“Completing the square” may seem like an arbitrary and hard to remember math trick but it is the algebraic analogue of something that is very intuitive in geometry. The Greeks solved quadratic equations by visually completing the square.
A new way to make quadratic equations easy (2019)
11–20 of 80 posts
Re: A new way to make quadratic equations easy (2019)
#12Three pop-ups or overlays when attempting to read the article.
If this is the new standard for the web, I want to get off the ride.
Re: A new way to make quadratic equations easy (2019)
#13I 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 Dece…
I think there is a different standard way in Europe, compared with the US. Dunno about other places.
Re: A new way to make quadratic equations easy (2019)
#14Leaving aside the fact that this isn't new -- it's how I was taught to solve quadratic equations in the 80s -- if you look closely he's still completing the square.
Re: A new way to make quadratic equations easy (2019)
#15This is neither new, nor easier. It can be interesting to think about and deriving the same results through different methods can help with understanding. But I really doubt that there will be any students who can understand and use this who at the same would struggle with the regular formula for the quadratic.
Re: A new way to make quadratic equations easy (2019)
#16Should have (2019) in the title, I think
Re: A new way to make quadratic equations easy (2019)
#17Leaving aside the fact that this isn't new -- it's how I was taught to solve quadratic equations in the 80s -- if you look closely he's still completing the square.
Surprisingly, the article seems to be written by a smart mathematician, not a crackpot or a social scientist. [1]
This kinda reminds me of that medical researcher who rediscovered trapezoid rule in 1994, and called it "Tai's Model". [2]
To be fair, this is slightly better as it is supposed to be pedagogical, but it is still quite dishonest to pretend that it is new.
[1]: https://en.wikipedia.org/wiki/Po-Shen_Loh
[2]: https://fliptomato.wordpress.com/2007/03/19/medical-research... (It's 459 citations now according to Google Scholar, btw :-))
Re: A new way to make quadratic equations easy (2019)
#18This is neither new, nor easier. It can be interesting to think about and deriving the same results through different methods can help with understanding. But I really doubt that there will be any students who can understand and use this who at the same would struggle with the regular formula for the quadratic.
It is 2019-new https://arxiv.org/abs/1910.06709
Re: A new way to make quadratic equations easy (2019)
#19I 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 a…
I have absolutely horrible memory, which I hold has served me fairly well academically. The need to occasionally re-derive basic formulae serves as a nice refresher -- at least in engineering (probably the weight of redoing everything would get too heavy if I'd ended up as a mathematician). Remembering too many rules makes math feel arcane. I tutored math for a bit, and as far as I can tell the quadratic formula serves as significant marker along the path of math as an exercise in just remembering things, unfortunately.
Re: A new way to make quadratic equations easy (2019)
#20The 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…
I taught it this way in homeschooling with heavy support of graph-geometric intuition, drawing and looking at various parabolas. Works great.