The sum of non-integeral roots of the equation $x^{4} - 3 x^{3} - 2 x^{2} + 3 x + 1 = 0$ is

3

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0

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- 3

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6

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Solution

The correct option is A $3$
Divide the given equation by $x^{2}$ $\Rightarrow x^{2} - 3 x - 2 + \frac{3}{x} + \frac{1}{x^{2}} = 0$
$\Rightarrow (x^{2} + \frac{1}{x^{2}}) - 3 (x - \frac{1}{x}) - 2 = 0$
Now put $x - \frac{1}{x} = y$
$\Rightarrow (y^{2} + 2) - 3 y - 2 = 0 \Rightarrow y^{2} - 3 y = 0 \Rightarrow y = 0, 3$
Case $I$ :
$x - \frac{1}{x} = 0 \Rightarrow x = \pm 1$
Case $I I$ : $x - \frac{1}{x} = 3 \Rightarrow x = \frac{1}{2} (3 \pm \sqrt{13})$

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