TITLE: Probability that a quadratic equation with random coefficients has real roots
QUESTION [12 upvotes]: Consider quadratic equations $Ax^2 + Bx + C = 0$ in which $A$, $B$, and $C$ are
  independently distributed $\mathsf{Unif}(0,1)$. What is the probability that the roots of such an equation are real?

This problem is from  Chapter 3 of Rice: Mathematical Statistics and Data Analysis (editions 1 through 3). Until recent printings of 3e, the incorrect answer 1/9 was given for this problem. 
However, Horton (2015) http://www3.amherst.edu/~nhorton/precursors/precursors.pdf
points out that the correct answer is slightly above 1/4, as can be verified by a simple simulation. (Horton and his colleagues are concerned with elements of an undergraduate curriculum to prepare students in the mathematical sciences to cope with modern data science.)
In a somewhat more practical setting, one might consider a discrete version of this problem.
A program that produces random drill problems on quadratic equations $Ax^2 + Bx + C = 0,$ selects values for $A, B,$ and $C$ at random and independently from among the ten equally likely values $0.1, 0.2, \dots, 1.0$. What proportion of such equations have real roots? And what proportion have only one root?
The initial Answer sketches the exact analytic solution of the original problem and shows numerical and graphical results from simulation. A simulated result for the discrete version is also shown.
Additional answers using other methods or discussing related topics are welcome.

REPLY [4 votes]: Someone asked almost this a few days ago. I did want to point out that a cube is not the most natural shape to consider for this problem, although it was the one chosen. Better, in some ways, to consider the ball $A^2 + B^2 + C^2 \leq R^2.$ In that case we take rotated coordinates
$$ u = B; v = (A - C)/ \sqrt 2; w = (A + C)/ \sqrt 2.  $$
Then the condition $B^2 \geq 4AC$ becomes $u^2 + 2 v^2 \geq 2 w^2.$ It makes no difference here whether we use volume or surface area, so we are asking for the total surface area of the two peculiar elliptical patches $u^2 + 2 v^2 \geq 2 w^2$ on the sphere $u^2 + v^2 + w^2 = 1,$ divided by $4 \pi.$ On second thought, the figure we want is one minus this, the funny annular region.
Not sure I know how to calculate this.