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Consider all quadratic equations with real coefficients:

$$y=ax^2+bx+c \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,, a,b,c \in \mathbb{R}, \, a≠0 $$

I was wondering whether if more of them have real roots, more have complex roots, or a same number of each?

I thought about this graphically, and wlog considered only the ones with positive $a$. By categorising the quadratics by their minimum point, there are an equal number of possible minima below the $x$-axis as above, so it seems that there are an equal number of quadratics with real roots as complex ones.

However, I then considered this problem algebraically by using the discriminant:

$$b^2-4ac$$

If $a$ and $c$ have opposite signs, then the discriminant will be positive, and so the quadratic will have real roots. This represents $50\%$ of the quadratic equations possible.

However, if $a$ and $c$ have the same sign, then some of these quadratics have real roots, whilst others have complex roots depending on whether $4ac$ is bigger than $b^2$ or not.

This suggests that there are more quadratics with real roots which contradicts the above answer.

Is the reason I've reached contradicting answers something to do with infinites, and that I can't really compare them in the way I've done above?

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  • $\begingroup$ $\mathbb{R}\subset \mathbb{C}$ $\endgroup$ – Qwerty Jun 17 '16 at 14:38
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    $\begingroup$ Is $\dfrac{1}{4}\times \infty$ strictly smaller than $\dfrac{3}{4}\times \infty$ ? $\endgroup$ – Henry Jun 17 '16 at 14:43
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    $\begingroup$ You can also argue that the number of polynomials with two real roots is excactly the same as the number of real polynomials with complex roots. This is because of the two polynomials $$ax^2 + bx + c,\quad -ax^2 - bx + c$$one will be of one kind and the other of the other, so you can pair them up, two by two. In the end, though, if you have two sets that are infinitely big, saying that one is twice as large as the other doesn't make sense. If one is actually bigger than the other, then the ratio will be so huge it doesn't make sense to put a number to it. $\endgroup$ – Arthur Jun 17 '16 at 14:49
  • $\begingroup$ Such a question only makes sense after you have chosen a probability measure on the set of all quadratic functions, because it amounts to the following: What is the probability that a "random" quadratic equation with real coefficients has real roots? $\endgroup$ – Christian Blatter Jun 17 '16 at 16:14
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There are two ways to write a quadratic equation.

1) $y = ax^2 + bx + c$ and 2) $y = a(x-h)^2 + k$

Both models have equal numbers of quadriatic formulas; i.e. the cardinality of $\mathbb R^3$.

However "random" distribution results in different concentrations when represented as corresponding elements of $\mathbb R^3$ and "randomness" is not at all unambiguously defined.

Model 2 has real roots if and only if k is non-positive. So that's a probability of 1/2.

Model 1 has roots if a and c are opposite signs OR if $b^2 - ab \ge 0$ which is ... I dunno... It's too early and my motivation isn't strong enough to figure it out although I know it can be done.... but it is significantly less than 1/2.

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Here's a way of thinking about it.

Consider $X = \mathbb R^3$ be euclidean three space and you can pick a point out of it at random.

Now let $f(a,b,c) = (a, -b/2, c-b^2/4)$. This is a bijective 1-1 map.

Let $Y = f(X) = \mathbb R^3$.

Both $X$ and $Y$ are the exact same shape with the exact same number of points. But if you pick a point from one and look at a corresponding point in the other it will seem that the space and concentration has been "warped" and "randomness" in one is not the same as randomness in the other.

The resolution seems to me that "random" and "probability" is not as simple or intuitive as we thought and, as stated, the "probability" of a quadratic having real roots is not actually a well-defined question. ... as yet.

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