TITLE: Probability that a quadratic equation has real roots
QUESTION [57 upvotes]: Problem
The premise is almost the same as in this question. I'll restate for convenience.

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

Note: The distribution is now $-1$ to $+1$ instead of $0$ to $1$.
My Attempt
Preparation
When the coefficients are sampled from $\mathcal{U}(0,1)$, the probability for the discriminant to be non-negative that is, $P(B^2-4AC\geq0) \approx 25.4\% $. This value can be obtained theoretically as well as experimentally. The link I shared above to the older question has several good answers discussing both approaches.
Changing the sampling interval to $(-1, +1)$ makes things a bit difficult from the theoretical perspective. Experimentally, it is rather simple. This is the code I wrote to simulate the experiment for $\mathcal{U}(0,1)$. Changing it from (0, theta) to (-1, +1) gives me an average probability of $62.7\%$ with a standard deviation of $0.3\%$
I plotted the simulated PDF and CDF. In that order, they are:

So I'm aiming to find a CDF that looks like the second image.
Theoretical Approach
The approach that I find easy to understand is outlined in this answer. Proceeding in a similar manner, we have
$$
f_A(a) = \begin{cases}
\frac{1}{2}, &-1\leq a\leq+1\\
0,           &\text{ otherwise}
\end{cases}
$$
The PDFs are similar for $B$ and $C$.
The CDF for $A$ is
$$
F_A(a) = \begin{cases}
\frac{a + 1}{2}, &-1\leq a\geq +1\\
0,&a<-1\\
1,&a>+1
\end{cases}
$$
Let us assume $X=AC$. I proceed to calculate the CDF for $X$ (for $x>0$) as:
$$
\begin{align}
F_X(x) &= P(X\leq x)\\
&= P(AC\leq x)\\
&= \int_{c=-1}^{+1}P(Ac\leq x)f_C(c)dc\\
&= \frac{1}{2}\left(\int_{c=-1}^{+1}P(Ac\leq x)dc\right)\\
&= \frac{1}{2}\left(\int_{c=-1}^{+1}P\left(A\leq \frac{x}{c}\right)dc\right)\\
\end{align}
$$
We take a quick detour to make some observations. First, when $0<c<x$, we have $\frac{x}{c}>1$. Similarly, $-x<c<0$ implies $\frac{x}{c}<-1$. Also, $A$ is constrained to the interval $[-1, +1]$. Also, we're only interested when $x\geq 0$ because $B^2\geq 0$.
Continuing, the calculation
$$
\begin{align}
F_X(x) &= \frac{1}{2}\left(\int_{c=-1}^{+1}P\left(A\leq \frac{x}{c}\right)dc\right)\\
&= \frac{1}{2}\left(\int_{c=-1}^{-x}P\left(A\leq \frac{x}{c}\right)dc + \int_{c=-x}^{0}P\left(A\leq \frac{x}{c}\right)dc + \int_{c=0}^{x}P\left(A\leq \frac{x}{c}\right)dc + \int_{c=x}^{+1}P\left(A\leq \frac{x}{c}\right)dc\right)\\
&= \frac{1}{2}\left(\int_{c=-1}^{-x}P\left(A\leq \frac{x}{c}\right)dc + 0 + 1 + \int_{c=x}^{+1}P\left(A\leq \frac{x}{c}\right)dc\right)\\
&= \frac{1}{2}\left(\int_{c=-1}^{-x}\frac{x+c}{2c}dc + 0 + 1 + \int_{c=x}^{+1}\frac{x+c}{2c}dc\right)\\
&= \frac{1}{2}\left(\frac{1}{2}(-x+x(\log(-x)-\log(-1)+1) + 0 + 1 + \frac{1}{2}(-x+x(-\log(x)-\log(1)+1)\right)\\
&= \frac{1}{2}\left(2 + \frac{1}{2}(-x+x(\log(x)) -x + x(-\log(x))\right)\\
&= 1 - x
\end{align}
$$
I don't think this is correct.
My Specific Questions

What mistake am I making? Can I even obtain the CDF through integration?
Is there an easier way? I used this approach because I was able to understand it well. There are shorter approaches possible (as is evident with the $\mathcal{U}(0,1)$ case) but perhaps I need to read more before I can comprehend them. Any pointers in the right direction would be helpful.

REPLY [41 votes]: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
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}