The sum of the moduli of the roots of the equation $x^{4} - 5 x^{3} + 6 x^{2} - 5 x + 1 = 0$ is:

2

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4

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6

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8

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Solution

The correct option is B $6$
$x^{4} - 5 x^{3} + 6 x^{2} - 5 x + 1 = 0$
Divide by $x^{2},$ $x^{2} - 5 x + 6 - \frac{5}{x} + \frac{1}{x^{2}} = 0$ $(x^{2} + \frac{1}{x^{2}}) - 5 (x + \frac{1}{x}) + 6 = 0$
Put $x + \frac{1}{x} = t$ $∴ t^{2} - 2 - 5 t + 6 = 0$
$\Rightarrow t^{2} - 5 t + 4 = 0$
$\Rightarrow t = 4, t = 1$
$x + \frac{1}{x} = 4,$ or $x + \frac{1}{x} = 1$
$x^{2} - 4 x + 1 = 0$ or $x^{2} - x + 1 = 0$
$x = \frac{4 \pm \sqrt{16 - 4}}{2} = 2 \pm \sqrt{3}$ or $x = \frac{1 \pm \sqrt{1 - 4}}{2} = \frac{1}{2} \pm i \frac{\sqrt{3}}{2}$
Sum of the moduli of the root $2 + \sqrt{3} + 2 - \sqrt{3} + 2 \sqrt{(\frac{1}{4} + \frac{3}{4})}$ $= 4 + 2 = 6$

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