Irreal

I did a poor job of making my intended point in Irreal’s Quadratic Equations Reconsidered post. The intended point was that Loh had discovered a method of solving quadratic equations without having to remember—or, heaven forefend, rederive—the quadratic formula or to invoke the usual guessing at the factors.

The post showed why the roots of \(x^{2}+bx+c\) sum to \(-b\) and have the product \(c\). The second observation was that since the sum of the roots is \(-b\) their average is \(-b/2\) and that therefore the roots must have the form \((-b/2-u)\) and \((-b/2+u)\) and that since their product is \(c\), we must have \(u^{2}=b^{2}/4 -c\).

I left the technical details there and judging from the comments, readers thought the method involved remembering \(b^{2}/4 -c\) instead of the quadratic formula. They even pointed out that \(b^{2}/4 -c\) is really just the usual discriminant, for the special case of \(a=1\), in the quadratic formula.

But the real point was the method so let me solve a couple of equations to illustrate. First up: \[x^{2}-8x+15=0\]
The roots sum to \(8\) so their average is \(4\) and therefore the roots are \((4-u)\) and \((4+u)\). Since their product is \(15\), we have \(16-u^{2}=15\) so \(u=1\) and the roots are \((4-1)=3\) and \((4+1)=5\).

When the \(x^{2}\) coefficient is not \(1\), we just divide by it first. Thus to solve \[2x^{2}+8x+6=0\] we divide both sides by \(2\) to get \[x^{2}+4x+3=0\]
Now, the roots sum to \(-4\) so their average is \(-2\) and the roots are \((-2-u)\) and \((-2+u)\). Since their product is \(3\) we have \(4-u^{2}=3\) so \(u=1\) and the roots are \((-2-1)=-3\) and \((-2+1)=-1\).

I chose those roots to be simple integers so as not to bog down in more complex arithmetic but the method works fine when the roots involve radicals or are complex. The point is that the only thing you have to remember is that the sum of the roots is \(-b\)—and that therefore the roots are of the form \((-b/2 \pm u)\)—and that their product is \(c\).

All of us remember quadratic equations, probably not fondly. If the need arises to solve one in our professional lives, our usual solution is to use the quadratic equation, which we may or may not have memorized through having had it drilled into our heads in Algebra I. The quadratic equation is what you get from applying the “complete the square” method to \(ax^{2}+bx+c\). Unless you’re a mathematician, that probably doesn’t mean much to you, even if you understood it at the time.

The mathematician Po-Shen Loh has discovered an incredibly simple method of solving quadratic equations that, astoundingly, doesn’t seem to have been known before. It’s based on the observation that if \(r\) and \(s\) are the roots of \(x^2+bx+c\) then \[x^2+bx+c = (x-r)(x-s) = x^2 -(r+s)x +rs\] so the sum of the roots is equal to \(-b\) and their product is \(c\). Therefore, the problem reduces to finding two numbers whose sum is \(-b\) and product is \(c\).

That may not seem a whole lot easier but Loh observed that if \(r+s=-b\) then their average is \(-b/2\). Two numbers will sum to \(-b\) precisely when their average is \(-b/2\). Those two numbers must be of the form \((-b/2-u)\) and \((-b/2+u)\) (because the average of two numbers lies mid way between them) so we need only find \(u\) such that \(b/2-u\) and \(-b/2+u\) have the product \(c\). That yields the equation \(b^{2}/4 -u^2 =c\). That’s easy to solve for \(u\) but the important point is there’s nothing to remember except that the roots sum to \(-b\) and multiply to \(c\).

You can find more details from Loh’s paper on the method, which is easy to read, but if you want to see a very simple demonstration of the method complete with examples take a look at his video. Watch the video and try a couple of examples to set the method and you’ll never have to worry about the quadratic equation again.

One thing that we programmers tend to obsess about is our code font. They’re mostly the same, of course, but there are a few characteristics that matter to individual programmers and there’s also the general appearance of the font.

For me, the important thing is a clear distinction between the letter o and the number 0, and the letter l and the number 1. I want the distinction to be immediately apparent. If the distinction is that the zero is skinnier than the letter o, I don’t want to use that font. Similarly for l and 1. I don’t want to think about it—the distinction should be immediately clear just by looking.

Equally important but more subtle is the appearance of the font. I use Inconsolata. I originally chose it because back in the old days it was one of the antialiased fonts and just looked much better than the others. These days, of course, most fonts are antialiased but I’ve stuck with Inconsolata mostly because of inertia: it’s good enough.

Still, every time I see a post about a new font, I have to take a look. Recently, I saw a pointer to a site that has screen shots of 111 programmer fonts. It’s a great way of exploring new fonts. On the left there’s a scrollable list of fonts. When you click on one of them, the corresponding font is displayed on the right. It’s a great way of exploring the fonts available. If you’re looking for a new font, this is a good resource. Even if you’re not looking for a new font, take a look. It will give you an idea of what’s available and may impel you to change.