Quadratic Functions: The Discriminant

Published: Dec 25th, 2020

Learning Objectives
By the end of this section, you should be able to:
1. Classify the roots of a function using the determinant

In the last lesson (Lesson #13: Solving Quadratic Functions), we learned how to find the zeroes, or roots, of a quadratic function — the input values for which the function is zero. We also looked at finding the max/min values of a function in our optimization problem. 

Now that we've met these concepts, let's get better acquainted. 

Classifying the Roots of an Equation

In the last lesson, we were introduced to the glorious quadratic equation:

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

Look at it. Memorize it. Love it.

Early on, we were introduced to the different types of functions. Maybe you might recognize a square root in that equation.

A property of square roots is that there is no real number square root of a negative number. In the last lesson, we solved this problem by factoring out \(\sqrt{-1}\), by stating that it is \(i\)

But what does this really mean? Let me tell you.

If you'll remember from Lesson #4: Types of Numbers, there are two main types of numbers: real numbers (\(\mathbb{R}\)) and complex numbers (\(\mathbb{C}\)). These are two separate sets of numbers that, together, form all numbers. 

Until this point, all of the graphs we've drawn have been in the real number plane; that is, both the \(x\) and \(y\) axes contained real numbers.

But that isn't always the case. For instance, we can represent \(2 + 3i\), which has a real component (3) and a complex component (\(3i\)) as: 

Are you surprised it's not a line? Well, thinking back, \(i = \sqrt{-1}\) (it's just a number), so this isn't a function; it's just a single point. 

If this isn't quite clicking yet, no worries, we'll cover it again in a few lessons. 

Now, let's say you used the quadratic equation, and get a negative number in the square root:

\(x = {2 \pm \sqrt{-4} \over 2}\) (a link to the original example from Lesson #13)

Last lesson, we brushed this off by just factoring out the \(i = \sqrt{-1}\). However, this means that our function does not have any real roots; that is, all of its roots are in the complex number plane. This means that we cannot graph the roots of this function on the real number plane. In fact, if you look at a graph of \(5x^2-10x+10\), you'll see that it never intersects the x-axis, meaning that the function is greater than 0 for all real numbers in the domain, or \(\{x\in\mathbb{R}\}\).

However, if we want to figure out whether a given function has any real roots, we don't always need to use the quadratic equation.

For that, we can just use one part: the discriminant. The equation for the discriminant is:

\(b^2 - 4ac\). Notice that it's the part inside the square root? Remember that, for there to be real roots, the value in the square root must be greater than or equal to zero.

Real roots\(b^2 - 4ac \geq 0\)

No real roots\(b^2 - 4ac \lt 0\)

Let's try a couple examples:

Determine the nature of the roots of \(x^2 + 8x + 16\)

Let's use the discriminant:


\(= (8^2) - 4(1)(16)\)

\(= 0\)

Dun dun duuun. Discriminant is zero! What do we do? Well, remember that real roots have a discriminant greater than or equal to zero, so this is a real root. 

But how many real roots? The discriminant, as you might remember, is part of the quadratic equation. When we set the discriminant to zero, we're saying:

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

\(x = {-b \pm \sqrt{0} \over 2a}\)

 \(x = {-b \over 2a}\)

In other words, there's only one real root: \(\frac{-b}{2a}\)

How about a more straightforward example?

Determine the nature of the roots of \(x^2-5x+6\)

Let's use the discriminant:


\(=(-5)^2 - 4(1)(6)\)

\(=25 - 24\)


Okay, so in this case, the discriminant is greater than zero. This means we have two real roots. 

One more example:

Determine the nature of the roots of \(2x^2+3x+2\)

Let's use the discriminant:


\(=(3)^2 - 4(2)(2)\)

\(=9 - 16\)


In this case, the discriminant is less than zero, so we have no real roots!

Aaaand that's it! How about some practice problems?


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