The sum of all the rational roots of the equation $6 x^{6} - 25 x^{5} + 31 x^{4} - 31 x^{2} + 25 x - 6 = 0$ is

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3.5

$\frac{25}{6}$

2.5

answer is D.

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Detailed Solution

$\begin{array}{l} G i v e n E q u a t i o n 6 x^{6} - 25 x^{5} + 31 x^{4} - 31 x^{2} + 25 x - 6 = 0 i s r e c i p r o c a l E q u a t i o n \\ ∴ x = \pm 1 a r e r o o t s \end{array}$

$\begin{array}{l} ∴ 6 x^{4} - 25 x^{3} + 37 x^{2} - 25 x + 6 = 0 \\ \Rightarrow 6 x^{2} - 25 x + 37 - \frac{25}{x} + \frac{6}{x^{2}} = 0 \\ \Rightarrow 6 (x^{2} + \frac{1}{x^{2}}) - 25 (x + \frac{1}{x}) + 37 = 0 \\ L e t x + \frac{1}{x} = y \Rightarrow x^{2} + \frac{1}{x^{2}} = y^{2} - 2 \\ \Rightarrow 6 (y^{2} - 2) - 25 (y) + 37 = 0 \\ \Rightarrow 6 y^{2} - 12 - 25 y + 37 = 0 \\ \Rightarrow 6 y^{2} - 25 y + 25 = 0 \end{array}$

$\begin{array}{l} \Rightarrow 6 y^{2} - 15 y - 10 y + 25 = 0 \\ \Rightarrow 3 y (2 y - 5) - 5 (2 y - 5) = 0 \\ \Rightarrow (2 y - 5) (3 y - 5) = 0 \\ \Rightarrow 2 y - 5 = 0 a n d 3 y - 5 = 0 \\ I f 2 y - 5 = 0 \\ T h e n 2 (x + \frac{1}{x}) - 5 = 0 \\ \Rightarrow 2 (\frac{x^{2} + 1}{x}) - 5 = 0 \\ \Rightarrow 2 x^{2} + 2 - 5 x = 0 \\ \Rightarrow 2 x^{2} - 5 x + 2 = 0 \\ \Rightarrow 2 x^{2} - 4 x - x + 2 = 0 \\ \Rightarrow 2 x (x - 2) - 1 (x - 2) = 0 \\ \Rightarrow x - 2 = 0, 2 x - 1 = 0 \\ \Rightarrow x = 2, x = \frac{1}{2} \\ I f 3 y - 5 = 0 \\ \Rightarrow 3 (x + \frac{1}{x}) - 5 = 0 \\ \Rightarrow 3 x^{2} - 5 x + 3 = 0 \\ N o r e a l r o o t s \\ S u m o f r a t i o n a l r o o t s = 1 - 1 + 2 + \frac{1}{2} \Rightarrow \frac{5}{2} \Rightarrow 2.5 \end{array}$