Question

Given `{a,b,c} in [-L,L]` What is the probability that the roots of `a x^2+b x + c = 0` be real?

Answer 1

`(41+6ln(2))/72~~0.6272`

Explanation:

We will operate on the assumptions `L>0` and that the probability is equal that `(a, b, c) = (x_1, x_2, x_3)` for all `(x_1, x_2, x_3) in [-L, L]^3`

Initial observations:

• `ax^2+bx+c = 0` has real roots if and only if the discriminant of the quadratic is nonnegative, that is, `b^2-4ac >= 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>=4ac|a, c<=0) = P(b^2>=4ac|a, 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>=4ac|a, c>=0) = P(b^2>=4ac|a, b, c>=0)`

With these, we can reformulate the problem as follows:

`P(ax^2+bx+c" has real roots") = P(b^2>=4ac)`

`=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>=4ac|a, c<0 or a, c >0)`

`=1/2+1/2P(b^2>=4ac|a, c>0)`

`=1/2+1/2P(b^2>=4ac|a, b, c>0)`

`=1/2+1/2P(b^2>=4ac|(a, b, c) in [0, L]^3)`

We will now calculate `P(b^2>=4ac|(a, b, c) in [0, L]^3)`

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If we consider a coordinate system in 3-space, then `[0, L]^3` is equivalent to a cube with side length `L` resting in the first octant and having sides colinear with the axes. Let `S` be the solid bounded by this cube and above by the surface `y^2 = 4xz`, and `V_S` be the volume of this solid. Then the probability that `b^2>=4ac` given an arbitrary `(a, b, c) in [0, L]^3` is equal to the probability that `(a, b, c)` falls within `S` given that `(a, b, c)` falls within the cube, i.e. `V_S/V_"cube" = V_S/L^3`

To find `V_S`, we will first consider the area `A_b` of the slice of `S` found by fixing `y=b` for some `b in [0, L]`, and then we will add up all of these areas by integrating `A_b` as `b` goes from `0` to `L`.

If we fix `y=b` for some `b`, then `A_b` is the area bounded by the lines `x=0, x=L, z=0, z=L` and the curve `4xz=b^2`, which we can rewrite as `z=b^2/(4x)`

![desmos.com]()

Notice that the curve intersects the square at the points where `x=L` or `z=L`, that is, at `(x, z) in{(L, b^2/(4L)), (b^2/(4L), L)}`

With that, we can now set up our integrals. The upper bound from `x=0` to `x=b^2/(4L)` is the line `z=L`. The upper bound from `x=b^2/(4L)` to `x=L` is the curve `z=b^2/(4x)`. Thus

`A_b = int_0^(b^2/(4L))Ldx + int_(b^2/(4L))^Lb^2/(4x)dx`

and

`V_S = int_0^LA_bdb`

Omitting the full process of integration to save space, we find the result

`V_S = (5+6ln(2))/36L^3`

Thus

`P(b^2>=4ac|a, b, c in [0, L]) = V_S/L^3= (5+6ln(2))/36`

Substituting this into our initial equation, we get our final result:

`P(ax^2+bx+c" has real roots")`

`=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`