Loh Po-Shen is a graduate of California Institute of Technology and an associate professor at Carnegie Mellon University.
In October 2019, Loh published a new way to solve quadratic equations. It is breathtakingly simple and easy to understand, and much faster in producing solutions. And yet if it so simple, why has it not been put together since Euclid first discovered quadratic equations? Loh searched extensively and found no other mathematicians had done this. (I said put together and not discovered because the techniques he uses are already known.)
Loh’s method involves simple distributive mathematics and pythagorean triples, though he does not used the latter term in this video. In his final example, he shows how his method produces the traditional quadratic equation (-b +/- √(b^2-4ac))/2a. As a fan of and dabbler in number theory, this is staggeringly beautiful.
Science Alert reported on this as well. The page contains another video explaining the technique.
That’s a very elegant explanation, and a method that, with a bit of practise, could be done all in the head in a minute. Thanks!
This — and that ScienceAlert link, and googling has found several other recent articles — make it sound like there’s something new here. Maybe Loh independently derived it, he has a nice way of teaching it to students, makes the proof clear, etc. But the basic idea isn’t new.
I’ve seen it called the “pq formula” as opposed to the “abc formula.” (The latter is “the quadratic formula” that most people memorized in school, the one you mentioned above.)
The a coefficient of x^2 doesn’t need to be 1 in the generic case of ax^2 + bx + c = 0. However, you can simply divide by a to get the new equation x^2 + b/a + c/a = 0.
That’s the same as x^2 + px + q = 0 when you rename the coefficients like so:
p = b/a
q = c/a
With a bit of algebra, you get the pq formula:
x = -p/2 ± √[(p^2)/4 – q]
Maybe Loh independently derived it I don’t think so. The slide says ancient Greeks and Babylonians knew it.
Sorry, there’s a typo. It should be obvious, but I left out the second “x” in bold:
x^2 + (b/a)x + c/a = 0
huh, that’s pretty clever.
A good math teacher is a rare occurrence, and can make a huge difference in not only learning the concepts, but also liking (or at least not hating) the subject in general.
jrkrideau:
“Independently” means the derivation doesn’t depend on such things. It doesn’t mean that none exist — even if they do, it could still be independent of them.
In any case, ancient approaches weren’t quite the same: no algebra, no complex numbers, etc.
I’m failing to see the improvement, or indeed any real difference, to completing the square;
x² + Bx + C = 0 (1)
can be written
y² – A = 0 (2)
where y = x + B/2, A = B²/4 − C
(2) can be written
(y − √A)(y + √A) = 0 (3)
So y = ±√A
@2 consciousness razor
You appear to be missing the point – this is *not* about new maths: as he was quite clear about on multiple times through that video, the components have been known for a long time. (And sure, the combined approach wouldn’t have been known to the ancients in exactly that form. How could it, being based on multiple advances over the following few thousand years? Hit point was clearly that bits are old, other bits are *very* old, but nothing about it is really new.)
The point is the pedagogy. The whole and only point.
I know when I was taught this there were two basic options: memorise and use the big awful quadratic formula for numeric solutions. We were expected to know it, but avoid it unless an exam question specifically asked for it. Alternatively solve it symbolically, and the method we were taught, as he points out, was to factor C and try to figure out which of those factors could be used to reconstruct B, with no guarantee that they would exist. You’re on your own with working out the right signs, or imaginary roots. (You could try plugging into the quadratic formula and solving that symbolically but then you’re in for a fair amount of tedious and complicated rearrangement.)
His point is that splitting the B rather than factoring C results in a short series of equations which are simple and obviously straightforwardly derived from the previous step, and provide a purely mechanical and quick way of recovering the roots, is guaranteed to work in every case. It requires no extraneous thinking or trial and error in which you can make a mistake.
All the approaches can get you the right answer. His only point is that this way is easier to teach, and easier to do, but it is not the way anyone appears to be teaching it at the moment.
@6 Rob Grigjanis
I don’t see how you can possibly think that is as simple, or not different. Firstly you’ve got the non-obvious transformation from (1) into (2): yes, you can derive, or verify it, but you have to put the extra effort into doing that. You’ve then got this mysterious B²/4 factor that pops up only in one half (because it is magically cancelled on expansion of the other half – again verifiable but tedious extra work.) Then at the end, you apparently miss that your y is not the same as the original x, so you’ve still got a bit of extra work to do to unpack y and A back into the original terms, which as a first step takes you to (x + B/2 – √(B²/4 – C))(x + B/2 + √(B²/4 – C)), where you can just about start to see his final factors dropping out, but you’ve still got to fix up those signs…
If I were teaching this to a class, I’d have added some justification. I assumed that in a thread about the supposed improvement over other methods, completing the square would be known. It’s a useful technique to know. Also, there’s no need to introduce y and A at all; I just have a fondness for substitution – it can save on typing.
x² + Bx + C = 0 (1)
The two variable terms can be reduced to one variable term by recognizing that
x² + Bx = (x + B/2)² − B²/4 (2)
Plugging into (1) gives
(x + B/2)² − B²/4 + C = 0 (3)
Move the last two terms to the right side;
(x + B/2)² = B²/4 − C (4)
Taking the square root of both sides;
x + B/2 = ±√(B²/4 − C) (5)
So
x = −B/2 ± √(B²/4 − C) (6)