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

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

That is, we have one real solution if
$$b^2 -4ac = 0,$$
we have two real solutions if
$$b^2 -4ac > 0,$$
and we have no real solutions if
$$b^2 -4ac < 0.$$

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.