If the numbers appeared on the two thrown of a fair six faced die are \( \alpha \) & \( \beta \), then the probability that \( x^{2}+\alpha x+\beta>0 \), for all \( x\in \mathbb{R} \) is:
\( \dfrac{17}{36} \)
\( \dfrac{4}{9} \)
\( \dfrac{1}{2} \)
\( \dfrac{19}{36} \)