Why does the discriminant in the quadratic formula reveal the number of real solutions to a quadratic equation?

That is, we have

onereal solution if

$$b^2 -4ac = 0,$$

we havetworeal solutions if

$$b^2 -4ac > 0,$$

and we havenoreal solutions if

$$b^2 -4ac < 0.$$

**Answer**

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.

**Attribution***Source : Link , Question Author : Community , Answer Author : M. Winter*