The difference of the irrational roots of the equation $x^{5} - 5 x^{4} + 9 x^{3} - 9 x^{2} + 5 x - 1 = 0$ is

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$\sqrt{5}$

$\sqrt{3}$

$2 \sqrt{5}$

answer is D.

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Detailed Solution

$\begin{array}{l} G i v e n E q u a t i o n x^{5} - 5 x^{4} + 9 x^{3} - 9 x^{2} + 5 x - 1 = 0 i s \\ r e c i p r o c a l E q u a t i o n ∴ x = 1 i s r o o t \end{array}$

$\begin{array}{l} ∴ x^{4} - 4 x^{3} + 5 x^{2} - 4 x + 1 = 0 \\ \Rightarrow x^{2} - 4 x + 5 - \frac{4}{x} + \frac{1}{x^{2}} = 0 \\ \Rightarrow (x^{2} + \frac{1}{x^{2}}) - 4 (x + \frac{1}{x}) + 5 = 0 \\ L e t x + \frac{1}{x} = y \Rightarrow x^{2} + \frac{1}{x^{2}} = y^{2} - 2 \\ ∴ y^{2} - 2 - 4 y + 5 = 0 \\ \Rightarrow y^{2} - 4 y + 3 = 0 \\ \Rightarrow y^{2} - 3 y - y + 3 = 0 \\ \Rightarrow (y - 1) (y - 3) = 0 \end{array}$

$\begin{array}{l} y = 1, y = 3 \\ I f y = 1 t h e n x + \frac{1}{x} = 1 \\ \Rightarrow x^{2} - x + 1 = 0 \\ x = \frac{1 \pm \sqrt{1 - 4}}{2} = \frac{1 \pm i \sqrt{3}}{2} \\ I f y = 3 \Rightarrow x + \frac{1}{x} = 3 \\ \Rightarrow x^{2} - 3 x + 1 = 0 \\ x = \frac{3 \pm \sqrt{9 - 4}}{2} = \frac{3 \pm \sqrt{5}}{2} \\ D i f f e r e n c e o f r o o t s = \frac{3 + \sqrt{5}}{2} - (\frac{3 - \sqrt{5}}{2}) = \sqrt{5} \end{array}$