## Thursday, 6 April 2017

### The quadratic formula is wrong.

No, of course the quadratic formula isn't incorrect. If you want the solutions to the equation \begin{equation} ax^2 + bx + c = 0 \label{tradQ} \end{equation} then they are, of course, given by the familiar formula \begin{equation} x= \frac{-b \pm \sqrt{b^2-4ac}}{2a}. \label{sol} \end{equation} You can't really argue with that. It's established using nothing more than very simple algebra, and I am not going to claim that basic algebra is incorrect.

But it is ugly: and I've seen people mix up where that $$4$$ and $$2$$ go. Fortunately, there's a neat way of having a simpler formula, and it's obtained by the equally neat trick of rephrasing the problem. (This is a standard trick in mathematics, and it doesn't get nearly enough publicity.)

So, instead of asking for solutions to \eqref{tradQ}, I ask for solutions to the equation \begin{equation} ax^2+2bx+c = 0. \label{newQ} \end{equation}

In other words, my new $$b$$ in \eqref{newQ} is half of the original $$b$$ in \eqref{tradQ}. There are two ways of seeing what the consequences are.

First, I can replace $$b$$ in the formula \eqref{sol} by $$2b$$. This gives \begin{equation*} \begin{split} x & = \frac{-2b \pm\sqrt{4b^2-4ac}}{2a} \\ &= \frac{-2b \pm 2\sqrt{b^2-ac}}{2a} \\ &= \frac{-b\pm\sqrt{b^2-ac}}{a} \end{split} \end{equation*}

Alternatively, you can go through the completion of the square argument: \begin{equation*} \begin{split} & \qquad ax^2+2bx+c = 0 \\ \Leftrightarrow & \qquad x^2+2\frac{b}{a}x+\frac{c}{a} = 0\\ \Leftrightarrow & \qquad (x+\frac{b}{a})^2 - \frac{b^2}{a^2} + \frac{c}{a} = 0 \\ \Leftrightarrow & \qquad (x+ \frac{b}{a})^2 = \frac{b^2-ac}{a^2}\\ \Leftrightarrow & \qquad x + \frac{b}{a} = \pm \frac{b^2-ac}{a}\\ \Leftrightarrow & \qquad x = \frac{-b \pm \sqrt{b^2-ac}}{a} \end{split} \end{equation*}

Both ways (of course!) give the same answer, \begin{equation} \label{newRoots} x = \frac{-b \pm \sqrt{b^2-ac}}{a} \end{equation} but now without the pesky coefficients of $$2$$ and $$4$$ to get in the right place.

But something else comes along for the ride. It's now very obvious that the discriminant \begin{equation*} \Delta=b^2-ac \end{equation*} looks an awful lot like a determinant: in fact \begin{equation*} \Delta = - \left| \begin{array}{cc} a & b \\ b & c \end{array} \right|. \end{equation*}

So in \eqref{newRoots} we have a simpler formula for the roots of a quadratic. But we also have a question, which is: what on earth is going on with the determinant? What is this matrix doing that the sign of its determinant should decide whether or not the quadratic has real roots?

The simpler formula for the roots is nice, but really this question (and its answer) are the reasons why I think that what I have just gone through is the 'right' way to state (and solve) a quadratic equation. The answer involves a journey to the fascinating world of conic sections, which deserves its own discussion, so I'll leave it to a future note.