Animated Solution for Mathematics - Probability: 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 :

Imagine you are standing before a vast, empty coordinate plane. You have a simple quadratic expression, $x^2+\alpha x+\beta$, and you are told that for any real number $x$ you choose, this expression must always result in a positive value.

This is not just an algebraic constraint; it is a beautiful geometric requirement. Think of the graph of this quadratic, $y=x^2+\alpha x+\beta$.

Because the coefficient of $x^2$ is $1$, which is positive, we know this graph is a parabola that opens upwards. For this parabola to be strictly greater than zero for all $x$, it must never touch or cross the x-axis. It must hover, like a bird in flight, entirely above the horizontal line $y=0$.

How do we translate this 'hovering' behavior into the language of algebra? We look to the discriminant, $D=b^2-4ac$.

The discriminant tells us about the roots of the quadratic equation $x^2+\alpha x+\beta =0$. If $D>0$, the parabola crosses the x-axis at two distinct points. If $D=0$, it kisses the x-axis at exactly one point.

But if $D<0$, the roots are complex, and the parabola never touches the x-axis at all. This is exactly what we need! Substituting our coefficients, where $a=1$, $b=\alpha$, and $c=\beta$, the condition $D<0$ becomes:

This is our golden inequality, the gatekeeper of our favorable outcomes.

Now, we enter the counting phase. We know $\alpha$ and $\beta$ are outcomes of two fair dice, so they can only be integers from $1$ to $6$. The total number of outcomes in our sample space is $6\times 6=36$.

We need to find how many pairs $(\alpha ,\beta )$ satisfy ${\alpha }^2<4\beta$. Let's hunt them down row by row, fixing $\beta$ and checking $\alpha$:

If $\beta =1$, then ${\alpha }^2<4$. The only integer $\alpha \in \{1,2,3,4,5,6\}$ satisfying this is $\alpha =1$. That is $1$ case. If $\beta =2$, then ${\alpha }^2<8$. Here, $\alpha$ can be $1$ ($1^2=1<8$) or $2$ ($2^2=4<8$). That is $2$ cases. If $\beta =3$, then ${\alpha }^2<12$. Here, $\alpha$ can be $1,2,3$ ($3^2=9<12$). That is $3$ cases. If $\beta =4$, then ${\alpha }^2<16$. Be careful! $\alpha$ can be $1,2,3$. If $\alpha =4$, then ${\alpha }^2=16$, which is not strictly less than $16$. So, we have $3$ cases. If $\beta =5$, then ${\alpha }^2<20$. Here, $\alpha$ can be $1,2,3,4$ ($4^2=16<20$). That is $4$ cases. If $\beta =6$, then ${\alpha }^2<24$. Here, $\alpha$ can be $1,2,3,4$ ($4^2=16<24$). That is $4$ cases.

Adding these up, we have $1+2+3+3+4+4=17$ favorable outcomes.

The probability is the ratio of these favorable outcomes to the total sample space:

It is a moment of pure satisfaction when the logic aligns, the cases are counted, and the fraction emerges, perfectly matching our expectations. You have navigated the geometry, mastered the discriminant, and conquered the counting. The final answer is 17/36.