A new way to make quadratic equations easy

https://www.technologyreview.com/s/614775/a-new-way-to-make-quadratic-equations-easy/

by uniqueid • 7 years ago

177 94 7 years ago

I don't know that it was ever hard.

yxhuvud• 7 years 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• 7 years ago

They are https://www.youtube.com/watch?v=o_zoHxC4y-U

mikorym• 7 years 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.

saagarjha2• 7 years 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• 7 years 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-appr...

robinhouston• 7 years 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• 7 years 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.

pflats4• 7 years 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/

leggomylibro• 7 years 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?"

ken• 7 years ago

Programming is full of these. Every day I see some new technique or style appear. Is it any good? We've got easier access to the biggest collection of free source code, the biggest and most diverse set of programmers, and the fastest automated testing of all time. Just do an experiment. You don't need to sell me on how cool it is. Show me how it's better.

- Does it lead to smaller programs? Don't show me a 10-line example. (I've heard medical researchers say: "Anyone can cure cancer in mice.") Convert a 10,000- or 100,000-line program.

- Does it lead to humans writing fewer bugs? Have a bunch of them try it, and measure their speed and defect rates. Everyone has designed a system that they themselves love, and nobody else can understand.

If you promote a new method which is logically equivalent to old ones, and not obviously much better, and without any data showing specific metrics that it improves, I'm forced to assume your metric is simply "I happen to like it better". Experimentation is cool, but it doesn't cause lasting long-term shifts.

sykick8• 7 years ago

I’ve taught elementary algebra for many years at a community college. The method described in the article is truly, in my opinion, very nice. I will be using this method from now on in elementary algebra. Students in elementary algebra don’t know what parabolas are. They haven’t been taught to complete the square and don’t know how to work with radicals. They have just been taught what square root is and basic binomial and trinomial factoring.

To me it is obvious that the method in the article is far superior than teaching completing the square. I’m teaching a pre-calculus course this semester and many of my students still can’t complete the square. Pre-calculus is 3 math courses beyond elementary algebra.

All of this is just my opinion of course and I have no data or studies to back up my opinion. I will be using the method described in the article from now on in my elementary algebra courses.

pflats4• 7 years ago

To me it is obvious that the method in the article is far superior than teaching completing the square.

I disagree. I would need some convincing that "two numbers that multiply to C and sum to B must have an average of B/2, so they must be B/2 + z and B/2 - z, so (B/2 + z)(B/2 - z) = C" is by any means obviously superior to completing the square. Neither is immediately intuitive; both will require prompting and teaching by the teacher. Completing the square has uses beyond proving the quadratic theorem; this does not.

I should say: I find this an incredibly cool and level-appropriate proof of the quadratic equation, but I think its merits as an improvement in pedagogy are dubious.

carapace4• 7 years ago

Apologies in advance for the harshness of my tone, but I nearly spit up my coffee...

Students in elementary algebra don’t know what parabolas are.

The mind boggles. Have you never shown them a sliced cone? ( https://en.wikipedia.org/wiki/Conic_section )

I mean, they know what parabolas are: they live on a planet, with gravity. Every ball, every jump: parabolic motion, yeah?

"Michael Jordan exerts his muscular power to enter a low-altitude earth orbit..." ~some Nike commercial in the 90's.

They know what parabolas are, you just have to connect the dots.

amluto• 7 years ago

I’m surprised that these students learn to solve quadratic equations before they learn what parabolas are.

If I were teaching quadratics, I would probably start with squares and square roots, and I would draw pictures. I would present some motivating examples (from kinematics, maybe? just the pictures may be okay, especially if the fun game where you try to zap the targets by hitting them with graphs is still around.). Then I would teach translations. After that comes the distributive law and polynomials written ax^2 + bx + c.

And now you can solve them! As far as I’m concerned, solving equations (polynomials, systems of equations, integrals, ODEs, PDEs, etc) is a puzzle, and learning a bag of tricks to solve them is just that: learning a bag of tricks. Quadratics are nice because the tricks always work. In more complicated math, it’s important to understand that the tricks can be very hard or even probably nonexistent, and accepting that is important.

But I don’t see why we should teach people to solve quadratics before teaching what they are.

edit: it’s not clear to me that the method in the article is dramatically different from completing the square. Assume a=1 for simplicity. Given the knowledge the the average of the roots is -b/2 (which one can deduce by any number of means), you can solve the equation in quite a few ways. One is the way in the article. Another is to say “the average is -b/2, so the vertex of the parabola is at x=-b/2, so the polynomial can be written (x+b/2)^2 + something”. Another is to just write down the solution x = -b/2 ± something and solve for “something” (which is more or less the same thing as in the article).

In high school, I used to have fun solving quadratics in my head by seeing which technique gave a quick answer.

jerf2• 7 years ago

I believe the objection is that deriving the formula itself from completing the square is the challenge, not the direct use of completing the square as you show in your example. (Although I wonder where you learned that, because I was certainly never taught that so directly.) The problem with deriving the quadratic equation that way is that for most people, that is a lot of symbols to keep track of, and you have to not only do an unintuitive "completing the square" step but you have to unintuitively do it fully generically.

For a professional mathematician this is barely a warmup, of course, but for the average person in middle school or early high school, this is distinctly nontrivial work.

(I'm not taking a side here, just trying to describe what the issue is, since you asked. I'm ambivalent. Neither of my kids are quite this far yet, so I'm not quite here yet.)

pflats4• 7 years ago

The problem with deriving the quadratic equation that way is that for most people, that is a lot of symbols to keep track of, and you have to not only do an unintuitive "completing the square" step but you have to unintuitively do it fully generically.

"Unintuitive" depends entirely on your introduction to the topic. If you're already completing the square, using it to solve quadractic equations you cannot factor is not unintuitive.

As far as doing it fully generically, well, how else do you get a generic formula? When teaching this, we would do some completing the square to solve quadractics, and then tell our students:

"You know, this is annoying to have to complete the square EVERY TIME. What if we just decided to solve it the really hard way once, with A, B, and C in the equation instead of the numbers, and see what we get?"

WhompingWindows• 7 years ago

I thought the article was trying to compare to the "typical" fractional formula, not to completing the squares?

saagarjha2• 7 years ago

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

mkl5• 7 years 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• 7 years ago

One would think an MIT technical publication (PR arm?) aimed at showing off their prowess would know how to typeset an equation.

ninjinxo• 7 years ago

/r/math discussion: https://www.reddit.com/r/math/comments/dzbmbu/a_new_way_to_s...

Original source: https://www.poshenloh.com/quadratic

thanhhaimai• 7 years 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• 7 years 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.

mkl5• 7 years ago

The a = 1 is not the important bit - that's probably nearly universal.

knzhou2• 7 years 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• 7 years 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?

mkl5• 7 years 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.)

mkl5• 7 years 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.

sykick8• 7 years 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.

mkl5• 7 years 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• 7 years 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• 7 years ago

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

jsjolen• 7 years ago

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

fyp3• 7 years 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.

LatteLazy2• 7 years 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...

fyp3• 7 years 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!

sethammons• 7 years ago

this is not new

While the mathematician claims to not find historical evidence of this, it is suuuuper similar to something we went over in high school in Calculus. It was related to finding the vertex of a parabola and noting the roots will be equally distant to both sides of the mid point. At the time, it was used as a "see? Neat. It all works out" type lesson.

This was around 2001.

carapace4• 7 years ago

*error prone

tom-thistime• 7 years ago

Perhaps people are skeptical about the claim of originality for a straightforward algebraic manipulation that can be carried out and understood by middle school students. (As far as I can see, that is what the article claims.) Bright people "reinvent the wheel" all the time. The rediscovery of the trapezoidal rule for numerical integration is a recent popular example.

Also, unrelated and probably unfair ... maybe a few cynical and overly skeptical people become a little cautious when they see a link to Tech Review.

shikoba• 7 years 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

LatteLazy2• 7 years 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 :)

gibspaulding2• 7 years 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.

dooglius2• 7 years 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• 7 years 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.

dooglius2• 7 years 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• 7 years 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• 7 years 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

6gvONxR4sf7o3• 7 years 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.

sykick8• 7 years 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.

6gvONxR4sf7o3• 7 years 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• 7 years ago

Surely this is a prank?

tel• 7 years 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.

edflsafoiewq2• 7 years 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• 7 years 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

wrycoder• 7 years 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)

knolax2• 7 years 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• 7 years ago

Are you perhaps thinking of 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• 7 years 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• 7 years 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• 7 years 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

Super Mario Quadratics.

orbifold• 7 years ago

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

abnry2• 7 years 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• 7 years ago

I agree. This is nothing novel and a fairly standard way of solving the equation.

jfengel• 7 years 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.

edflsafoiewq2• 7 years ago

Algebraically, my intuition is that the x^2 + bx part of x^2 + bx + c = 0 "looks" pretty close to the expansion of (x+b)^2. If you relabel b so this becomes x^2 + 2b, it's even clearer. So you can try to fiddle with the constant term to make the LHS exactly that square, and you get completing the square.

This is similar to how you solve a first order linear DE, y' + f(x) y = g(x). The idea is that the LHS "looks" like the derivative of a product, (hy)' = h y' + h' y, so you fiddle with an integration factor to make the LHS exactly that derivative.

bigred100• 7 years ago

Once you get to quadratic forms you can use it again.

ajarmst• 7 years 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• 7 years ago

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

eyeundersand• 7 years 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!

knzhou2• 7 years 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.

fyp3• 7 years 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• 7 years 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#Integrat... or worked example here 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 ), 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• 7 years 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• 7 years 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_...

A new way to make quadratic equations easy

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