$\sec 36^{0}$ is root of the equation is

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$x^{2} - 2 x + 4 = 0$

$x^{2} - 2 x - 4 = 0$

$x^{2} + 2 x - 4 = 0$

$4 x^{2} + 2 x - 1 = 0$

answer is A.

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Detailed Solution

Given $x = \sec 36^{0} = \frac{1}{\cos 36^{0}}$
$x = \frac{4}{\sqrt{5} + 1} \times \frac{\sqrt{5} - 1}{\sqrt{5} - 1}$
$x = \frac{4 (\sqrt{5} - 1)}{4}$
$x + 1 = \sqrt{5} \Rightarrow x^{2} + 1 + 2 x = 5$
$x^{2} + 2 x - 4 = 0$

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