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 $x \in R$, is :

• A. $\frac{17}{36}$
• B. $\frac{4}{9}$
• C. $\frac{1}{2}$
• D. $\frac{19}{36}$

Answer 1

Answer: (A)

$\frac{17}{36}$

Answer 2

The correct option is(A): $\frac{17}{36}$.