Given #{a,b,c} in [L,L]# What is the probability that the roots of #a x^2+b x + c = 0# be real?
1 Answer
Explanation:
We will operate on the assumptions
Initial observations:

#ax^2+bx+c = 0# has real roots if and only if the discriminant of the quadratic is nonnegative, that is,#b^24ac >= 0# , or#b^2>=4ac# . 
We can partition the probability space into the two cases in which
#"sgn"(a) = "sgn"(c)# or#"sgn"(a)!="sgn"(c)# , and
#P("sgn"(a)="sgn"(c)) = P("sgn"(a)!="sgn"(c)) = 1/2# 
If
#"sgn"(a) != "sgn"(c)# , then#4ac <= 0# , meaning# b^2>=4ac# . 
By symmetry:
#P(b^2>=4aca, c<=0) = P(b^2>=4aca, c>=0)# 
As
#b^2=(b)^2# for all#b# , we can restrict#b# to#[0, L]# without changing the probability of it falling within a certain range. Thus
#P(b^2>=4aca, c>=0) = P(b^2>=4aca, b, c>=0)#
With these, we can reformulate the problem as follows:
#=1/2P(b^2>=4ac"sgn"(a)!="sgn"(c))#
#+1/2P(b^2>=4ac"sgn"(a)="sgn"(c))#
#=1/2(1)+1/2P(b^2>=4aca, c<0 or a, c >0)#
#=1/2+1/2P(b^2>=4aca, c>0)#
#=1/2+1/2P(b^2>=4aca, b, c>0)#
#=1/2+1/2P(b^2>=4ac(a, b, c) in [0, L]^3)#
We will now calculate
If we consider a coordinate system in 3space, then
To find
If we fix
Notice that the curve intersects the square at the points where
With that, we can now set up our integrals. The upper bound from
and
Omitting the full process of integration to save space, we find the result
Thus
Substituting this into our initial equation, we get our final result:
#=1/2+1/2P(b^2>=4ac(a, b, c) in [0, L]^3)#
#=1/2+1/2((5+6ln(2))/36)#
#=(41+6ln(2))/72#
#~~0.6272#