I thought I would take a crack at deriving the quadratic formula as discussed in last week’s post, Thinking Like a Mathematician. I wanted to discuss last week’s post with my fourteen-year-old daughter. If I could walk her through the steps to derive the quadratic formula like my tutor did all those years ago, would she agree that math was more interesting and easier to remember if you derived things out like a mathematician, rather than memorizing facts?

Well, I’d better make sure I still can derive the quadratic formula, before I do that, so I’m going to work through that here. Consider these notes for a conversation with Flynn later this weekend.

So the quadratic formula is basically solving

\[ax^2 + bx + c = 0\]

in terms of \(x\). That is, we want to rewrite the above as

\[x = Q\]

where \(Q\) is going to be the quadratic formula.

It has been nearly forty years since high-school algebra, but I kind of remember this being about something something complete the square mumble mumble. Ok, I had to review a bit about completing the square. Let’s say we have

\[x^2 + bx = 0\]

Now notice that

\[\big{(}x^2+\frac{b}{2}\big{)}^2 = x^2 + bx + \frac{b^2}{4}\]

To complete the square we simply add \(\frac{b^2}{4}\) to each side of \(x^2 + bx = 0\):

\[\begin{align} x^2 + bx & = 0 \\ x^2 + bx + \frac{b^2}{4} & = \frac{b^2}{4} \\ \big{(}x + \frac{b}{2}\big{)}^2 & = \frac{b^2}{4} \end{align}\]

Getting back to our quadratic equation:

\[\begin{align} ax^2 + bx + c & = 0 \\ ax^2 + bx & = -c \\ x^2 + \frac{b}{a}x & = -\frac{c}{a} \end{align}\]

Now we can complete the square on the left side of that equation:

\[\begin{align} x^2 + \frac{b}{a}x & = 0 \\ x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} & = -\frac{b^2}{4a^2} \\ \big{(}x + \frac{b}{2a}\big{)}^2 & = -\frac{b^2}{4a^2} \end{align}\]

Plugging that back in to our previous equation yields

\[\begin{align} x^2 + \frac{b}{a}x & = -\frac{c}{a} \\ \big{(}x + \frac{b}{2a}\big{)}^2 & = -\big{(}\frac{c}{a}+\frac{b^2}{4a^2}\big{)} \\ & = -\frac{4ac+b^2}{4a^2} \end{align}\]

Now we can turn that into

\[\begin{align} x + \frac{b}{2a} & = \pm\frac{\sqrt{4ac + b^2}}{2a} \\ x & = \frac{-b\pm\sqrt{4ac + b^2}}{2a} \end{align}\]

And there is our quadratic formula.

Ok, I’m sure I’ve amazed you with my elite high-school algebra skills. Still, at least I feel ready to discuss this with an actual high-school algebra student!