If the numbers appeared on the two throws of a fair six faced die...

Total sample space of two dice will be equal to& $6 \times 6 = 36$

Now for~$x^{2} + \alphax + \beta > 0$~its discriminant& $D$~must be less than zero,

So& $D = b^{2} - 4ac < 0$~or& ${α}^{2} - 4\beta < 0⇒ {α}^{2} < 4\beta$~

Now, if $\alpha = 1, \beta$ can take values $1, 2, 3, 4, 5, 6$

If $\alpha = 2, \beta$ can take values $2, 3, 4, 5, 6$

If $\alpha = 3, \beta$ can take values $3, 4, 5, 6$

If $\alpha = 4, \beta$ can take values $5, 6$

If $\alpha = 5$~and $6$, no value of $\beta$ possible

So number of favourable ways $= 17$

Required probability$= \dfrac{favourable ways}{total sample space}= \dfrac{17}{36}$