What is the probability that the equation x 2 + bx + c = 0 has two real solutions?

Question

Suppose two real numbers, b and c, are randomly chosen from the interval [−1, 1]. In each case, any real number in the interval is as likely to be chosen as any other, and it is possible for the two numbers to be equal.

Answer 1

There are two Real solutions whenever b² > 4c in this case (since a=1).

 

Since 0 ≤  b² ≤ 1 under the terms specified, any c value that is negative will satisfy the above (bold) inequality. This will account for probability 0.5, "so far."

 

Next, consider cases where c is positive.  c would have to be > ¼•b².  Can you find a probability for this (to be added to the aforementioned 0.5)?