TITLE: Probability that a quadratic polynomial with random coefficients has real roots
QUESTION [34 upvotes]: The following is a homework question for which I am asking guidance.

Let $A$, $B$, $C$ be independent random variables uniformly distributed between $(0,1)$. What is the probability that the polynomial $Ax^2 + Bx + C$ has real roots?

That means I need $P(B^2 -4AC \geq 0$). I've tried calling $X=B^2 -4AC$ and finding $1-F_X(0)$, where $F$ is the cumulative distribution function.
I have two problems with this approach. First, I'm having trouble determining the product of two uniform random variables. We haven't been taught anything like this in class, and couldn't find anything like it on Sheldon Ross' Introduction to Probability Models.
Second, this strategy just seems wrong, because it involves so many steps and subjects we haven't seen in class. Even if I calculate the product of $A$ and $C$, I'll still have to square $B$, multiply $AC$ by four and then subtract those results. It's too much for a homework question. I'm hoping there might be an easier way.

REPLY [4 votes]: Here's another approach which utilizes the total law of probability directly. $$P(B^2\geq 4AC)=\int_{0}^{1}P(B^2\geq4AC|C=c)f_{C}(c)dc$$ Using independence the above simplifies to $$P(B^2\geq 4AC)=\int_{0}^{1}P(B^2\geq4Ac)dc$$ When $c\in\Big(0,\frac{1}{4}\Big)$ we have $$P(B^2 \geq 4Ac)=\int_0^1 \int_{\sqrt{4ac}}^1dbda=1-\frac{4\sqrt{c}}{3}$$ When $c\in \Big[\frac{1}{4},1\Big)$ we have $$P(B^2 \geq 4Ac)=\int_0^1 \int_0^{b^2/4c}dadb=\frac{1}{12c}$$ Putting everything together $$P(B^2\geq 4AC)=\int_0^{1/4}\Bigg(1-\frac{4\sqrt{c}}{3}\Bigg)dc+\int_{1/4}^1\frac{dc}{12c}=\frac{5+3\ln(4)}{36}$$