# Are Parabolas With Two x-Intercepts More Numerous Than Parabolas With No x-Intercepts?

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:

1. $4ac<0$,
2. $4ac=0$ & $0<b^2$, or
3. $0<4ac<b^2$

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


No?

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?

• I think instead of thinking about this interms of cases, which you have not proven are equally likely, it would be more helpful to think about the problem in terms of sets. – Aaron Quitta Dec 18 '17 at 18:41
• @Aaron Quitta Which cases did I hastily assume to be equally likely? – Chaim Dec 18 '17 at 18:42
• What do you mean by "randomly"? If $a$, $b$, and $c$ can be any real numbers, how are you assigning the probabilities of (say) $a = 2$, $b= -\pi^e$, and $c = g_{64}$? Keep in mind that you can't say "all values are equally likely", since a uniform probability distribution over all real numbers doesn't exist. – Michael Seifert Dec 18 '17 at 18:45
• $|4ac|<b2$ and $|4ac|=b2$, as well as $4ac<0$ and $4ac>0$, etc. It may seem intuitive to make such equalities but proving such may be helpful to you – Aaron Quitta Dec 18 '17 at 18:46
• See this page. – BruceET Dec 18 '17 at 21:12

## 4 Answers

In a 3-space with coordinates $a$, $b$, $c$, equation $b^2-4ac=0$ is that of a double cone. After a rotation in the $ac$ plane ($u=(a+c)/\sqrt2$, $v=(a-c)/\sqrt2$), this can be rewritten as $$b^2=2u^2-2v^2$$ which is the equation of an elliptical cone centered about the $u$-axis. Points with $b^2-4ac>0$ correspond to points outside the cone. The probability that $b^2-4ac<0$ is then the ratio between the solid angle subtended by the double cone and $4\pi$.

That solid angle can be readily computed with an integral. If $b=r\cos\theta$, $u=r\sin\theta\cos\phi$ and $v=r\sin\theta\sin\phi$, then the equation of the cone becomes $\cos^2\theta=2\sin^2\theta\cos2\phi$. Lets perform the integration on a single sheet of the cone, corresponding to $-\pi/4\le\phi\le\pi/4$. For a given value of $\phi$, we have $\arctan{1\over2\cos2\phi}\le\theta\le\pi-\arctan{1\over2\cos2\phi}$ and the solid angle delimited by this single cone is $$\int_{-\pi/4}^{\pi/4}\int_{\arctan{1\over2\cos2\phi}}^{\pi-\arctan{1\over2\cos2\phi}}\sin\theta\,d\theta\,d\phi=2\arctan2.$$

The total solid angle is the double of that and the probability that $b^2-4ac<0$ is then $${4\arctan2\over4\pi}\approx 0.352416.$$ NOTE 1.

This result was derived assuming that $abc$ are chosen randomly in a spherical region centered at $(0,0,0)$. Choosing different geometries will probably lead to different results.

NOTE 2.

I made two simulations with Mathematica:

1) choosing $abc$ in a cube centered at $(0,0,0)$, I obtained a probability of about $0.373$ for $b^2-4ac<0$, with ten million parabolas;

2) choosing $abc$ in a sphere centered at $(0,0,0)$, I obtained a probability of about $0.351$ for $b^2-4ac<0$, with five million parabolas.

• I don't really follow this, but I think that you're saying that the discriminant will be negative, and the parabola will have no x-intercepts, in 35.3416% of randomly generated quadratic equations. Are you agreeing with the conclusion that most parabolas have two x-intercepts? What's wrong with my rebuttal of that view, in the last grey table? – Chaim Dec 18 '17 at 19:57
• I'm not convinced that the vertex can be above or below $x$-axis with equal probability. Because we are not choosing the vertex at random (uniformly) but rather coefficients $a$, $b$, $c$. – Aretino Dec 18 '17 at 20:24
• @Chaim: This is precisely the type of problem I was alluding to in my comment on the main question. Choosing the vertex to lie in some region of the plane with "equal probability" is not the same as choosing the coefficients $a$, $b$, and $c$ with "equal probability", since the vertex location is related to $a$, $b$, and $c$ in a non-linear way. (To be precise, the vertex of the parabola $y = ax^2 + bx + c$ is located at $(x_0, y_0) = (b/2a, (4ac - b^2)/4a)$. – Michael Seifert Dec 18 '17 at 22:13
• I think that this calculation will hold for any probability distribution for which the selection method is symmetric between $a$, $b$, and $c$, i.e., the probability is constants for all triples $(a,b,c)$ that have the same "distance from the origin" $\sqrt{a^2 + b^2 + c^2}$. For example, we'd get the same result from a 3D normal distribution (with the same $\sigma$ in all three directions) centered at $(0,0,0)$ – Michael Seifert Dec 18 '17 at 22:19
• @Aretino You said: I'm not convinced that the vertex can be above or below x-axis with equal probability. I wonder: Suppose that above-the-axis vertices are more likely. Still, if every randomly selected vertex is equally likely to OPEN up or down, we would get an equal probability of two x-intercepts or none. Right? – Chaim Dec 19 '17 at 12:44

In order to "randomly assign values to $a$, $b$, and $c$," we need a probability distribution. Since there is no uniform distribution over all real numbers (or over any unbounded subset of the real numbers, or over all integers, or over any unbounded subset of the integers), the probability distribution must

• be restricted to a finite interval;
• be non-uniform in probability; or
• be non-uniform and restricted to a finite interval.

Let's make the following assumptions, which seem to be indicated by the notions behind the question:

• The random variables $a$, $b$, and $c$ are independent.
• Each of the variables $a$, $b$, and $c$ has a probability distribution that is symmetric around zero.

If we further assume that each of the variables $a$, $b$, and $c$ has a standard normal distribution, their joint probability distribution is spherically symmetric. As shown in another answer, if the distribution of $(a,b,c)$ is spherically symmetric, the probability that the parabola given by $y=ax^2+bx+c$ intersects the $x$-axis is exactly defined and is approximately $0.648.$ If each of the variables $a$, $b$, and $c$ is uniformly distributed over $[-1,1],$ however, their joint distribution is not spherically symmetric, and the probability that $y=ax^2+bx+c$ intersects the $x$-axis is (by simulation) nearer to $0.627.$

Other choices of probability distributions may give yet other values for the probability that the parabola intersects the $x$-axis. For example, suppose $a$, $b$, and $c$ are each equally likely to be either $-1$ or $1$, and suppose all three variables are independent. This technically satisfies the assumptions laid out above, and the probability that the parabola intersects the $x$-axis is exactly $\frac12$; we have either $b^2 < 4ac$ or $b^2 > 4ac$ depending on whether the signs of $a$ and $c$ are the same or opposite.

Within any cuboid of points $(a,b,c)$ bounded by $-a_0 \leq a \leq a_0,$ $-b_0 \leq b \leq b_0,$ and $-c_0 \leq c \leq c_0,$ however, more than half the volume is occupied by points such that $b^2 > 4ac.$ Any part of the probability distribution in the region where $b^2 < 4ac$ is matched by the distribution in the region where $-b^2 > 4ac.$ That is, the point $(a,b,c)$ is equally likely to fall in either of those two regions. The only way to have a $\frac12$ probability that the parabola intersects the $x$-axis is for there to be a zero probability that $-b^2 < 4ac < b^2.$ Since that case includes all points where $|a|<2|b|$ and $|c|<2|b|,$ we have to completely rule out values of $a$ and $c$ in some neighborhood of $0.$ If, on the other hand, each of the variables $a,$ $b,$ and $c$ has an identical probability distribution with a non-zero density around $0$ (which seems like a reasonable requirement to make), the parabola will intersect the $x$-axis with a probability greater than $\frac12.$

Under the basic assumptions of this answer (independent variables each symmetric around zero), and assuming there is zero probability that any of the variables will be exactly zero (for example, if their distributions are continuous), it is true that the vertex of the parabola is above or below the $x$-axis with probability $\frac12,$ and it also is true that the parabola opens upward or downward with probability $\frac12.$ It is not necessarily true that those events are independent, however. Under certain reasonable assumptions, when the vertex is below the $x$-axis the parabola is more likely to open up than down.

Wlog consider

$$x^2 + 2 b x + c =0$$

Intercepted parabolas occur when $b^2 > c,x-$ axis touching parabolas when $b^2 = c,$ and non- intersecting parabolas when $b^2 < c.$

As you pointed out, you can use the location of the vertex to deduce the number of real roots. Your question is therefore analogous to asking what is the probability of a random real number to be positive. There is no answer to that question if by "random" you are thinking about a uniform distribution, because there is no such distribution over all real numbers. The question is ill-posed.

Consider a uniform distribution with endpoints $[-d,3d]$. The probability to choose a positive number from that distribution is $3/4$, independent of $d$. By taking the limit $d\rightarrow\infty$ you may think that this describes a uniform distribution over all real numbers, but the choice of $3$ is arbitrary and so the probability calculation is arbitrary. Put another way, try visualize a uniform distribution over an infinite range, do you think you can divide this distribution in half so that half the probability is on each side? What about the range $[0,\infty]$?

If, on the other hand your distribution is symmetric about zero, then the probability to choose a positive random number is $1/2$, even for a distribution that extends over all real numbers, like the standard normal distribution, for example.

• Everyone seems to agree that this an important flaw in my thinking, but apparently my confusion goes beyond the scope of a Stack-Exchange-question-and-answer. Is there a 50% probability that a random integer is even? – Chaim Dec 21 '17 at 17:18