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A new way to make quadratic equations easy

technologyreview.com

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Re: A new way to make quadratic equations easy

#81

Post by WhompingWindows »

robinhouston wrote: post #6

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 yo…

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

Re: A new way to make quadratic equations easy

#82

Post by jfengel »

abnry wrote: post #63

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 mat…

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.

Re: A new way to make quadratic equations easy

#83

Post by knolax »

czrnb wrote: post #38

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 in…

Why is MIT technology review on the front page so often. It's literally a student newspaper with the expected quality of one.

Re: A new way to make quadratic equations easy

#84

Post by jessaustin »

sykick wrote: post #65
6gvONxR4sf7o wrote: post #58

Earlier quoted context omitted.

I freely admit that my viewpoint is clouded by fluency, but I didn't come up with this today. This is how I've done it since I was still in school and I remember being annoyed by the opaque "just memorize this formula" approach from early on, where the concepts just seemed much clearer. I admit to not being the typical math student. But I think my approach was my competitive advantage, not something that makes my exp…

In elementary algebra students can do this for the most part: Evaluate x^2 + 2x + 4 for x = 3. Quite a few will struggle with: Evaluate x^2 + 2x + 4 for x = –3 Almost all will struggle with: Evaluate -x^2 - 2x + 4 for x = -3 I don't think they'd handle replacing x with z+2. In elementary algebra they hate fractions. Many struggle with 4 – (–5). I like the approach in the article because of how it relates to factoring…

'sykick you have my sincere appreciation for your tireless efforts in this thread to communicate both the existence and particulars of real differences in math aptitude among humans. Those on HN who have never attempted to help someone else with math would have had no idea...

Re: A new way to make quadratic equations easy

#85

Post by mkl »

sykick wrote: post #36
mkl wrote: post #13

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 aw…

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 arti…

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.

Re: A new way to make quadratic equations easy

#86

Post by sykick »

jessaustin wrote: post #84
sykick wrote: post #65

Earlier quoted context omitted.

In elementary algebra students can do this for the most part: Evaluate x^2 + 2x + 4 for x = 3. Quite a few will struggle with: Evaluate x^2 + 2x + 4 for x = –3 Almost all will struggle with: Evaluate -x^2 - 2x + 4 for x = -3 I don't think they'd handle replacing x with z+2. In elementary algebra they hate fractions. Many struggle with 4 – (–5). I like the approach in the article because of how it relates to factoring…

'sykick you have my sincere appreciation for your tireless efforts in this thread to communicate both the existence and particulars of real differences in math aptitude among humans. Those on HN who have never attempted to help someone else with math would have had no idea...

Thank you. That’s very nice of you to say.

Re: A new way to make quadratic equations easy

#87

Post by pfedak »

knolax wrote: post #83
czrnb wrote: post #38

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 in…

Why is MIT technology review on the front page so often. It's literally a student newspaper with the expected quality of one.

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.

Re: A new way to make quadratic equations easy

#88

Post by ivansavz »

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.

Re: A new way to make quadratic equations easy

#89

Post by amluto »

sykick wrote: post #41
robinhouston wrote: post #6

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 yo…

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 binomi…

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.

Re: A new way to make quadratic equations easy

#90

Post by edflsafoiewq »

jfengel wrote: post #82
abnry wrote: post #63

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 mat…

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 med…

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.

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