Why does the discriminant in the Quadratic Formula reveal the number of real solutions?

Think about it geometrically $-$ then compute.

Everyone knows $x^2$ describes a parabola with its apex at $(0,0)$. By adding a parameter $\alpha$, we can move the parabola up and down: $x^2+\alpha$ has its apex at $(0,\alpha)$. Looking at the graph as it moves up and down you immediately see how the number of zeros depends on $\alpha$:

• for $\alpha>0$ we have no zeros.
• for $\alpha=0$ we have a single zero.
• for $\alpha<0$ we have two zeros.

Now we can introduct a second paramter $\beta$ to move the parabola left and right: $(x-\beta)^2+\alpha$ has its apex at $(\beta,\alpha)$.

^{Note: we used the fact that given a function $f(x)$, the graph of the function $f(x-\beta)$ looks exactly like the one of $f$ but shifted to the right by an amount $\beta$.}

But of course, shifting a function left and right does not alter the amount of zeros. So it still only depends on $\alpha$. We expand the term a bit:

$$(x-\beta)^2+\alpha=x^2-2\beta x+\beta^2+\alpha.$$

Would your quadratic equation be given in this form, you would immediately see the amount of zeros as describes above. Unfortunately it is mostly given as

$$ x^2+\color{red}px+\color{blue}q=0$$

So instead, you have to look at what parts of the $\alpha$-$\beta$-form above corresponds to these new parameters $p$ and $q$:

$$ x^2\color{red}{-2\beta} x+\color{blue}{\beta^2+\alpha} = 0.$$

So we have $p=-2\beta$ and $q=\beta^2+\alpha$. If we only could extract $\alpha$ from these new parameters, we would immediately see the amount of zeros. But wait! We can!

$$\alpha=q-\beta^2=q-\left(\frac p2\right)^2.$$

This is exactly what you know as (the negative of) the discriminant.

I used the form $x^2+px+q=0$ and you used $ax^2+bx+c=0$. I hope this is not confusing you. Just divide by $a$ (if $a$ is non-zero):

$$x^2+ \color{red}{\frac ba}x+\color{blue}{\frac ca}=0$$

If you set $p=b/a$ and $q=c/a$ and plug this into my discriminant from above you obtain the one you know:

$$\left( \frac {\color{red}p}2 \right )^2-\color{blue}q = \frac{(\color{red}{b/a})^2}4-\color{blue}{\frac ca}=\frac{b^2}{4a^2}-\frac{4ac}{4a^2} = \frac{b^2-4ac}{4a^2}.$$

Because $4a^2$ is always positive it suffices to look at $b^2-4ac$ as you did in your question.

Because for $a\neq0$ we obtain: $$ax^2+bx+c=a\left(x^2+\frac{b}{a}x+\frac{c}{a}\right)=a\left(\left(x+\frac{b}{2a}\right)^2-\frac{b^2-4ac}{4a^2}\right).$$ Now, we see that if $b^2-4ac<0$ then $\left(x+\frac{b}{2a}\right)^2-\frac{b^2-4ac}{4a^2}>0$,

which says that the equation $ax^2+bx+c=0$ has no solutions.

For $b^2-4ac=0$ we have one root only: $$x_1=-\frac{b}{2a}$$ and for $\Delta=b^2-4ac>0$ our equation has two distinct roots: $$x_1=\frac{-b+\sqrt{\Delta}}{2a}$$ and $$x_2=\frac{-b-\sqrt{\Delta}}{2a}$$ because in this case we obtain: $$ax^2+bx+c=a\left(x^2+\frac{b}{a}x+\frac{c}{a}\right)=a\left(\left(x+\frac{b}{2a}\right)^2-\frac{b^2-4ac}{4a^2}\right)=$$ $$=a\left(x+\frac{b}{2a}-\frac{\sqrt{\Delta}}{2a}\right)\left(x+\frac{b}{2a}+\frac{\sqrt{\Delta}}{2a}\right)=$$ $$=a\left(x-\frac{-b+\sqrt{\Delta}}{2a}\right)\left(x-\frac{-b-\sqrt{\Delta}}{2a}\right)=a(x-x_1)(x-x_2).$$

The discriminant tells you the number of real solutions if $a,b,c$ are real.

That is because $\pm\sqrt{b^2-4ac}$ is real if and only if $b^2-4ac\ge 0.$

If $a,b,c$ are not real then there's more to say than that.

Other answers do calculations and it's probably what you asked for, but the answer to "why?" can be deeper, more subtle than that. The discriminant reveals the number of real solutions because:

• it's useful to know the number of real solutions
• and at least one formula (function of $a$, $b$ and $c$) that leads to this number does exist.

We discovered (derived) the simplest such formula and said: 'It's so useful it deserves a name, let it be discriminant.' At this point it doesn't matter what the formula is in terms of $a$, $b$ and $c$. I mean, if it were something different than $b^2-4ac$ then this other function would be called 'discriminant' in the first place.

I checked Wikipedia and the first thing I saw was:

In algebra, the discriminant of a polynomial is a polynomial function of its coefficients, which allows deducing some properties of the roots without computing them. […] for a polynomial of an arbitrary degree, the discriminant is zero if and only if it has a multiple root, and, in the case of real coefficients, it is positive if and only if the number of non-real roots is a multiple of 4.

Note few things:

• If you take a polynomial of degree 2 (so there are two roots, equal or not, either one real or not) with real coefficients, then the above property translates exactly to what you're asking about.
• Property is the right word here. The above is not a strict definition for sure. $2(b^2-4ac)$ or $(b^2-4ac)^3$ have the same property.
• Wikipedia gives this rather simple property first, the rather complex general definition later. This suggests it's the property that is important, not the definition itself. The desired property is a reason to have a definition.

In this context, if the question was "why does the value of $b^2-4ac$ reveal the number of real solutions?" then some plain calculations would be the full answer. But the question is "why does the discriminant reveal… ?", so the deeper answer is:

Because we wanted it to reveal this in the first place. We deliberately defined the discriminant to have this property.