Written by Sarah Stecher published 8 months ago

The quadratic formula is one of the most quintessential topics in high school mathematics. Many adults can still sing the quadratic formula song taught to them by their Algebra teacher. The problem? Nobody really understands why it works.

The quadratic functions unit is a staple of both an Algebra 1 and Algebra 2 course. We introduce important ideas about their rates of change, their maxima or minima, and their graphs. We explore quadratics graphically, analytically, and numerically. We practice writing quadratics in equivalent forms to reveal key features like intercepts, vertices, or transformations. We solve quadratic equations using inverse operations for simple quadratics with no linear term, using graphs and finding intersections, factoring and applying the zero product property, and then…with the quadratic formula. Identify a, b, and c. Plug them in. Get your answers. They might be imaginary.

With all the big conceptual ideas intrinsic to quadratic functions, I struggled with the procedural, plug-and-chug approach of the quadratic formula. It felt like a departure from all the other rich ideas we had been working hard to develop. I started to wonder if there was a better way to teach the principle behind the quadratic formula (a sure way to find all solutions, real or imaginary, to any quadratic equation) without resorting to just giving students a formula they don’t understand. How was the quadratic formula connected to the ideas we had already studied? It turns out that it’s *incredibly* closely linked, but the current form of it completely obscures those connections. Enter: the **new** quadratic formula.

Big idea: If we know the vertex of a parabola and the distance between the vertex and an x-intercept, we can find both zeros using symmetry.

### How to build up to the quadratic formula

Before we can teach students anything related to the quadratic formula, we must establish the important idea of symmetry. The graphs of quadratic functions are symmetric across their axis of symmetry, which passes through the vertex. This implies that if you know where the axis of symmetry is and one other point on the parabola, you can automatically determine another point on the parabola. This idea is helpful for finding zeros and x-intercepts, but it extends far beyond this. Students can use symmetry to find missing values of a quadratic function, identify an axis of symmetry, and even determine the vertical stretch. This idea is heavily developed in Lesson 2.4 Features of Quadratic Functions. Note how the Check Your Understanding questions require students to *understand* a parabola’s symmetry, not just execute a known procedure.

### Establish the idea before you generalize with an algebraic expression.

By the time students get to Lesson 7.8: Solving Quadratics using Symmetry, they have familiarity with solving quadratics using factoring or inverse operations, when possible. They also understand how to leverage the symmetry of a parabola to identify various features of the quadratic. But now it’s time to put it all together.

Because the quadratic formula tends to be used when zeros aren’t integers, we use the question below early on in the activity to again drive home the point that a distance between vertex and an intercept, regardless of how $\text{\textquotedblleft}$ugly$\text{\textquotedblright}$ that number is, is sufficient to determine all zeros.

Every question up to this point is driving students toward the essential question: how do we determine the distance between the axis of symmetry and an intercept? That’s what the rest of the lesson seeks to answer.

Students understand intuitively that for parabolas with large vertical stretches, the distance to an intercept will be smaller. Each one unit change in x produces a large change in y, so getting back to the x-axis happens in fewer steps. Similarly, parabolas with a vertex far from an axis will generally require a greater distance to get to an intercept. It makes sense, then, that the distance between the axis of symmetry and an intercept would be determined by the vertex of the parabola, as well as the vertical stretch.

Returning to Janyce’s parabola, now with its equation given in standard form, students determine the axis of symmetry and then calculate the value of the expression $\sqrt{h^2-\frac{c}{a}}$ where *h* is the x-coordinate of the vertex (just like in vertex form) and *c* and *a* are the coefficients of the standard form equation. They notice that this expression represents the distance they found in question 2b. Applying this to a few more quadratics, students are able to generalize that the x-intercepts can always be found by adding and subtracting the distance from the axis of symmetry. In other words, the solutions to a quadratic equation are given by

$x=h±\sqrt{h^2-\frac{c}{a}}$.

In an Algebra 1 course, we chose to have students discover the $\text{\textquotedblleft}$distance$\text{\textquotedblright}$ equation using inductive reasoning. In a Precalculus course, you may choose to have students derive the formula from scratch!

### What about the discriminant?

When we think about the word $\text{\textquotedblleft}$discriminant$\text{\textquotedblright}$, our mind jumps immediately to $b^2 -4ac$. But what does the term $\text{\textquotedblleft}$discriminant$\text{\textquotedblright}$ actually mean?

$\text{\textquotedblleft}$*In mathematics, the discriminant of a polynomial is a quantity that depends on the coefficients and allows deducing some properties of the roots without computing them.*$\text{\textquotedblright}$

A key feature of the quadratic formula is being able to determine the number of solutions of a quadratic equation and whether they are real or imaginary. The value of the expression $b^2 -4ac$ gives this information. However, students are left again to evaluate and apply a rule. If $b^2-4ac$ > 0 then the quadratic has two real solutions, and so on.

Using the new quadratic formula, students can use the value of $h^2-\frac{c}{a}$ to get at the same information but in a more intuitive way.

If $h^2-\frac{c}{a}$ is positive, then there is a known distance between the AOS and an intercept. Adding or subtracting that distance from the x-coordinate of the vertex will provide both solutions.

If $h^2-\frac{c}{a}$ is zero, then there is NO distance between the vertex and the x-intercepts! The vertex must be the x-intercept! The shape of a parabola and its symmetry means that the x-coordinate of the vertex is the only solution.

If $h^2-\frac{c}{a}$ is negative, then it’s impossible to find a distance between the axis of symmetry and an intercept. Thus, the parabola must be strictly above or strictly below the x-axis. We can use the same formula to find both complex zeros but they won’t be x-intercepts. In the Math Medic curriculum we save the idea of complex zeros for Algebra 2.

**At this point you may be wondering: why have students first calculate $\text{\textquotedblleft}$h$\text{\textquotedblright}$ when the traditional quadratic formula is based only on values of a, b, and c? Isn’t this extra work?**

The $\text{\textquotedblleft}$extra step$\text{\textquotedblright}$ of finding *h* first is intentional and, we believe, well worth it. This is because it is very challenging to make any kind of meaning out of the quadratic formula. Students generally just plug and chug values of $\text{\textquotedblleft}$a$\text{\textquotedblright}$, $\text{\textquotedblleft}$b$\text{\textquotedblright}$, and $\text{\textquotedblleft}$c$\text{\textquotedblright}$ and then magically arrive at an answer. Since our emphasis is on making sense of mathematics and not just getting answers, we wanted to use an approach that is more based on the underlying concept of symmetry, connecting to what students already know about quadratics. $\text{\textquotedblleft}$Seeing$\text{\textquotedblright}$ the axis of symmetry in the expression is helpful toward this end.
Additionally, by finding *h* first, students that have made a computational error have an opportunity to self-correct, because they can think about if their value for the vertex makes sense based on the equation. When students use the quadratic formula, they rarely are able to catch mistakes because the meaning of each part of the formula is arbitrary to students.

The quadratic formula is complicated because it insists on making common denominators and rationalizing the denominator, not because the process of finding zeros is actually that complex. An $\text{\textquotedblleft}$unsimplified$\text{\textquotedblright}$ version of the formula actually would shed a lot more light on the symmetry of the quadratic and the relationship between factored and standard form. The expression for *h* is baked into the quadratic formula with the $\frac{-b}{2a}$ expression, but students aren’t usually able to identify this.

Here’s a full proof for how the **new** quadratic formula was derived and why it’s equivalent to the traditional formula.

### But my state standards explicitly mention the traditional quadratic formula and its discriminant!

We have found this to be true in a few cases. If your state requires the use of the traditional formula, there’s still a way to help students make better connections to symmetry. Simply split the quadratic formula into two parts!

$\frac{-b}{2a}±\frac{\sqrt{b^2-4ac}}{2a}$

The expression $\frac{-b}{2a}$ represents the x-coordinate of the vertex (or the axis of symmetry) and the expression $\frac{\sqrt{b^2-4ac}}{2a}$ represents the distance between the axis of symmetry and an intercept! Note, however, that squaring the discriminant does NOT give the distance between the axis of symmetry and an x-intercept in this scenario.

Ready to give this lesson a try? Check out the full lesson in our Algebra 1 course!