Quadratic Method: Detailed Explanation (2020) | Hacker News Reader

Quadratic Method: Detailed Explanation (2020)

poshenloh.com

81 pointsaleyan19 comments

If you don't remember the quadratic formula and don't want to look it up, the graphical intuition is super simple, if you're comfortable connecting the graphical to the algebraic bits.

A quadratic equation is the same as finding where a parabola crosses zero. The formula for that is messy unless the parabola is centered around x=0, in which case it's easy. Then it's just a "standard" parabola (y=x^2), stretched horizontally and/or moved up and down. The equation for that shape is 0=ax^2-k, so x^2=k/a, which is super simple.

So then you just have to figure out how to turn your original problem (a parabola that's not centered at x=0) into the easy one (a parabola that is centered at x=0).

The answer is to just slide your axes over until it's centered. That is, define a coordinate z = x-c such that your parabola is centered at z=0. To do that, just plug in z+c=x into the original equation, and solve for a c that makes the middle term disappear (you want to end up with y=az^2 + 0z + b, per my second paragraph). Now you have the problem that's easy to solve. And there's the added benefit that whatever problem this comes up in is likely easier to think about in terms of this z than in x.

This method is just some trivial steps, which makes it really easy to remember. It's easy to slide a shape left and right, and it's easy to solve x^2=k.

You just explained why the quadratic formula is the way it is better than any of the discussions of completing the square that all my middle school textbooks gave.

Maybe for you. Different explanations work for different people.

The quadratic formula is simpler to remember and to derive if you first put your equation into the form: x² – 2ax = b

Then (x – a)² = a² + b

So x = a ± √(a² + b)

The method is interesting, but what exactly is the intended use? I don't get what the OP is trying to do.

- Normal people are going to forget any kind of proof you show them exactly eight seconds after you're done. The pedant in me wants to note that plugging the standard formula in and veryfing it works is also in fact a proof, but technically correct is, as always, the most useless kind of correct.

- Everyone solves quadratics by (a) eyeballing either zero, or (b) \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. It just doesn't make any sense to use any other method when the standard formula is simple arithmetic you can do in your head.

- Students who are into kinky stuff like competitions already know several formulas for speed.

- If this is just pedagogical, it's basically a way to adamantly refuse to complete the square by using a slightly different way to complete the square

am I missing something? I'm open to change my mind.

> Everyone solves quadratics by (a) eyeballing either zero, or (b) \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. It just doesn't make any sense to use any other method when the standard formula is simple arithmetic you can do in your head.

You have to remember the formula. I'm a maths student, but I can never remember it (and I also hate to memorize stuff). I remember the shape of it so that I could generally recognise it, but that's about it. I I need the general form, it's easy (for someone with practice) to rederive it from scratch.

In practice, when given a quadratic equation, I can either guess its roots fairly quickly or I use completing the square. There are probably some other tricks, such as the one from the article, I could use and others undoubtedly use, but my method works for me, and it's not like I need to solve quadratic equations all the time.

A related question is how to solve quadratic equations robustly using IEEE floating point. When solving ax² + bx + c, the most common problem is when a is very tiny, or even zero (then it becomes a linear equation, which is easy, but the classic solution becomes undefined). There are other things that can go wrong.

At some point I might write up a solution, but in the meantime I'll point to code that implements it[1].

[1]: https://docs.rs/kurbo/0.8.1/src/kurbo/common.rs.html#105-149

I really like this new way of doing factoring, since 90% of the steps of the explanation are more intuitive and direct than the usual guessing techniques and the complete-the-square technique that leads to the quadratic formula. I was so excited by this new approach that I started rewriting the parts in the math book that explain quadratics, and looking forward to getting rid of the complete-the-square procedure, which many readers find confusing.

However, the Po-Shen Loh method (a.k.a the Viète method) has one big conceptual hurdle at the beginning (the missing 10%), which I had trouble explaining, since it seems to come out of nowhere and not obvious:

> "Two numbers have a sum of 14 exactly when their average is 7."

> "Two numbers have an average is 7 when those numbers are 7-u and 7+u"

via https://www.youtube.com/watch?v=XKBX0r3J-9Y&t=592s

^ both these statements are true and correct, but they seem to come out of nowhere, like one of those "Suppose u = <complicated expression> ..." substitution tricks that are needed to solve certain calculus problems, which are otherwise impossible if you don't know the needed substitution.

The substitution 7-u and 7+u is related to the general idea that any two numbers (or functions) can be written in terms of a symmetric component m (half-sum, or mean) and a half-difference part d:

   given any a, b
   define:
       m = (a+b)/2     (the half-sum of a and b)
       d = (a-b)/2     (half-difference of a and b)
   then:
       a = m + d
       b = m - d

Again, this is totally true (and a very useful math idea that comes up in other places), but not intuitive for beginners.

In the end I decided to stick with the complete-the-square approach: complete-the-square is not that bad, especially when you show the picture: https://minireference.com/static/excerpts/noBSmath_v5_previe...

It's easy to motivate using a picture. If you draw the parabola x^2 + bx + c then the minimum is at x = -b/2 and you can visually see that the average of the x-values of the roots is that minimum x = -b/2. So then u is simply the x-distance from the minimum to the roots.

The more problematic step is the assumption that you can write the parabola as (x - r1)(x - r2). It isn't obvious why that is possible.

Replace the word "average" with the word "middle" and it might be a bit more obvious?

Previous thread: https://news.ycombinator.com/item?id=21720656

But the explanation doesn't easily generalize to higher order polynomials.

There's actually an entire branch of math (galois theory) that spawned from the fact that it doesn't generalize!

In particular, there's no formula for polynomials of degree 5 or higher.

https://en.wikipedia.org/wiki/Abel%E2%80%93Ruffini_theorem

Sure, but there are analogous cubic and quartic formulas that follow somewhat similar derivations. Here's a fairly complete derivation of the cubic formula and a slightly abbreviated derivation of the quartic formula: http://math.sfsu.edu/smith/Documents/Cubic&Quartic.pdf

The (classical) derivation of the cubic solution involves square roots of negative numbers. Hence it either requires unsatisfying hand-waving, or knowledge of complex numbers. These things come up quite a bit later to the much simpler "solve a quadratic formula" question.

Knowledge of complex numbers are required to solve quadratics too (when the discriminant is negative).

Right.

Any vaguely rigorous derivation of the cubic formula requires dipping into complex numbers, because the discriminant of the cubic can be negative. You can just choose to ignore that case and focus only on nonnegative discriminants, but that's no fun, is it? :-) Although, that is how it was "classically" done.

If you want to find the complex roots, but don't want to deal with them in the derivation, you can just take the real root and multiply it by (-1 ± i √3)/2. Knowing that all you have to do is this, of course involves some much heavier algebra than would have been available classically, but, at least it's easy for students to understand (if they have the proper group theory and/or Galois theory background), or, at least to understand a plausibility argument.

After looking at Wikipedia, I found that Lagrange had an interesting method[0] I didn't know about, in which he considers the discrete Fourier transform of the roots rather than the roots themselves. That approach makes some sense, because the whole power of the Fourier transform is to turn gnarly multiplication problems into simple addition problems.

I also like this method, because it's honest about needing to dip into the complex numbers to find all roots. Also, it's kind of a cool historical note, because he was trying to solve the general problem of finding all roots of a polynomial, and hoped this method might generalize.

Edit: here's a lesson plan from Stanford that walks through all the necessary algebra in 5 class periods: https://web.stanford.edu/~aaronlan/assets/symmetries-and-pol...

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[0]: https://en.wikipedia.org/wiki/Cubic_equation#Lagrange's_meth...

Let me give you the method to solve x^2 + bx + c = 0.

Answer: solve (x + b/2)^2 - b^2/4 = - c instead.

Why, you ask? Well, first of all because some long-dead bloke figured out this neat trick and it would be a shame not to use it. And secondly it is easy arithmetic to figure out that this is an equivalent problem, so if you solve it then it certainly gives the right answer.

Maybe it is just me, but if you cannot grok this then I think that your difficulties with mathematics are not curable by spending more time trying to develop intuition on this particular problem. And therefore I fail to see what Professor Loh brings to the table here.

Edit: it appears that my position is being misunderstood. I do not think that someone who does not grok this argument will never be able to properly learn mathematics. I think that they should instead focus on things like logic and arithmetic and, for this particular subject, on playing with quadratics (plotting them and finding their extrema, for example). Any dressing up of the above argument for the quadratic formula is, again in my view, akin to trying to 'develop intuition' for something like the third digit of pi - it is just what it is (4 I think).

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