Suppose we randomly assign values to a, b and c in the equation $y=ax^2 + bx + c$.
Whenever the discriminant $(b^2-4ac)$ is positive, the parabola will have two x-intercepts. This will happen whenever $4ac<b^2$, or more explicitly, whenever:
- $4ac=0$ & $0<b^2$, or
Let’s distinguish the question of whether $4ac=0$. If our discussion is limited to quadratic equations (in which a cannot equal $0$), $4ac=0$ only when $c=0$. I don’t know what probability to assign to that case, but I don’t think that the question of probability will be important.
If $4ac=0$, a parabola usually has two x-intercepts:
When $b=0$, $b^2=0$, discriminant=$0$, and the parabola has 1 x-intercept
When b is not $0$, $b^2>0$ and the parabola has 2 x-intercepts
If 4ac does not equal $0$, a parabola usually has two x-intercepts:
4ac < 0 4ac > 0 |4ac|< b2 Two x-intercepts Two x-intercepts |4ac|= b2 Two x-intercepts One x-intercept |4ac|> b2 Two x-intercepts No x-intercept
(I'm assuming that these columns are equally likely, not that the rows are. There are two x-intercepts whenever $4ac<0$ (50% of the time) and sometimes even if $4ac>0$ (some positive percent of the time). So that seems to be a greater-than-50% chance.)
So regardless of whether $4ac=0$, parabolas usually have two x-intercepts.
But that must be wrong. A randomly selected parabola must have the same probability of two x-intercepts as of no x-intercepts. Parabolas can open up or down (with equal probability), with vertex above or below the x-axis (with equal probability).
Vertex is above x-axis Vertex is below x-axis Parabola opens up No x-intercepts Two x-intercepts Parabola opens down Two x-intercepts No x-intercept
SOME REFLECTIONS ON THE RESPONSE SO FAR
As happens with painful regularity, the response is a bit too sophisticated for me to understand totally. But I believe there are two major lines of analysis going: (1) doubt that randomly selected values of a, b and c create an equal likelihood that the parabola's vertex lies above or below the x-axis; and (2) doubt about the concept of selection "at random."
Let's consider that first question first.
In an equation in the form $y=ax^2 + bx + c$, the axis of symmetry is the line $x=-b/2a$, and the y-coordinate of the vertex is the y-value associated with that x-value:
$y = a(x)^2 + b(x) + c$
$y = a(-b/2a)^2 + b(-b/2a) + c$
$y = ab^2/4a^2 – b^2/2a + c$
$y = b^2/4a – b^2/2a + c$
$y = b^2/4a – 2b^2/4a + 4ac/4a$
$y = (b^2 – 2b^2 + 4ac)/4a$
$y = (– 1b^2 + 4ac)/4a$
When will y be positive?
What seems fairly clear to me is that those four alternatives should be equally numerous. I admit to some confusion about the sign of y in those cases where a and c are either both positive or both negative, but it does seem to me that the top-left case should supply some number of positive-y results, the bottom-right case should supply an equal number of negative-y results, and as a whole the table suggests an equal probability that y is positive or negative.
But in any event, even if this analysis is wrong, and a random selection of a, b and c DOES NOT allow an equal likelihood of the vertex above or below the x-axis, doesn't it remain the case that the parabola is equally likely to open up or down, so that half of all parabolas (whether vertex-above or vertex-below) will open towards the x-axis and create two x-intercepts?
As to the second concern, about random selection, I just don't understand the issue. I read the referenced page about probability distributions, and the only thing that strikes me is the paucity of examples with neither greatest nor least possible value, cases like mine in which a, b and c can be any number. Would it improve the question to consider the random selection of an integer, instead of all real numbers? Do I need to constrain the question to values within a certain interval?
What is to be done?