With respect the the roots of $x^{2} - x - 2 = 0$ , we can say that

both of them are integers

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both of them are natural numbers

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the latter of the two is negative

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None of the above

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Solution

The correct option is C both of them are integers
$F o r m u l a u s e d$
$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$
$G i v e n$
$x^{2} - x - 2 = 0$
$x = \frac{1 \pm \sqrt{1^{2} - 4 (- 2) (1)}}{2}$
$= \frac{1 \pm 3}{2}$
$= 2, - 1$
$Hence both the roots are integers$

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