Discriminant Problems

Question

We all know that the discriminant is the part $b^2-4ac$ of the equation

$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

that we use to find the roots of a quadratic equation eg: $ax^2+bx+c=0$ or the part in a trinomial expression like $ax^2+bx+c$.

What I want to know is what kind of ideas we can conclude for a given inequality of a discriminant. We know that in quadratic formulas the number of real roots depends on the case whether the discriminant is <0, 0, or >0. But if given only the discriminant and the inequality of it, for examples like this:

$b^2-4ac=0$

$b^2-4ac>0$

$ b^2-4ac<0 $

What can we conclude by the given information for each case? How can it help to reveal certain problems involving trinomial expressions?

Thank you in advance. It will be really nice if you could solve my problem and provide links that can lead me to study further on this topic.

Answer 1

Let me rewrite the quadratic equation as $$x = -\frac{b}{2a} \pm \sqrt{-\frac{1}{a} \left(\frac{4ac - b^{2}}{4a}\right)}.$$

Notice that I wrote it in terms of the coordinates of the vertex. Now, let us consider some cases about the vertex $(h,k)$, assuming that the quadratic opens upward.

1.) $k < 0$.

In this case, the vertex is below the $x$-axis and the quadratic must have two real roots. Now, \begin{align*} \frac{4ac - b^{2}}{4a} < 0 \\ 4ac - b^{2} < 0 \\ -(4ac - b^{2}) > 0 \\ b^{2} - 4ac > 0 \end{align*}

which matches our criteria that a quadratic has two real roots if and only if $b^{2} - 4ac > 0$.

2.) $k = 0$.

This means that the vertex is on the $x$-axis and the quadratic has a double root. Now, making a similar solution from the first case, \begin{align*} \frac{4ac - b^{2}}{4a^{2}} &= 0 \\ 4ac - b^{2} &= 0 \\ -4ac + b^{2} &= 0 \\ b^{2} - 4ac &= 0 \end{align*}

This matches our criteria that a quadratic has a double root if and only if $b^{2} - 4ac = 0$.

3.) $k > 0$.

In this case, the vertex is above the $x$-axis and since it is the minimum of the function, the quadratic does not have any real root. Now, \begin{align*} \frac{4ac - b^{2}}{4a} > 0 \\ 4ac - b^{2} > 0 \\ -4ac + b^{2} < 0 \\ b^{2} - 4ac < 0 \end{align*}

and this matches our criteria that a quadratic does not have any real root if and only if $b^{2} - 4ac < 0$.

I'll leave the case where a quadratic opens downward to you.

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TL;DR Based on the discriminant of a quadratic, you can get some information about the vertex.