Probability of real roots for $x^2 + Bx + C = 0$ Question: The numbers $B$ and $C$ are chosen at random between $-1$ and $1$, independently of each other. What is the probability that the quadratic equation $$x^2 + Bx + C = 0$$ has real roots? Also, derive a general expression for this probability when B and C are chosen at random from the interval $(-q, q)$ for and $q>0$.
My approach: since we're trying to find the probability of real roots. We should first realize when it has imaginary roots. So, $$B^2 - 4aC < 0$$ $$a = 1$$ $$B^2 < 4C $$ I'm not sure where to go from here. How do I now find the probability of $B$ being greater than $4C$?

I assume you mean that $B,\,C\sim U(-1,\,1)$. We'll get the answer as a function of a fixed value for $C$, then average it out. For $C<0$ (which has probability $1/2$), the result is $0$; for $C> 1/4$ (which has probability $3/8$), the result is $1$; for $0\le C\le\frac{1}{4}$ (which has probability $\frac{1}{8}$), the condition $-2\sqrt{C}\le B\le 2\sqrt{C}$ has probability $4\sqrt{C}$. So the final result is $$\frac{3}{8}+\frac{1}{8}\int_0^{1/4} 4\sqrt{C}dC=\frac{3}{8}+\frac{1}{3}\bigg(\frac{1}{4}\bigg)^{3/2}=\frac{5}{12}.$$
Edit: the above is the probability of non-real complex roots; the probability of real roots is $1-\frac{5}{12}=\frac{7}{12}$.
