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 $$x \in \mathbb{R}$$, is

Given quadratic expression

$$x^2+\alpha x+\beta$$

For

$$x^2+\alpha x+\beta>0\quad \forall x\in\mathbb R,$$

the quadratic must have a positive leading coefficient and no real roots.

Since the coefficient of $$x^2$$ is already positive, we only require

$$\Delta<0$$

Therefore,

$$\alpha^2-4\beta<0$$

$$\alpha^2<4\beta$$

Now,

$$\alpha,\beta\in\{1,2,3,4,5,6\}$$

We count favorable cases.

For $$\alpha=1,$$

$$1<4\beta$$

which is true for all

$$\beta=1,2,3,4,5,6$$

Hence,

$$6$$ cases.

For $$\alpha=2,$$

$$4<4\beta$$

$$\beta>1$$

Hence,

$$\beta=2,3,4,5,6$$

giving

$$5$$ cases.

For $$\alpha=3,$$

$$9<4\beta$$

$$\beta\ge3$$

Hence,

$$4$$ cases.

For $$\alpha=4,$$

$$16<4\beta$$

$$\beta>4$$

Hence,

$$\beta=5,6$$

giving

$$2$$ cases.

For $$\alpha=5,$$

$$25<4\beta$$

is impossible.

Hence,

$$0$$ cases.

For $$\alpha=6,$$

$$36<4\beta$$

is impossible.

Hence,

$$0$$ cases.

Therefore, total favorable outcomes are

$$6+5+4+2=17$$

Total possible outcomes are

$$6\times6=36$$

Hence, the required probability is

$$\boxed{\frac{17}{36}}$$