A new way to make quadratic equations easy (2019)

The B/2 term places the center of the parabola, which you can verify by either inspecting the derivative or fiddling with a graphing calculator. Then, because the parabola is symmetric, the zeros must be a pair of points mirrored across the center, with the distance from the center determined by the vertical offset (the C term) versus the narrowness (the A term). Which I definitely agree is an easier way to deal with the problem! But I think that, if you're going to do it that way, it might be easier to work from geometric intuition instead of an algebra trick that represents but obscures the same relationship.

Re: A new way to make quadratic equations easy (2019)

#7

Post by finite_jest » Sat, Nov 27, 2021, 9:31 PM UTC

I don't think this is new at all. This looks like just a standard proof of the quadratic formula.

I believe most people probably have seen a variation of this in high school. Depending on your preferences, this might be a slightly better or worse presentation than what you have seen before.

The fact that they couldn't (or haven't) publish it in a journal also supports this. [1]. (The ArXiv pre-print [2] is dated December 16, 2019)

[1]: https://scholar.google.com/scholar?hl=en&as_sdt=0%2C5&q=%22A...

[2]: https://arxiv.org/pdf/1910.06709.pdf

Re: A new way to make quadratic equations easy (2019)

#8

Post by coyotespike » Sat, Nov 27, 2021, 9:34 PM UTC

I learned this technique while reviewing high school math to take the GRE, and really enjoyed it. I find it elegant and easier to derive than the usual formula - I prefer not to memorize equations. And with just a little practice, it was easy to work through manually on my whiteboard (no scratch paper allowed).

Of course it is not some big discovery in fundamental knowledge - but it is a helpful pedagogical advance and I am happy Po-Shen Loh has advertised it.

Re: A new way to make quadratic equations easy (2019)

#9

Post by codeflo » Sat, Nov 27, 2021, 9:39 PM UTC

I just hand-derived it both ways to make a comparison, the standard way was from memory.

So. The standard way to complete the square uses a trick to rearrange the equation so that one can use the formula (a+b)^2 = a^2 + 2ab + b^2.

This uses a different trick to rearrange the equation to make use of (a+b)(a-b) = a^2 - b^2.

Frankly, I don’t see how this is any easier. If anything, I think it’s harder to do from memory because you need to introduce a new variable in a very specific way to make the simplification work. The standard way feels a lot more systematic.

Re: A new way to make quadratic equations easy (2019)

#10

Post by prof-dr-ir » Sat, Nov 27, 2021, 9:52 PM UTC

From 2019.

And, dare I say it, still a waste of time. In my not-so-humble opinion you either know enough algebra to understand that, for non-zero a, 0 = a x^2 + b x + c 0 = x^2 + b/a x + c/a 0 = (x + b/2a)^2 - b^2/4a^2 + c/a (x + b/2a)^2 = (b^2 - 4 a c) / 4a^2 or you need to improve your basic algebra skills - not search for a better derivation of this particular formula.