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Question

If the numbers appeared on the two throws of a fair six faced die are $\alpha$ and $\beta$, then the probability that $x^{2}+\alpha x+\beta>0$, for all $x \in \mathbf{R}$, is :

$\frac{17}{36}$
$$ \frac{4}{9} $$
$\frac{1}{2}$
$\frac{19}{36}$

Solution

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