Probability that a quadratic equation has real roots
I would probably start by breaking into cases based on $A$ and $C$.
Conditioned on $A$ and $C$ having different signs, there are always real roots (because $4AC\leq 0$, so that $B^2-4AC\geq0$). The probability that $A$ and $C$ have different signs is $\frac{1}{2}$.
Conditioned on $A\geq0$ and $C\geq 0$, you return to the problem solved in the link above. Why? Because $B^2$ has the same distribution whether you have $B$ uniformly distributed on $(0,1)$ or on $(-1,1)$. At the link, they computed this probability as $\frac{5+3\log4}{36}\approx0.2544134$. The conditioning event here has probability $\frac{1}{4}$.
Finally, if we condition on $A<0$ and $C<0$, we actually end up with the same probability, as $4AC$ has the same distribution in this case as in the case where $A\geq0$ and $C\geq 0$. So, this is an additional $\frac{5+3\log 4}{36}\approx0.2544134$ conditional probability, and the conditioning event has probability $\frac{1}{4}$.
So, all told, the probability should be $$ \begin{align*} P(B^2-4AC\geq0)&=1\cdot\frac{1}{2}+\frac{1}{4}\cdot\frac{5+3\log4}{36}+\frac{1}{4}\cdot\frac{5+3\log 4}{36}\\ &=\frac{1}{2}+\frac{5+3\log4}{72}\\ &\approx0.6272... \end{align*} $$
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ Hereafter, $\ds{\bracks{P}}$ is an Iverson Bracket. Namely, $\ds{\bracks{P} = \color{red}{1}}$ whenever $\ds{P}$ is $\ds{\tt true}$ and $\ds{\color{red}{0}}$ $\ds{\tt otherwise}$. They are very convenient whenever we have to manipulate constraints.
\begin{align} &\bbox[5px,#ffd]{\int_{-1}^{1}{1 \over 2}\int_{-1}^{1} {1 \over 2}\int_{-1}^{1}{1 \over 2}\bracks{b^{2} - 4ac > 0} \dd c\,\dd a\,\dd b} \\[5mm] = &\ {1 \over 4}\int_{0}^{1}\int_{-1}^{1} \int_{-1}^{1}\bracks{b^{2} - 4ac > 0} \dd c\,\dd a\,\dd b \\[5mm] = &\ {1 \over 4}\int_{0}^{1}\int_{-1}^{1} \int_{0}^{1}\braces{\bracks{b^{2} - 4ac > 0} + \bracks{b^{2} + 4ac > 0}} \dd c\,\dd a\,\dd b \\[5mm] = &\ {1 \over 4}\int_{0}^{1}\int_{0}^{1} \int_{0}^{1}\left\{\bracks{b^{2} - 4ac > 0} + \bracks{b^{2} + 4ac > 0}\right. \\[2mm] &\ \phantom{{1 \over 4}\int_{0}^{1}\int_{-1}^{1} \int_{0}^{1}} \left. + \bracks{b^{2} + 4ac > 0} + \bracks{b^{2} - 4ac > 0}\right\}\dd c\,\dd a\,\dd b \\[5mm] = &\ {1 \over 2} + {1 \over 2}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1} \bracks{b^{2} - 4ac > 0}\dd c\,\dd a\,\dd b \\[5mm] = &\ {1 \over 2} + {1 \over 2}\int_{0}^{1}\int_{0}^{1}{1 \over a}\int_{0}^{a} \bracks{b^{2} - 4c > 0}\dd c\,\dd a\,\dd b \\[5mm] = &\ {1 \over 2} + {1 \over 2}\int_{0}^{1}\int_{0}^{1}\bracks{b^{2} - 4c > 0} \int_{c}^{1}{1 \over a}\,\dd a\,\dd c\,\dd b \\[5mm] = &\ {1 \over 2} - {1 \over 2}\int_{0}^{1}\int_{0}^{1} \bracks{c < {b^{2} \over 4}}\ln\pars{c}\,\dd c\,\dd b \\[5mm] = &\ {1 \over 2} - {1 \over 2}\int_{0}^{1}\int_{0}^{b^{2}/4} \ln\pars{c}\,\dd c\,\dd b \\[5mm] = &\ {1 \over 2} - {1 \over 2}\int_{0}^{1}\bracks{% -\,{1 + 2\ln\pars{2} \over 4}\,b^{2} + {1 \over 2}\,b^{2}\ln\pars{b}}\,\dd b \\[5mm] = & \bbx{{\ln\pars{2} \over 12} + {41 \over 72}} \approx 0.6272 \\ & \end{align}
We know from the quadratic formula that the polynomial $Ax^2 + Bx + C$ has real roots if $B^2 - 4AC \geq 0$. We can think of this problem in terms of volumes. To do so, it's easier if we rename the coefficients as $x \equiv A$, $y \equiv C$, and $z \equiv B$. Hence, in order to have real roots we require that $z^2 \geq 4xy$ for $x,y,z \in (-1,1)$. The probability we are after is the ratio between the volume of the region for which this inequality is true and the volume of the containing cube, which is 8. Begin by observing that if $x$ and $y$ have opposite signs then this inequality is trivially satisfied. The volume of the region for which they have opposite signs 4. Now consider the case where $x$ and $y$ have the same signs. In this case, we want to compute the volume above the surface $z^2 = 4xy$ and below the containing cube. There are four cases to consider:
- $-1 < x \leq -\frac{1}{4}$ and $\frac{1}{4x} \leq y \leq 0$.
- $-\frac{1}{4} \leq x \leq 0$ and $-1 < y \leq 0$.
- $0 \leq x \leq \frac{1}{4}$ and $0 \leq y < 1$.
- $\frac{1}{4} \leq x < 1$ and $0 \leq y \leq \frac{1}{4x}$.
By symmetry we can just consider cases 1 and 2 and then multiply that volume by 2. In each case we have to compute the integral: \begin{align*} \int_a^b\int_c^d 2 - 4\sqrt{xy}\,dy\,dx, \end{align*} where the limits of integration are defined above. Evaluating cases 1 and 2 we find that the volume is $5/18 + (1/6)\ln(4)$. Hence, the total volume that satisfies the inequality is \begin{align*} 4 + 2\left(\frac{5}{18} + \frac{1}{6}\ln(4)\right) = \frac{41}{9} + \frac{1}{3}\ln(4) \end{align*} which leads to a probability of \begin{align*} \frac{1}{8}\left(\frac{41}{9} + \frac{1}{3}\ln(4)\right) \approx 0.62721 \end{align*}