Irrational roots of the equation $2 x^{4} + 9 x^{3} + 8 x^{2} + 9 x + 2 = 0$ are

- 2 - \sqrt{3}, 2 + \sqrt{3}

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2 - \sqrt{3}, 2 + \sqrt{3}

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2 + \sqrt{3}, - 2 - \sqrt{3}

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Solution

The correct option is A $- 2 - \sqrt{3}, 2 + \sqrt{3}$
$2 x^{4} + 9 x^{3} + 8 x^{2} + 9 x + 2 = 0$
Divide by $x^{2}$
$2 x^{2} + 9 x + 8 + 9 (\frac{1}{x}) + 2 (\frac{1}{x^{2}}) = 0$
$2 (x^{2} + 1 / x^{2}) + 9 (x + 1 / x) + 8 = 0$
Put $(x + \frac{1}{x}) = y$
So $(x^{2} + \frac{1}{x^{2}}) = y^{2} - 2$
$2 (y^{2} - 2) + 9 y + 8 = 0$
$2 y^{2} + 9 y + 4 = 0$
By solving $y = - 4, - \frac{1}{2}$
When $y = - 4$
$x + \frac{1}{x} = - 4$ ; $x^{2} + 4 x + 1 = 0$
So $x = - 4 \pm \sqrt{16 - 4 / 2}$
$x = - 2 \pm \sqrt{3}$ .

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