[HN Gopher] Quadratic Method: Detailed Explanation (2020)
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Quadratic Method: Detailed Explanation (2020)
Author : aleyan
Score : 77 points
Date : 2021-05-20 13:36 UTC (1 days ago)
(HTM) web link (www.poshenloh.com)
(TXT) w3m dump (www.poshenloh.com)
| 6gvONxR4sf7o wrote:
| 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.
| BeFlatXIII wrote:
| 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.
| concreteblock wrote:
| Maybe for you. Different explanations work for different
| people.
| jacobolus wrote:
| The quadratic formula is simpler to remember and to derive if
| you first put your equation into the form: _x_ 2 - 2 _ax_ =
| _b_
|
| Then ( _x_ - _a_ )2 = _a_ 2 + _b_
|
| So _x_ = _a_ +- [?]( _a_ 2 + _b_ )
| qsort wrote:
| 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.
| Tainnor wrote:
| > 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.
| raphlinus wrote:
| 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
| steve76 wrote:
| Here's a trick. Whenever someone discusses the quadratic formula,
| do a quick search on "discriminant". If it's not mentioned, move
| on.
| prof-dr-ir wrote:
| 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).
| ivan_ah wrote:
| 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 Viete 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...
| jules wrote:
| 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.
| rocqua wrote:
| Replace the word "average" with the word "middle" and it might
| be a bit more obvious?
| amelius wrote:
| But the explanation doesn't easily generalize to higher order
| polynomials.
| fyp wrote:
| 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
| actually_a_dog wrote:
| 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
| rocqua wrote:
| 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.
| onhn wrote:
| Knowledge of complex numbers are required to solve
| quadratics too (when the discriminant is negative).
| actually_a_dog wrote:
| 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...
|
| ---
|
| [0]: https://en.wikipedia.org/wiki/Cubic_equation#Lagrang
| e's_meth...
| fyp wrote:
| Previous thread: https://news.ycombinator.com/item?id=21720656
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