I really don’t like the quadratic formula. As a teacher, it feels like one of the absolute worst examples of what’s wrong with mathematics education: An arbitrary formula with weird aspects, including what is often the first appearance for students of the plus-minus sign. Sure, we can do algebraic manipulation to show why it is what it is, but it still feels like magic. In high school, I worked so hard to memorize it that, after over twenty years out of math classrooms, it came back to me instantly: \[x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\]
I’ve been teaching Algebra II now for a few years. A year or two ago, I wrote the quadratic formula using the vertex instead of the standard form, and got a far more logical looking thing: \[x=h\pm\sqrt{-\frac{k}{a}}\]
That looks so much better, but I still wasn’t quite sure how to motivate that negative sign in the radical. I could do so logically, but that logical argument felt circular to me.
A few nights ago, I dreamed of explaining this version, and the explanation had an element to it that I hadn’t realized before. It even allows for a more general method than just finding roots.
Let’s say we have a given quadratic function. For its graph, we know the coordinates of the vertex (\((h, k)\)) and its vertical stretch (\(a\)). We’d like to know the input values (\(x_0\)) that will generate a given output value (\(y_0\)).
Given the information we know, we can write the quadratic function this way: \[f(x)=a(x-h)^{2}+k\]
Since quadratic function graphs are symmetrical around their vertices, we know that any value of \(y_0\) will correspond to exactly two input values \(x_0\), and that these two input values will be equidistant from \(h\).
But we can demonstrate this fairly readily regardless: Let \(d\) equal the distance from \(x_0\) to \(h\), that is, \(d=|x_0-h|\).
We want to know \(x_0\) such that \(f(x_0)=y_0\), that is: \[a(x_0-h)^{2}+k=y_0\]
Since \(x_0-h\) is squared, it doesn’t matter whether it’s positive or negative: The result will be positive, and equal to \(d^2\). So \[ad^2+k=y_0\]
Now we can solve for \(d\): \[d^2=\frac{y_0-k}{a}\\d=\sqrt{\frac{y_0-k}{a}}\]
This is the horizontal distance on the graph between the vertex and the target input values. Starting at \(h\), we go \(d\) in either direction, and our resulting value will yield \(y_0\) as its output: \[x_0=h\pm\sqrt{\frac{y_0-k}{a}}\]
For instance, if we want to know for what values of \(x, f(x)=4(x-3)^2+6=10\), those are \(x_0=3\pm\sqrt{\frac{10-6}{4}}=\{2,4\}\). This is more useful than the standard Quadratic Formula because it allows us to find the inputs corresponding to any arbitrary output, not just the roots.
We could do the same thing for the standard Quadratic Formula, of course. The revised formula is \[x=\frac{-b\pm\sqrt{b^{2}+4a(y_0-c)}}{2a}\]
… which is even more complicated than the standard form.
Let’s rearrange the general vertex version a little: \[x_0-h=\pm\sqrt{\frac{y_0-k}{a}}\] This shows the relationship plainly: Regarding the graph, we are relating the horizontal distance between an input value and the vertex to the vertical distance between the corresponding output value and the vertex. This has even more discovery value than \(x=h\pm\sqrt{-\frac{k}{a}}\) does.
The roots can be seen as a special case where \(y_0=0\), and so \(x_0=h\pm\sqrt{\frac{0-k}{a}}\).
If a quadratic function is currently in standard form, we have two options for finding its roots: We could use the quadratic formula directly, or we could convert the function to vertex form and then use this easier form.
There are two mitigating factors for converting to vertex form first. One is that the Vertex Form of the Quadratic Formula does have a radical in the denominator, something we could address with a minor change: \(x_0=h\pm\frac{\sqrt{-ka}}a\). The other, though, is that \(k=\frac{4ac-b^2}{4a}\) can be a fraction even when the roots themselves are integers.
More specifically, a quadratic with integer roots will have \(h\) being measured in halves (either an integer or a half-integer), and hence \(k\) being measured in quarters. We could force this issue even further with \(x_0=h\pm\frac{\sqrt{-4ka}}{2a}\), but at some point it gets complex enough that there’s not a noticeable benefit to just using the original version of the Quadratic Formula directly.
The key here, though, is that a quadratic function with integer roots won’t ever have an irrational denominator in \(\sqrt{\frac{-k}{a}}\). There may be fractions in the intermediate forms, but the denominator in that situation will always simplify to 1 or 4. So if the concern is for students who struggle with decimal values, keeping the roots to integers will keep the fractions in line. (The vertical stretch can even be a non-integer without creating an irrational denominator in the radicand!)
And if you really want to avoid fractions, create quadratics with integer roots that differ by an even amount, as well as an integer vertical stretch. That way, both coordinates of the vertex will always be integers.