2$ then the value $e f g h$ is

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Solution

$(2 x + 1)^{2} a + (2 x + 1) (x - 1) b + (x - 1)^{2} c = 0 \Rightarrow {(\frac{2 x + 1}{x - 1})}^{2} a + (\frac{2 x + 1}{x - 1}) b + c = 0 \Rightarrow \frac{2 x + 1}{x - 1} = m, n \Rightarrow 2 + \frac{3}{x - 1} = m, n \Rightarrow x = \frac{m + 1}{m - 2}, \frac{n + 1}{n - 2} \Rightarrow e = 1, f = - 2, g = 1, h = - 2$

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