10 ms·

A new way to make quadratic equations easy

177 points by uniqueid 7y ago

jimhefferon 7y ago
I don't know that it was ever hard.
- yxhuvud 7y ago
  It is hard for the people that isn't comfortable doing symbolic algebra. I don't see how this method improves it for those.
- user271828 7y ago
  They are https://www.youtube.com/watch?v=o_zoHxC4y-U https://www.youtube.com/watch?v=o_zoHxC4y-U
- mikorym 7y ago
  >> to make the quadratic equations easy They mean: to make the quadratic equation easy to remember. However, I don't think this will have any impact on the average high school student. The key is: >> Loh points out that the two roots, R and S, add up to -B when their average is -B/2. >> “So we seek two numbers of the form -B/2±z, where z is a single unknown quantity,” he says. We can then multiply these numbers together to get an expression for C. But you still have to derive the formula and their first example assumes A is fixed.
  - saagarjha 7y ago
    > their first example assumes A is fixed This is usually how completing the square is taught, in my experience. The leading coefficient is handled later with "just divide everything by a".
montalbano 7y ago
It would be interesting to have math historian compare this with other derivations of quadratic roots to assess its originality. My favourite bit of knowledge about quadratic equations is that its roots can always be visualised as the intersection between a simple parabola (x^2) and a straight line (m*x + c). In fact, the above is why imaginary numbers did not arise from needing to solve quadratic equations. Because in the case of complex roots, the line and the parabola simply do not interesect. So it was originally thought that there was no worthwhile solution anyway. The real 'need' for complex numbers arose from solving cubic equations. [1] [1] https://www.goodreads.com/book/show/19161684-a-friendly-approach-to-complex-analysis https://www.goodreads.com/book/show/19161684-a-friendly-appr...
robinhouston 7y ago
This is bizarre. It’s no less a mathematical trick than completing the square, and it doesn’t seem to be any simpler to use. The example given is to find the roots of x² - 2x + 4 = 0. Completing the square gives (x - 1)² + 3 = 0, from which you can immediately see that the roots are 1 ± √3 i. If anything this seems easier than the method of the article. Am I missing something? Added: The argument seems to be that young students will find this method easier to understand than completing the square. I have no experience of teaching mathematics to children, so this may be true for all I know. It would be interesting to test this hypothesis experimentally, because I don’t think it’s obviously true.
- banachtarski 7y ago
  Was going to reply precisely this. Completing the square is, I believe, already taught in standard curricula (at least, it was when I was in grade school), and is a trivial way to derive the quadratic formula.
  - pflats 7y ago
    You are correct. It is generally expected to be part of an Algebra I curriculum (as per Common Core Appendix A's Traditional Pathway). > Common Core High School: Algebra » Reasoning with Equations & Inequalities » Solve equations and inequalities in one variable. » 4 » a > Use the method of completing the square to transform any quadratic equation in x into an equation of the form (x - p)^2 = q that has the same solutions. Derive the quadratic formula from this form. http://www.corestandards.org/Math/Content/HSA/REI/B/4/a/ http://www.corestandards.org/Math/Content/HSA/REI/B/4/a/
  - leggomylibro 7y ago
    Depends. I took a mid-level calculus class recently because it had been awhile since I'd integrated anything, and the professor had to spend a day reviewing how to factor and divide polynomials when they got to integrals which used trig identities. "Completing the square" was a new phrase to most students. Everyone knew the quadratic formula, but most people don't pay attention to how formulas are derived, much less remember that sort of thing a few years later. If you want to guess whether an average student knows something, ask yourself, "would it be on the test?"
saagarjha 7y ago
Is it just me, or are the images in the article extremely pixelated?
- mkl 7y ago
  Yes, they're terrible. Katex or MathJax would be both easier and better. There's also at least one formula mistake: a cx^2 shows up briefly.
  - imglorp 7y ago
    One would think an MIT technical publication (PR arm?) aimed at showing off their prowess would know how to typeset an equation.
ninjinxo 7y ago
/r/math discussion: https://www.reddit.com/r/math/comments/dzbmbu/a_new_way_to_solve_quadratic_equations_poshen_loh/ https://www.reddit.com/r/math/comments/dzbmbu/a_new_way_to_s... Original source: https://www.poshenloh.com/quadratic https://www.poshenloh.com/quadratic
thanhhaimai 7y ago
I have trouble understanding why this deserves a paper. This is what I learned in middle school back in Vietnam (although I did take advanced Math). For equation Ax^2 + Bx^2 + C = 0, the roots are: x1 = (-B + sqrt(B^2 - 4AC))/2A x2 = (-B - sqrt(B^2 - 4AC))/2A Now the author looks at a special case where A = 1. The equation becomes x^2 + Bx^2 + C = 0. Of course the roots simply become: x1 = -B/2 + sqrt(B^2 - 4C)/2 = -B/2 + sqrt((B^2)/4 - C) x2 = -B/2 - sqrt((B^2)/4 - C) > "The author would actually be very surprised if this approach has entirely eluded human discovery until the present day, given the 4,000 years of history on this topic, and the billions of people who have encountered the formula and its proof. Yet this technique is certainly not widely taught or known (the author could find no evidence of it in English sources)" I certainly don't think it has eluded humany discovery until the present day. It's known to middle school students in Asia that multiplying the original equation so that A == 1 would greatly simplify the roots formula.
- wolfgke 7y ago
  > I have trouble understanding why this deserves a paper. This is what I learned in middle school back in Vietnam (although I did take advanced Math). > [...] > It's known to middle school students in Asia that multiplying the original equation so that A == 1 would greatly simplify the roots formula. I think the same having attended a regular grammar school in Germany.
- mkl 7y ago
  The a = 1 is not the important bit - that's probably nearly universal.
- deleted 7y ago
  [deleted]
- knzhou 7y ago
  Yeah, half the elegance of the linked expressions just comes from setting A = 1. This is definitely well-known. I learned this under the name of "Vieta's formulas" back in middle school. Vieta lived in the 1500s.
bograt 7y ago
I have been using a technique that's almost identical to this for years, based simply on observing that the roots are symmetric around the min/maximum: differentiate, set to zero, then difference of squares. This cannot be an unrecorded technique can it?
- mkl 7y ago
  No, it's basically what the quadratic formula does. (Well, all methods solve the same equation and get the same roots, so they have to be equivalent.)
mkl 7y ago
This seems more complicated and roundabout than completing the square. The average of the roots shows up that way too: We want to write x^2 + bx + c = 0 in the form (x+m)^2 + n = 0, so there's only one x left and we can rearrange for it. Expanding, (x+m)^2 = x^2 + 2mx + m^2, so we get the x^2 we want, and the coefficients of x tell us b = 2m, so m = b/2. We also get an m^2 (= b^2/4) we don't want, so let's take it away: (x + b/2)^2 - b^2/4 = x^2 + bx That x + b/2 is x - (-b/2), x minus the average of the roots, which is the x value the parabola is centred on. Then we add c: (x + b/2)^2 - b^2/4 + c = x^2 + bx + c To find the roots, set it to 0 and rearrange for the one x that's left: (x + b/2)^2 - b^2/4 + c = 0 (x + b/2)^2 = b^2/4 - c x + b/2 = ±√(b^2/4 - c) x = -b/2 ± √(b^2/4 - c) Note that this is the average of the roots ± the article's z. Then combine: x = -b/2 ± √((b^2-4c)/4) x = -b/2 ± √(b^2-4c)/2 x = (-b ± √(b^2-4c))/2 If you have ax^2+bx+c = 0, divide the equation by a first, then do the same steps and you get the normal quadratic formula: x = (-b ± √(b^2-4ac))/(2a) I think the linked post misstates the purpose of the article: it's not about new maths, but about pedagogy and ways of explaining the quadratic formula.
- sykick 7y ago
  Having taught elementary algebra for many years at a community college I think your perception of what is complicated and easy are not correct. The method in the article is far easier for elementary algebra students than what you describe. You’ve proved the quadratic formula. This is a formula students in elementary algebra will struggle with. It’s a formula whose proof will be lost on them. What is shown in the article is an easy to apply mechanism for finding the roots. The method in the article is one that I can use in the classroom. There’s no way I’d attempt your explanation in an elementary algebra class.
  - mkl 7y ago
    I wouldn't try to teach it to elementary algebra students either; I was explaining for maths-weak HN readers. It would be worth trying with some of my first year university students (but definitely not all). What's shown in the article is also a derivation of the quadratic formula. I'm not convinced the less general solving method at the end of the article is any easier than completing the square, and it's certainly more mystifying, as completing the square can be checked by expanding, but this is a bit out of nowhere. I will try the article's method if I get an opportunity, but I think the introduction of an extra variable z is quite an obstacle, and that the fact that -b/2 is the average of the roots will be quite a leap for most students.
nestorD 7y ago
Note that he critisize the classical formula but that he, then, suppose that A=1 when he gives his formulation (making it simpler). Adding A back gives : -B/2A ± sqrt(B²/4A - C/A) which is not obviously simpler than : (-B ± sqrt(B² - 4AC)) / 2A
verandacoffee 7y ago
Isn't this the same as the pq-formula? https://www.matteboken.se/lektioner/matte-2/andragradsekvationer/pq-formeln https://www.matteboken.se/lektioner/matte-2/andragradsekvati... (Link in Swedish).
- jsjolen 7y ago
  Yes, and apparently it's also used in Indian schools (see comments found here: https://www.facebook.com/poshenloh https://www.facebook.com/poshenloh ).
fyp 7y ago
People are criticizing this because it is still the same quadratic formula. But of course it is! Math is consistent. But representation matters. A good chunk of mathematics is just about rewriting the same mathematical fact in a different way. For example the equation of a line could be written with coefficients or in slope/intercept form or in polar coordinates or in homogeneous coordinates or etc etc. Here the claim is that explicitly giving a name to the variable -b/2 makes the equation easier to think about. I see nothing wrong with that.
- LatteLazy 7y ago
  Full disclosure: I critized the article in a comment below... Respectfully, that's not the reason people are critiquing the article. I fully agree mathematics is what works and many methods use identical underpinning logic, just expressed in different ways. I'm fine with that. But that doesn't mean all methods are equally good. This method is no quicker or easier or less error prown than the quadratic formula it "replaces". Even in the authors chosen example, it's no better. In many other cases it's harder (if B or C are not divisible by A, dividing by A to force A=1 just spreads and increases the complexity). That makes it a bad method because now, a user has to not only know both methods but also pick the right one. And for this extra time and risk, the gain nothing the standard Quadratic Formula didn't give them. We could equally "simplify" the quadratic formula by forcing B=1 or C=1. Are those methods new and useful? No. They're trivial and have limited use cases. They're never better than just using the full formula. My issues with the article are a bit wider: this is not new. I was taught this as a limited version of quadratics in 2000 in a run of the mill school in London. I also think it's derivative. Anyone smart enough to be solving quadratics should also be smart enough to apply basic algebra to simplify quadratics. But the article presents this, assuming the audience knows no better, like it's a breakthrough. That feels dishonest to me...
  - fyp 7y ago
    I think it's useful to think of this as refactoring code to make it more readable (and therefore more teachable). If you're a memorizer, your code might as well be obfuscated code: def quadratic_formula(a, b, c): return [ (-b - math.sqrt(b ** 2 - 4 * a * c)) / (2 * a), (-b + math.sqrt(b ** 2 - 4 * a * c)) / (2 * a) ] If you understand the importance of the expression under the square root (i.e. the sign of the discriminant) you can rename one subexpression: def quadratic_formula(a, b, c): discriminant = b ** 2 - 4 * a * c return [ (-b - math.sqrt(discriminant)) / (2 * a), (-b + math.sqrt(discriminant)) / (2 * a) ] Of course you can refactor it further with pointless stuff like denom = 2a, but that doesn't add much semantic value. So the above is more or less the vocabulary we have about quadratic equations today. Loh's contribution is a specific way of refactoring the code by first dividing by a: def quadratic_formula(a, b, c): b = b / a c = c / a discriminant = b ** 2 - 4 * c return [ -b / 2 - math.sqrt(discriminant) / 2, -b / 2 + math.sqrt(discriminant) / 2 ] Which then unlocks the ability to talk about the subexpressions in relation to the roots (a la Vieta's formula): def quadratic_formula(a, b, c): b = b / a c = c / a sumOfRoots = -b productOfRoots = c averageRoot = sumOfRoots / 2 # Want roots [averageRoot - delta, averageRoot + delta] # such that: # productOfRoots == (averageRoot - delta) * (averageRoot + delta) # == averageRoot ** 2 - delta ** 2 delta = math.sqrt(averageRoot ** 2 - productOfRoots) return [ averageRoot - delta, averageRoot + delta ] Your code is no longer using single-letter variable names!
shikoba 7y ago
(r-s)^2=r^2-2rs+s^2=(r+s)^2-4rs=b^2-4c r=((r-s)+(r+s))/2=(sqrt(b^2-4c)-b)/2
LatteLazy 7y ago
I must be missing something here... The "standard" quadratic formula at the top of the article is just as quick and painless to solve the equation he uses as an example. And its easier for many other versions (basically any time B/A or C/A are not integers). Plus, this isn't new: I was taught to do exactly this IF it simplified the whole equation. That was in 2000 in London in a pretty standard secondary school. (Also, what's with taking "z^2 = 3" at font 10, treating it as an image and then displaying it 20 times the original size?! Is it meant to look more maths-ey?) I am excited to reveal something though: I recently discovered a whole new way to make a percentages! Instead of multiplying the number by 100, you just multiply it by 10 twice! So much better :)
- gibspaulding 7y ago
  The focus is not on getting students to be able to solve quadratics (if it were, then the quadratic formula is a great choice!), but rather to understand how solving a quadratic works. That is, how to derive the quadratic formula for themselves. Generally this is done by completing the square. This is another method which some students may find more intuitive.
dooglius 7y ago
I disagree that this proof is better pedagogically; it assumes the quadratic case of the Fundamental Theorem of Algebra (the correspondence between factorization and solutions), which at this stage would have to be taken on faith by students, whereas completing the square is fully justified.
- boxfire 7y ago
  Not sure why you think it's easier to take that on faith than complete the square, but I'm sure you never tutored new students to mathematics. At this point we are taking all of abstract algebra and some number theory for granted in anyone's education. Pedagogically it's easier to understand, the thesis of the article.
  - dooglius 7y ago
    Granted, I have not tutored new students. I am making the assumption that someone encountering this would be familiar with basic algebraic manipulations, and the solutions to x^2=a. This is all that is necessary to justify completing the square. The assertion that quadratic equations must have at most two solutions R and S, and that it can be equivalently written (x-R)(x-S) is what has to be taken on faith.
- pdonis 7y ago
  > it assumes the quadratic case of the Fundamental Theorem of Algebra No, it doesn't; it proves that case. The key step in the proof is the fact that, over the complex numbers, every number has a square root: from that it follows that the factorization used in the proof must always exist.
spicymaki 7y ago
Here is a link to the paper the story was based on: "A Simple Proof of the Quadratic Formula" https://arxiv.org/pdf/1910.06709.pdf https://arxiv.org/pdf/1910.06709.pdf
6gvONxR4sf7o 7y ago
This still seems way more complicated than it needs to be to teach it. I always used a much simpler way to avoid memorizing anything. Imagine you have a parabola y - c = k x^2 and want to solve for y = 0. Dead easy, right? To turn any other parabola into this form, you only need to scroll left or right on x until the minimum is at x'=0 (algebraically, this means eliminating any b*x' term). Teach students how to do change of coordinates and how to solve this trivial problem, and they don't need to memorize any formulas. It also sets students up for the useful math mindset of solving new problems by reducing them to previously solved ones and relies on conceptual understanding. Seems way better than the "memorize this formula" approach.
- sykick 7y ago
  I teach mathematics at a community college. This includes teaching a lot of elementary algebra courses. These are pre-college level math courses. In elementary algebra we introduce solving quadratic equations by factoring. In the next course intermediate algebra we teach the quadratic formula. Students in elementary algebra are not equipped to understand change of coordinates. This is too hard of a concept at that stage in my opinion. The method described in the article is a very nice one and is appropriate for elementary algebra. It ties in nicely with factoring and makes solving quadratic equations quite easy. I will be using this method in the future.
  - 6gvONxR4sf7o 7y ago
    How aren't they equipped for it? Do you mean change of coordinates in its full generality? I'm talking about simple substitution of x' = x + k, connected with the intuition sketched below. They know how to perform simple substitution by then, and they can surely follow this logic: x 0---1---2---3--> x' 0---1---2---3---4--> x' = x + 1 ==> x = x' - 1 y = f(x) ==> y = f(x' - 1) If you can talk about Napa being an hour north of San Francisco and San Francisco being an hour north of San Jose, this kind of student should be able to tell how far north of San Jose Napa is. Can they not even do that? If not, I'd questioning why we teach them to solve quadratic equations before they can do substitution.
prvc 7y ago
Surely this is a prank?
tel 7y ago
I think this is nice. It's a method of calculation derived from an important fact—that quadratic equations have two solutions and those solutions can be written in such a form. I don't have the experience or interest to debate whether it's better or worse than any other method pedagogically. I can definitely imagine that I would have appreciated learning this method as well as completing the square as a child. But honestly, the real interesting thing here is the (a) gathering information about what you think the solution ought to look like and (b) working backwards from there. That's a good general trick and worth having in your back pocket. It also emphasizes the importance of visualizing and examining your belief about what should occur. People should suspect the fundamental theorem of algebra long before they prove it.
edflsafoiewq 7y ago
If you're going to rename the constants to make a=1, you might as well also replace b with b/2. The solution to x^2 + 2bx + c = 0 is x = ± √(b^2 - c) - b
czrnb 7y ago
There is an error in the (pixelated) example that is given int the MIT tech review article (z² = 3 instead of -3). I think the fact that the author of the news missed this, that he probably took a screenshot of the formulas rewritten in Word, and that he was compelled to write such a long article on such a simple topic speaks for his level on the topic. The fact that he only lists the formal article as a reference instead of the announcement, video, and accessible blog post by Po-Shen Loh really baffles me. The original "disclosure" by Po-Shen Loh [0] is much less sensational and gives some context for his work (teaching middle school students). In the formal article, he is also stating that the method is very likely not __new__, but that he wants to popularize it in teaching. I think, as many other commenters pointed out, that there is no great breakthrough here. However "his" method may have the advantage of training the intuition of young students, by helping them understand the concepts of average and "deviation" (I'm not really sure how to call it in that case), and maybe visualizing them. [0]: https://www.poshenloh.com/quadratic https://www.poshenloh.com/quadratic
- wrycoder 7y ago
  And the third pixelated equation has an error - the final term on the left side should be C, not Cx^2. I think the point of the arXiv article is this is a more straightforward /proof/ of the quadratic formula. That article has some interesting historical commentary that shows exactly where the author thinks his contribution fits - he is not naively coming up with something "new". As far as computation of the roots goes, it is a slightly streamlined approach: First, put the quadratic into canonical form by dividing by A, if necessary. Then, take B/2 into a new variable, call it F. Get F^2. The roots are then -F +- sqrt(F^2 - C)
- knolax 7y ago
  Why is MIT technology review on the front page so often. It's literally a student newspaper with the expected quality of one.
  - pfedak 7y ago
    Are you perhaps thinking of https://thetech.com/ https://thetech.com/ ? I'm not sure in what sense the Technology Review is a student newspaper - to my knowledge (and checking a few bylines), the journalists are not students.
tomp 7y ago
Um... what's the difference? x = (-b ± √(b^2 - 4ac)) / 2a Ok, remove a assuming it's 1 - well, yeah, less general math is simpler. x = (-b ± √(b^2 - 4c)) / 2 Move around 2 & 4... voila. x = -b/2 ± √(b^2 / 4 - c) Like, whatever. :)
bikenaga 7y ago
The old (Baudhayana, Omar Khayyam) geometric approach to completing the square (and so solving [some] quadratics) is very nice because it's visual, and it's what I show students in many classes. It explains why it's "completing the square". Suppose you want to complete the square for x^2 + 6 x. Represent this as an x-by-x square and a 6-by-x rectangle: x ..... x ..... ..... ***** ***** ***** 6 ***** ***** ***** Cut the 6 x rectangle into two 3-by-x rectangles: x ..... x ..... ..... ***** 3 ***** ***** ***** 3 ***** ***** Move the lower 3-by-x rectangle up next to the square. The L-shaped figure still has area x^2 + 6 x. x 3 ..... ***** x ..... ***** ..... ***** ***** 3 ***** ***** What do you need to add (what is the size of the small square on the lower right) to complete the (large) square? The small square is 3-by-3, so it has area 9: x 3 ..... ***** x ..... ***** ..... ***** ***** +---+ 3 ***** | | ***** +---+ You get x^2 + 6 x + 9 = (x + 3)^2. If the original x^2 + 6 x was on one side of an equation, you add 9 to both sides.
aphextim 7y ago
One of my favorite things to break the ice with a younger gamer and the Quadratic Formula. https://www.youtube.com/watch?v=ewAHYVzMobw https://www.youtube.com/watch?v=ewAHYVzMobw Super Mario Quadratics.
orbifold 7y ago
There is nothing new about this formula or derivation. In fact this precise formula (including that derivation) is taught in Germany.
abnry 7y ago
Let's get the criticisms of the article out of the way: --It has terrible formatting and typos. --It puffs up something more important than it is. --It is more about pedagogy than a mathematical idea. --It suggests the idea is original, when it almost certainly is not. --It doesn't link to the original (and better source). Okay, here is the good things about the approach: --It is good to shift your thinking about mathematical derivations and proofs and think about them as code that runs on people's brains. You input a derivation into someone's brain and they return a boolean value (this is true, it makes sense, etc.). Pedagogy is trying to optimize the code for less powerful architectures. Just like when you are optimizing code tiny little details of instruction orders matter, the same with mathematical derivations. --Fundamentally, algebraic manipulations are uncomfortable and nonintuitive for students. They feel like tricks. Going forwards from (x+a)^2=x^2+2ax+a^2 makes sense but going backwards as in the case of completing the square is hard. It's not the same case for x^2=a vs sqrt(x)=a. This is kind of a similar case to math students feeling confused by adding and subtracting the same quantity when doing calculus limits. For any trained mathematician, this is obvious, but it really feels like a trick at first. The nice thing about this approach is that it avoids this issue and gives you a good reason WHY the -b/2 term shows up. Additionally, it avoids the problem of substituting, which tends to bog students down (try teaching the chain rule someday). Students should still understand completing the square but I don't think this is a bad way to introduce them to the quadratic formula. It highlights the symmetric of the roots (at least for real values), which makes sense if you plot a quadratic.
- infocom6502 7y ago
  I agree. This is nothing novel and a fairly standard way of solving the equation.
- jfengel 7y ago
  I have never understood "completing the square". I mean, I get why and how it works, but it's completely unintuitive to me, or why you'd do it that way. I think it is a historical hangover. It makes more sense viewed geometrically, and the technique is attributed to the same al-Khwarizmi for whom algorithms are named. But it's less intuitive as part of algebra, and we focus a lot more on algebra today than in the medieval times, when geometry ruled.
ajarmst 7y ago
I think there's an unacknowledged premise here that is fundamentally misguided: that the point of teaching math is to minimize the complexity of the formulas and recipes the students have to memorize, or that skill in finding the roots of a quadratic is so important that we must optimize the experience of learning it. I work with mathematics teachers and engineers and I know of no one who has willingly factored a quadratic by hand since we put our slide rules away. The goal, fundamentally, is not the skill of factoring a quadratic. The goal is understanding of the relationships of numbers, operations, and domains that allow algebra to be a powerful set of tools for solving huge classes of problems. I _never_ teach the quadratic equation. I teach completing the square, because it's an illustration of a useful way of algebraically manipulating a relationship into a form (square of a linear binomial) that they recognize and can easily factor. I usually do a quick proof of the quadratic equation using completing the square, but generally as an illustration that if you properly understand the associated algebra, you don't need to memorize formulas and algorithms. The use of the field axioms to manipulate polynomials, and the goal of manipulating them to a tractable form, is what they need---not the quadratic equation. This 'new' technique (which appears to be a simple riff on the standard technique of looking for a pair of numbers whose sum is B and whose product is C that is taught as a matter of course in every middle school on the planet) is missing the actual goal of the lesson. In particular, the reason I even teach completing the square is because it's a precursor of me bringing up the roots of $x^2 + 1$, which takes us in to an introduction to complex numbers, the fundamental theorem of algebra, and the discovery that the same old algebra over the Reals they learned in high school can now be used to do things like solve differential equations. Later, we introduce matrices and I can return to our old friends the algebraic field axioms to solve whole systems of equations. If they're lucky, they also get modular arithmetic and can be shown Galois fields and start using their understanding of algebra for problems in logic and set theory. All of that is easily within the reach of students in their first or second year of college, and I think if we bothered to try, we'd discover it's easily in the reach of secondary school students---or would be if we'd stop pretending that teaching them that "mathematics" is making change and pushing numbers from a word problem into the blanks of some generic formula they've memorized. The task isn't teaching them to factor quadratics. The task is teaching them algebra.
mogadish 7y ago
To me this seems like a special case of the Vieta formulas (https://en.wikipedia.org/wiki/Vieta%27s_formulas https://en.wikipedia.org/wiki/Vieta%27s_formulas) which in my case I learned in highschool
eyeundersand 7y ago
In Nepal, you are taught around the sixth grade that you can solve ax^2 + bx + c = 0 by splitting b into m and n such that mn = ac and m + n = b ( or m -n = b, if a and c have opposing signs). This "discovery" the author claims and some other corollaries were quite commonly known. I find it absurd and, frankly, laughable that this is being heralded as something new!
knzhou 7y ago
This kind of approach is familiar to me from competition math. Back in middle school math club we were taught this exact approach (under the name of "Vieta's formulas"), i.e. that thinking about the sum and product of the roots could be faster in some cases. Po-Shen Loh is the director of the US IMO team, so it makes sense he would like this approach. However, I don't think it makes logical sense to teach it only this way. Here you start by assuming that a quadratic has two roots, which is not at all obvious the first time a kid sees a quadratic equation. (Especially because those roots can be complex numbers!) Completing the square tells you why there are two roots, and also naturally leads you to the necessity of complex numbers, i.e. when the "square" you end up making is negative. You can use the nice Vieta's formula tricks only after establishing that.
- deleted 7y ago
  [deleted]
- fyp 7y ago
  I am also familiar with Vieta from math competitions but I think this is more than that. Given two numbers, r1 and r2, if you know their arithmetic mean: m = (r1 + r2) / 2 and geometric mean: g = sqrt(r1 * r2) then following Loh's derivation you get a very cute formula: r1 = m - sqrt(m^2 - g^2) r2 = m + sqrt(m^2 - g^2) Vieta gives an easy way of finding those means from a quadratic equation: r1+r2=-b and r1*r2=c. So you can plug in m=-b/2 and g=sqrt(c) in the equation above. The fact that you can state roots in terms of their means is the more novel insight to me. (note: Loh doesn't talk about geometric means but I thought using just product of the roots isn't as "symmetric")
ivan_ah 7y ago
That's very cool. To the best of my knowledge the only purpose of learning how to complete the square in high school math is to be able to understand the derivation of the quadratic formula, so if we can use this alternative proof then students don't need to know "completing the square" anymore. Good riddance! The only other application of "completing the square" I can think of is the trig substitution in calculus, see https://en.wikipedia.org/wiki/Completing_the_square#Integration https://en.wikipedia.org/wiki/Completing_the_square#Integrat... or worked example here https://www.khanacademy.org/math/ap-calculus-ab/ab-integration-new/ab-6-10/v/integration-using-completing-the-square-arctan https://www.khanacademy.org/math/ap-calculus-ab/ab-integrati... In sum, completing the square is definitely a cool trick (see https://en.wikipedia.org/wiki/File:Completing_the_square.gif https://en.wikipedia.org/wiki/File:Completing_the_square.gif ), but maybe we can skip it... or present it as extra/optional material? I'm going to think about dropping it from my books. It will save 5+ pages of suffering for readers, which is a clear win.
- hollandheese 7y ago
  Nope, completing the square is primarily used to translate a standard quadratic function into a vertex form quadratic so it can be graphed using transformations instead of guessing. It is an essential multi-use tool in precalculus. The thing we should dump is the quadratic formula.
Tiruneh 7y ago
I have read the paper and it gives a good educational context from which the quadratic equation may be looked at. I have also worked out simpler way to solve quadratic equations using function evaluation without much memorization of symbolic formula: for the equation : f(x) = ax^2 +bx + c = 0 X = Z +- Sqrt [-f(Z)/a] ; Z = -b/2a for the equation f(x) = x^2 +bx+c = 0 it is even a bit simpler: X = Z +- Sqrt[ -f(z) ] where Z = -b/2. Example 1: f(x) = 3x^2 -8x-35 = 0 Z = -b/2a = - (-8)/(2.3) = 4/3 F(Z) = -121/3 X = 4/3 + or - Sqrt ( -(1/3)*(-121/3) = 4/3 + or - 11/3 = {5, -7/3) Example 2: (Simpler form): f(x) = x^2 - 4x+ 3 = 0 Z = -b/2 = -(-4)/2 = 2 f(Z) = 2^2 - 4.2 + 3 = -1 X = 2 + or - Sqrt( -(-1) = 2 +- 1 = { 3, 1} For detail of this method, please see the following pre-print https://www.researchgate.net/publication/337829551_A_simple_formula_for_solving_quadratic_equations_using_function_evaluation?fbclid=IwAR1yoUtuTQAc8Mpc-BRME5vdKkBZ4QbXZj741lGhUdvc7s7R3rAugZE6VGg https://www.researchgate.net/publication/337829551_A_simple_...

177 points by uniqueid 7y ago

jimhefferon 7y ago

I don't know that it was ever hard.

yxhuvud 7y ago
It is hard for the people that isn't comfortable doing symbolic algebra. I don't see how this method improves it for those.
user271828 7y ago
They are https://www.youtube.com/watch?v=o_zoHxC4y-U https://www.youtube.com/watch?v=o_zoHxC4y-U
mikorym 7y ago
>> to make the quadratic equations easy They mean: to make the quadratic equation easy to remember. However, I don't think this will have any impact on the average high school student. The key is: >> Loh points out that the two roots, R and S, add up to -B when their average is -B/2. >> “So we seek two numbers of the form -B/2±z, where z is a single unknown quantity,” he says. We can then multiply these numbers together to get an expression for C. But you still have to derive the formula and their first example assumes A is fixed.
- saagarjha 7y ago
  > their first example assumes A is fixed This is usually how completing the square is taught, in my experience. The leading coefficient is handled later with "just divide everything by a".

montalbano 7y ago

It would be interesting to have math historian compare this with other derivations of quadratic roots to assess its originality. My favourite bit of knowledge about quadratic equations is that its roots can always be visualised as the intersection between a simple parabola (x^2) and a straight line (m*x + c). In fact, the above is why imaginary numbers did not arise from needing to solve quadratic equations. Because in the case of complex roots, the line and the parabola simply do not interesect. So it was originally thought that there was no worthwhile solution anyway. The real 'need' for complex numbers arose from solving cubic equations. [1] [1] https://www.goodreads.com/book/show/19161684-a-friendly-approach-to-complex-analysis https://www.goodreads.com/book/show/19161684-a-friendly-appr...

robinhouston 7y ago

This is bizarre. It’s no less a mathematical trick than completing the square, and it doesn’t seem to be any simpler to use. The example given is to find the roots of x² - 2x + 4 = 0. Completing the square gives (x - 1)² + 3 = 0, from which you can immediately see that the roots are 1 ± √3 i. If anything this seems easier than the method of the article. Am I missing something? Added: The argument seems to be that young students will find this method easier to understand than completing the square. I have no experience of teaching mathematics to children, so this may be true for all I know. It would be interesting to test this hypothesis experimentally, because I don’t think it’s obviously true.

banachtarski 7y ago
Was going to reply precisely this. Completing the square is, I believe, already taught in standard curricula (at least, it was when I was in grade school), and is a trivial way to derive the quadratic formula.
- pflats 7y ago
  You are correct. It is generally expected to be part of an Algebra I curriculum (as per Common Core Appendix A's Traditional Pathway). > Common Core High School: Algebra » Reasoning with Equations & Inequalities » Solve equations and inequalities in one variable. » 4 » a > Use the method of completing the square to transform any quadratic equation in x into an equation of the form (x - p)^2 = q that has the same solutions. Derive the quadratic formula from this form. http://www.corestandards.org/Math/Content/HSA/REI/B/4/a/ http://www.corestandards.org/Math/Content/HSA/REI/B/4/a/
- leggomylibro 7y ago
  Depends. I took a mid-level calculus class recently because it had been awhile since I'd integrated anything, and the professor had to spend a day reviewing how to factor and divide polynomials when they got to integrals which used trig identities. "Completing the square" was a new phrase to most students. Everyone knew the quadratic formula, but most people don't pay attention to how formulas are derived, much less remember that sort of thing a few years later. If you want to guess whether an average student knows something, ask yourself, "would it be on the test?"

saagarjha 7y ago

Is it just me, or are the images in the article extremely pixelated?

mkl 7y ago
Yes, they're terrible. Katex or MathJax would be both easier and better. There's also at least one formula mistake: a cx^2 shows up briefly.
- imglorp 7y ago
  One would think an MIT technical publication (PR arm?) aimed at showing off their prowess would know how to typeset an equation.

ninjinxo 7y ago

/r/math discussion: https://www.reddit.com/r/math/comments/dzbmbu/a_new_way_to_solve_quadratic_equations_poshen_loh/ https://www.reddit.com/r/math/comments/dzbmbu/a_new_way_to_s... Original source: https://www.poshenloh.com/quadratic https://www.poshenloh.com/quadratic

thanhhaimai 7y ago

I have trouble understanding why this deserves a paper. This is what I learned in middle school back in Vietnam (although I did take advanced Math). For equation Ax^2 + Bx^2 + C = 0, the roots are: x1 = (-B + sqrt(B^2 - 4AC))/2A x2 = (-B - sqrt(B^2 - 4AC))/2A Now the author looks at a special case where A = 1. The equation becomes x^2 + Bx^2 + C = 0. Of course the roots simply become: x1 = -B/2 + sqrt(B^2 - 4C)/2 = -B/2 + sqrt((B^2)/4 - C) x2 = -B/2 - sqrt((B^2)/4 - C) > "The author would actually be very surprised if this approach has entirely eluded human discovery until the present day, given the 4,000 years of history on this topic, and the billions of people who have encountered the formula and its proof. Yet this technique is certainly not widely taught or known (the author could find no evidence of it in English sources)" I certainly don't think it has eluded humany discovery until the present day. It's known to middle school students in Asia that multiplying the original equation so that A == 1 would greatly simplify the roots formula.

wolfgke 7y ago
> I have trouble understanding why this deserves a paper. This is what I learned in middle school back in Vietnam (although I did take advanced Math). > [...] > It's known to middle school students in Asia that multiplying the original equation so that A == 1 would greatly simplify the roots formula. I think the same having attended a regular grammar school in Germany.
mkl 7y ago
The a = 1 is not the important bit - that's probably nearly universal.
deleted 7y ago
[deleted]
knzhou 7y ago
Yeah, half the elegance of the linked expressions just comes from setting A = 1. This is definitely well-known. I learned this under the name of "Vieta's formulas" back in middle school. Vieta lived in the 1500s.

bograt 7y ago

I have been using a technique that's almost identical to this for years, based simply on observing that the roots are symmetric around the min/maximum: differentiate, set to zero, then difference of squares. This cannot be an unrecorded technique can it?

mkl 7y ago
No, it's basically what the quadratic formula does. (Well, all methods solve the same equation and get the same roots, so they have to be equivalent.)

mkl 7y ago

This seems more complicated and roundabout than completing the square. The average of the roots shows up that way too: We want to write x^2 + bx + c = 0 in the form (x+m)^2 + n = 0, so there's only one x left and we can rearrange for it. Expanding, (x+m)^2 = x^2 + 2mx + m^2, so we get the x^2 we want, and the coefficients of x tell us b = 2m, so m = b/2. We also get an m^2 (= b^2/4) we don't want, so let's take it away: (x + b/2)^2 - b^2/4 = x^2 + bx That x + b/2 is x - (-b/2), x minus the average of the roots, which is the x value the parabola is centred on. Then we add c: (x + b/2)^2 - b^2/4 + c = x^2 + bx + c To find the roots, set it to 0 and rearrange for the one x that's left: (x + b/2)^2 - b^2/4 + c = 0 (x + b/2)^2 = b^2/4 - c x + b/2 = ±√(b^2/4 - c) x = -b/2 ± √(b^2/4 - c) Note that this is the average of the roots ± the article's z. Then combine: x = -b/2 ± √((b^2-4c)/4) x = -b/2 ± √(b^2-4c)/2 x = (-b ± √(b^2-4c))/2 If you have ax^2+bx+c = 0, divide the equation by a first, then do the same steps and you get the normal quadratic formula: x = (-b ± √(b^2-4ac))/(2a) I think the linked post misstates the purpose of the article: it's not about new maths, but about pedagogy and ways of explaining the quadratic formula.

sykick 7y ago
Having taught elementary algebra for many years at a community college I think your perception of what is complicated and easy are not correct. The method in the article is far easier for elementary algebra students than what you describe. You’ve proved the quadratic formula. This is a formula students in elementary algebra will struggle with. It’s a formula whose proof will be lost on them. What is shown in the article is an easy to apply mechanism for finding the roots. The method in the article is one that I can use in the classroom. There’s no way I’d attempt your explanation in an elementary algebra class.
- mkl 7y ago
  I wouldn't try to teach it to elementary algebra students either; I was explaining for maths-weak HN readers. It would be worth trying with some of my first year university students (but definitely not all). What's shown in the article is also a derivation of the quadratic formula. I'm not convinced the less general solving method at the end of the article is any easier than completing the square, and it's certainly more mystifying, as completing the square can be checked by expanding, but this is a bit out of nowhere. I will try the article's method if I get an opportunity, but I think the introduction of an extra variable z is quite an obstacle, and that the fact that -b/2 is the average of the roots will be quite a leap for most students.

nestorD 7y ago

Note that he critisize the classical formula but that he, then, suppose that A=1 when he gives his formulation (making it simpler). Adding A back gives : -B/2A ± sqrt(B²/4A - C/A) which is not obviously simpler than : (-B ± sqrt(B² - 4AC)) / 2A

verandacoffee 7y ago

Isn't this the same as the pq-formula? https://www.matteboken.se/lektioner/matte-2/andragradsekvationer/pq-formeln https://www.matteboken.se/lektioner/matte-2/andragradsekvati... (Link in Swedish).

jsjolen 7y ago
Yes, and apparently it's also used in Indian schools (see comments found here: https://www.facebook.com/poshenloh https://www.facebook.com/poshenloh ).

fyp 7y ago

People are criticizing this because it is still the same quadratic formula. But of course it is! Math is consistent. But representation matters. A good chunk of mathematics is just about rewriting the same mathematical fact in a different way. For example the equation of a line could be written with coefficients or in slope/intercept form or in polar coordinates or in homogeneous coordinates or etc etc. Here the claim is that explicitly giving a name to the variable -b/2 makes the equation easier to think about. I see nothing wrong with that.

LatteLazy 7y ago
Full disclosure: I critized the article in a comment below... Respectfully, that's not the reason people are critiquing the article. I fully agree mathematics is what works and many methods use identical underpinning logic, just expressed in different ways. I'm fine with that. But that doesn't mean all methods are equally good. This method is no quicker or easier or less error prown than the quadratic formula it "replaces". Even in the authors chosen example, it's no better. In many other cases it's harder (if B or C are not divisible by A, dividing by A to force A=1 just spreads and increases the complexity). That makes it a bad method because now, a user has to not only know both methods but also pick the right one. And for this extra time and risk, the gain nothing the standard Quadratic Formula didn't give them. We could equally "simplify" the quadratic formula by forcing B=1 or C=1. Are those methods new and useful? No. They're trivial and have limited use cases. They're never better than just using the full formula. My issues with the article are a bit wider: this is not new. I was taught this as a limited version of quadratics in 2000 in a run of the mill school in London. I also think it's derivative. Anyone smart enough to be solving quadratics should also be smart enough to apply basic algebra to simplify quadratics. But the article presents this, assuming the audience knows no better, like it's a breakthrough. That feels dishonest to me...
- fyp 7y ago
  I think it's useful to think of this as refactoring code to make it more readable (and therefore more teachable). If you're a memorizer, your code might as well be obfuscated code: def quadratic_formula(a, b, c): return [ (-b - math.sqrt(b ** 2 - 4 * a * c)) / (2 * a), (-b + math.sqrt(b ** 2 - 4 * a * c)) / (2 * a) ] If you understand the importance of the expression under the square root (i.e. the sign of the discriminant) you can rename one subexpression: def quadratic_formula(a, b, c): discriminant = b ** 2 - 4 * a * c return [ (-b - math.sqrt(discriminant)) / (2 * a), (-b + math.sqrt(discriminant)) / (2 * a) ] Of course you can refactor it further with pointless stuff like denom = 2a, but that doesn't add much semantic value. So the above is more or less the vocabulary we have about quadratic equations today. Loh's contribution is a specific way of refactoring the code by first dividing by a: def quadratic_formula(a, b, c): b = b / a c = c / a discriminant = b ** 2 - 4 * c return [ -b / 2 - math.sqrt(discriminant) / 2, -b / 2 + math.sqrt(discriminant) / 2 ] Which then unlocks the ability to talk about the subexpressions in relation to the roots (a la Vieta's formula): def quadratic_formula(a, b, c): b = b / a c = c / a sumOfRoots = -b productOfRoots = c averageRoot = sumOfRoots / 2 # Want roots [averageRoot - delta, averageRoot + delta] # such that: # productOfRoots == (averageRoot - delta) * (averageRoot + delta) # == averageRoot ** 2 - delta ** 2 delta = math.sqrt(averageRoot ** 2 - productOfRoots) return [ averageRoot - delta, averageRoot + delta ] Your code is no longer using single-letter variable names!

shikoba 7y ago

(r-s)^2=r^2-2rs+s^2=(r+s)^2-4rs=b^2-4c r=((r-s)+(r+s))/2=(sqrt(b^2-4c)-b)/2

LatteLazy 7y ago

I must be missing something here... The "standard" quadratic formula at the top of the article is just as quick and painless to solve the equation he uses as an example. And its easier for many other versions (basically any time B/A or C/A are not integers). Plus, this isn't new: I was taught to do exactly this IF it simplified the whole equation. That was in 2000 in London in a pretty standard secondary school. (Also, what's with taking "z^2 = 3" at font 10, treating it as an image and then displaying it 20 times the original size?! Is it meant to look more maths-ey?) I am excited to reveal something though: I recently discovered a whole new way to make a percentages! Instead of multiplying the number by 100, you just multiply it by 10 twice! So much better :)

gibspaulding 7y ago
The focus is not on getting students to be able to solve quadratics (if it were, then the quadratic formula is a great choice!), but rather to understand how solving a quadratic works. That is, how to derive the quadratic formula for themselves. Generally this is done by completing the square. This is another method which some students may find more intuitive.

dooglius 7y ago

I disagree that this proof is better pedagogically; it assumes the quadratic case of the Fundamental Theorem of Algebra (the correspondence between factorization and solutions), which at this stage would have to be taken on faith by students, whereas completing the square is fully justified.

boxfire 7y ago
Not sure why you think it's easier to take that on faith than complete the square, but I'm sure you never tutored new students to mathematics. At this point we are taking all of abstract algebra and some number theory for granted in anyone's education. Pedagogically it's easier to understand, the thesis of the article.
- dooglius 7y ago
  Granted, I have not tutored new students. I am making the assumption that someone encountering this would be familiar with basic algebraic manipulations, and the solutions to x^2=a. This is all that is necessary to justify completing the square. The assertion that quadratic equations must have at most two solutions R and S, and that it can be equivalently written (x-R)(x-S) is what has to be taken on faith.
pdonis 7y ago
> it assumes the quadratic case of the Fundamental Theorem of Algebra No, it doesn't; it proves that case. The key step in the proof is the fact that, over the complex numbers, every number has a square root: from that it follows that the factorization used in the proof must always exist.

spicymaki 7y ago

Here is a link to the paper the story was based on: "A Simple Proof of the Quadratic Formula" https://arxiv.org/pdf/1910.06709.pdf https://arxiv.org/pdf/1910.06709.pdf

6gvONxR4sf7o 7y ago

This still seems way more complicated than it needs to be to teach it. I always used a much simpler way to avoid memorizing anything. Imagine you have a parabola y - c = k x^2 and want to solve for y = 0. Dead easy, right? To turn any other parabola into this form, you only need to scroll left or right on x until the minimum is at x'=0 (algebraically, this means eliminating any b*x' term). Teach students how to do change of coordinates and how to solve this trivial problem, and they don't need to memorize any formulas. It also sets students up for the useful math mindset of solving new problems by reducing them to previously solved ones and relies on conceptual understanding. Seems way better than the "memorize this formula" approach.

sykick 7y ago
I teach mathematics at a community college. This includes teaching a lot of elementary algebra courses. These are pre-college level math courses. In elementary algebra we introduce solving quadratic equations by factoring. In the next course intermediate algebra we teach the quadratic formula. Students in elementary algebra are not equipped to understand change of coordinates. This is too hard of a concept at that stage in my opinion. The method described in the article is a very nice one and is appropriate for elementary algebra. It ties in nicely with factoring and makes solving quadratic equations quite easy. I will be using this method in the future.
- 6gvONxR4sf7o 7y ago
  How aren't they equipped for it? Do you mean change of coordinates in its full generality? I'm talking about simple substitution of x' = x + k, connected with the intuition sketched below. They know how to perform simple substitution by then, and they can surely follow this logic: x 0---1---2---3--> x' 0---1---2---3---4--> x' = x + 1 ==> x = x' - 1 y = f(x) ==> y = f(x' - 1) If you can talk about Napa being an hour north of San Francisco and San Francisco being an hour north of San Jose, this kind of student should be able to tell how far north of San Jose Napa is. Can they not even do that? If not, I'd questioning why we teach them to solve quadratic equations before they can do substitution.

prvc 7y ago

Surely this is a prank?

tel 7y ago

I think this is nice. It's a method of calculation derived from an important fact—that quadratic equations have two solutions and those solutions can be written in such a form. I don't have the experience or interest to debate whether it's better or worse than any other method pedagogically. I can definitely imagine that I would have appreciated learning this method as well as completing the square as a child. But honestly, the real interesting thing here is the (a) gathering information about what you think the solution ought to look like and (b) working backwards from there. That's a good general trick and worth having in your back pocket. It also emphasizes the importance of visualizing and examining your belief about what should occur. People should suspect the fundamental theorem of algebra long before they prove it.

edflsafoiewq 7y ago

If you're going to rename the constants to make a=1, you might as well also replace b with b/2. The solution to x^2 + 2bx + c = 0 is x = ± √(b^2 - c) - b

czrnb 7y ago

There is an error in the (pixelated) example that is given int the MIT tech review article (z² = 3 instead of -3). I think the fact that the author of the news missed this, that he probably took a screenshot of the formulas rewritten in Word, and that he was compelled to write such a long article on such a simple topic speaks for his level on the topic. The fact that he only lists the formal article as a reference instead of the announcement, video, and accessible blog post by Po-Shen Loh really baffles me. The original "disclosure" by Po-Shen Loh [0] is much less sensational and gives some context for his work (teaching middle school students). In the formal article, he is also stating that the method is very likely not __new__, but that he wants to popularize it in teaching. I think, as many other commenters pointed out, that there is no great breakthrough here. However "his" method may have the advantage of training the intuition of young students, by helping them understand the concepts of average and "deviation" (I'm not really sure how to call it in that case), and maybe visualizing them. [0]: https://www.poshenloh.com/quadratic https://www.poshenloh.com/quadratic

wrycoder 7y ago
And the third pixelated equation has an error - the final term on the left side should be C, not Cx^2. I think the point of the arXiv article is this is a more straightforward /proof/ of the quadratic formula. That article has some interesting historical commentary that shows exactly where the author thinks his contribution fits - he is not naively coming up with something "new". As far as computation of the roots goes, it is a slightly streamlined approach: First, put the quadratic into canonical form by dividing by A, if necessary. Then, take B/2 into a new variable, call it F. Get F^2. The roots are then -F +- sqrt(F^2 - C)
knolax 7y ago
Why is MIT technology review on the front page so often. It's literally a student newspaper with the expected quality of one.
- pfedak 7y ago
  Are you perhaps thinking of https://thetech.com/ https://thetech.com/ ? I'm not sure in what sense the Technology Review is a student newspaper - to my knowledge (and checking a few bylines), the journalists are not students.

tomp 7y ago

Um... what's the difference? x = (-b ± √(b^2 - 4ac)) / 2a Ok, remove a assuming it's 1 - well, yeah, less general math is simpler. x = (-b ± √(b^2 - 4c)) / 2 Move around 2 & 4... voila. x = -b/2 ± √(b^2 / 4 - c) Like, whatever. :)

bikenaga 7y ago

The old (Baudhayana, Omar Khayyam) geometric approach to completing the square (and so solving [some] quadratics) is very nice because it's visual, and it's what I show students in many classes. It explains why it's "completing the square". Suppose you want to complete the square for x^2 + 6 x. Represent this as an x-by-x square and a 6-by-x rectangle: x ..... x ..... ..... ***** ***** ***** 6 ***** ***** ***** Cut the 6 x rectangle into two 3-by-x rectangles: x ..... x ..... ..... ***** 3 ***** ***** ***** 3 ***** ***** Move the lower 3-by-x rectangle up next to the square. The L-shaped figure still has area x^2 + 6 x. x 3 ..... ***** x ..... ***** ..... ***** ***** 3 ***** ***** What do you need to add (what is the size of the small square on the lower right) to complete the (large) square? The small square is 3-by-3, so it has area 9: x 3 ..... ***** x ..... ***** ..... ***** ***** +---+ 3 ***** | | ***** +---+ You get x^2 + 6 x + 9 = (x + 3)^2. If the original x^2 + 6 x was on one side of an equation, you add 9 to both sides.

aphextim 7y ago

One of my favorite things to break the ice with a younger gamer and the Quadratic Formula. https://www.youtube.com/watch?v=ewAHYVzMobw https://www.youtube.com/watch?v=ewAHYVzMobw Super Mario Quadratics.

orbifold 7y ago

There is nothing new about this formula or derivation. In fact this precise formula (including that derivation) is taught in Germany.

abnry 7y ago

Let's get the criticisms of the article out of the way: --It has terrible formatting and typos. --It puffs up something more important than it is. --It is more about pedagogy than a mathematical idea. --It suggests the idea is original, when it almost certainly is not. --It doesn't link to the original (and better source). Okay, here is the good things about the approach: --It is good to shift your thinking about mathematical derivations and proofs and think about them as code that runs on people's brains. You input a derivation into someone's brain and they return a boolean value (this is true, it makes sense, etc.). Pedagogy is trying to optimize the code for less powerful architectures. Just like when you are optimizing code tiny little details of instruction orders matter, the same with mathematical derivations. --Fundamentally, algebraic manipulations are uncomfortable and nonintuitive for students. They feel like tricks. Going forwards from (x+a)^2=x^2+2ax+a^2 makes sense but going backwards as in the case of completing the square is hard. It's not the same case for x^2=a vs sqrt(x)=a. This is kind of a similar case to math students feeling confused by adding and subtracting the same quantity when doing calculus limits. For any trained mathematician, this is obvious, but it really feels like a trick at first. The nice thing about this approach is that it avoids this issue and gives you a good reason WHY the -b/2 term shows up. Additionally, it avoids the problem of substituting, which tends to bog students down (try teaching the chain rule someday). Students should still understand completing the square but I don't think this is a bad way to introduce them to the quadratic formula. It highlights the symmetric of the roots (at least for real values), which makes sense if you plot a quadratic.

infocom6502 7y ago
I agree. This is nothing novel and a fairly standard way of solving the equation.
jfengel 7y ago
I have never understood "completing the square". I mean, I get why and how it works, but it's completely unintuitive to me, or why you'd do it that way. I think it is a historical hangover. It makes more sense viewed geometrically, and the technique is attributed to the same al-Khwarizmi for whom algorithms are named. But it's less intuitive as part of algebra, and we focus a lot more on algebra today than in the medieval times, when geometry ruled.

ajarmst 7y ago

I think there's an unacknowledged premise here that is fundamentally misguided: that the point of teaching math is to minimize the complexity of the formulas and recipes the students have to memorize, or that skill in finding the roots of a quadratic is so important that we must optimize the experience of learning it. I work with mathematics teachers and engineers and I know of no one who has willingly factored a quadratic by hand since we put our slide rules away. The goal, fundamentally, is not the skill of factoring a quadratic. The goal is understanding of the relationships of numbers, operations, and domains that allow algebra to be a powerful set of tools for solving huge classes of problems. I _never_ teach the quadratic equation. I teach completing the square, because it's an illustration of a useful way of algebraically manipulating a relationship into a form (square of a linear binomial) that they recognize and can easily factor. I usually do a quick proof of the quadratic equation using completing the square, but generally as an illustration that if you properly understand the associated algebra, you don't need to memorize formulas and algorithms. The use of the field axioms to manipulate polynomials, and the goal of manipulating them to a tractable form, is what they need---not the quadratic equation. This 'new' technique (which appears to be a simple riff on the standard technique of looking for a pair of numbers whose sum is B and whose product is C that is taught as a matter of course in every middle school on the planet) is missing the actual goal of the lesson. In particular, the reason I even teach completing the square is because it's a precursor of me bringing up the roots of $x^2 + 1$, which takes us in to an introduction to complex numbers, the fundamental theorem of algebra, and the discovery that the same old algebra over the Reals they learned in high school can now be used to do things like solve differential equations. Later, we introduce matrices and I can return to our old friends the algebraic field axioms to solve whole systems of equations. If they're lucky, they also get modular arithmetic and can be shown Galois fields and start using their understanding of algebra for problems in logic and set theory. All of that is easily within the reach of students in their first or second year of college, and I think if we bothered to try, we'd discover it's easily in the reach of secondary school students---or would be if we'd stop pretending that teaching them that "mathematics" is making change and pushing numbers from a word problem into the blanks of some generic formula they've memorized. The task isn't teaching them to factor quadratics. The task is teaching them algebra.

mogadish 7y ago

To me this seems like a special case of the Vieta formulas (https://en.wikipedia.org/wiki/Vieta%27s_formulas https://en.wikipedia.org/wiki/Vieta%27s_formulas) which in my case I learned in highschool

eyeundersand 7y ago

In Nepal, you are taught around the sixth grade that you can solve ax^2 + bx + c = 0 by splitting b into m and n such that mn = ac and m + n = b ( or m -n = b, if a and c have opposing signs). This "discovery" the author claims and some other corollaries were quite commonly known. I find it absurd and, frankly, laughable that this is being heralded as something new!

knzhou 7y ago

This kind of approach is familiar to me from competition math. Back in middle school math club we were taught this exact approach (under the name of "Vieta's formulas"), i.e. that thinking about the sum and product of the roots could be faster in some cases. Po-Shen Loh is the director of the US IMO team, so it makes sense he would like this approach. However, I don't think it makes logical sense to teach it only this way. Here you start by assuming that a quadratic has two roots, which is not at all obvious the first time a kid sees a quadratic equation. (Especially because those roots can be complex numbers!) Completing the square tells you why there are two roots, and also naturally leads you to the necessity of complex numbers, i.e. when the "square" you end up making is negative. You can use the nice Vieta's formula tricks only after establishing that.

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fyp 7y ago
I am also familiar with Vieta from math competitions but I think this is more than that. Given two numbers, r1 and r2, if you know their arithmetic mean: m = (r1 + r2) / 2 and geometric mean: g = sqrt(r1 * r2) then following Loh's derivation you get a very cute formula: r1 = m - sqrt(m^2 - g^2) r2 = m + sqrt(m^2 - g^2) Vieta gives an easy way of finding those means from a quadratic equation: r1+r2=-b and r1*r2=c. So you can plug in m=-b/2 and g=sqrt(c) in the equation above. The fact that you can state roots in terms of their means is the more novel insight to me. (note: Loh doesn't talk about geometric means but I thought using just product of the roots isn't as "symmetric")

ivan_ah 7y ago

That's very cool. To the best of my knowledge the only purpose of learning how to complete the square in high school math is to be able to understand the derivation of the quadratic formula, so if we can use this alternative proof then students don't need to know "completing the square" anymore. Good riddance! The only other application of "completing the square" I can think of is the trig substitution in calculus, see https://en.wikipedia.org/wiki/Completing_the_square#Integration https://en.wikipedia.org/wiki/Completing_the_square#Integrat... or worked example here https://www.khanacademy.org/math/ap-calculus-ab/ab-integration-new/ab-6-10/v/integration-using-completing-the-square-arctan https://www.khanacademy.org/math/ap-calculus-ab/ab-integrati... In sum, completing the square is definitely a cool trick (see https://en.wikipedia.org/wiki/File:Completing_the_square.gif https://en.wikipedia.org/wiki/File:Completing_the_square.gif ), but maybe we can skip it... or present it as extra/optional material? I'm going to think about dropping it from my books. It will save 5+ pages of suffering for readers, which is a clear win.

hollandheese 7y ago
Nope, completing the square is primarily used to translate a standard quadratic function into a vertex form quadratic so it can be graphed using transformations instead of guessing. It is an essential multi-use tool in precalculus. The thing we should dump is the quadratic formula.

Tiruneh 7y ago

I have read the paper and it gives a good educational context from which the quadratic equation may be looked at. I have also worked out simpler way to solve quadratic equations using function evaluation without much memorization of symbolic formula: for the equation : f(x) = ax^2 +bx + c = 0 X = Z +- Sqrt [-f(Z)/a] ; Z = -b/2a for the equation f(x) = x^2 +bx+c = 0 it is even a bit simpler: X = Z +- Sqrt[ -f(z) ] where Z = -b/2. Example 1: f(x) = 3x^2 -8x-35 = 0 Z = -b/2a = - (-8)/(2.3) = 4/3 F(Z) = -121/3 X = 4/3 + or - Sqrt ( -(1/3)*(-121/3) = 4/3 + or - 11/3 = {5, -7/3) Example 2: (Simpler form): f(x) = x^2 - 4x+ 3 = 0 Z = -b/2 = -(-4)/2 = 2 f(Z) = 2^2 - 4.2 + 3 = -1 X = 2 + or - Sqrt( -(-1) = 2 +- 1 = { 3, 1} For detail of this method, please see the following pre-print https://www.researchgate.net/publication/337829551_A_simple_formula_for_solving_quadratic_equations_using_function_evaluation?fbclid=IwAR1yoUtuTQAc8Mpc-BRME5vdKkBZ4QbXZj741lGhUdvc7s7R3rAugZE6VGg https://www.researchgate.net/publication/337829551_A_simple_...