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

~~~
sykick
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.

~~~
pflats
> 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.

~~~
sykick
I doubt I can convince you. I’m just going by my experience teaching the
topic. At the time students first learn solving such equations they have just
been taught factoring and what it means to factor a trinomial. They know the
product of the constant terms in the binomials must be c. It’s also easy to
explain that the average of two numbers is the midpoint. And thus if I start
with the midpoint then to get to the numbers I took the average of I add and
then subtract some number from the midpoint. The geometry makes this easier to
explain over using completing the square.

I’ve seen a shocking number of calculus students struggle with completing the
square. The merits of the approach in the article are entirely obvious to me
but like everyone else I’ve had my share of obvious beliefs turn out to be
false.

~~~
Isamu
>but like everyone else I’ve had my share of obvious beliefs turn out to be
false.

Refreshing candor! Wish it held true that more people saw it through to
discover their obvious truths didn't hold up.

------
abnry
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.

~~~
jfengel
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.

~~~
edflsafoiewq
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.

~~~
abnry
It's not intuitive for someone new to mathematics, but as someone with two
math degrees, it is just the strategy of "do stuff to something until
something about it looks similar to something you've seen before" that
mathematicians use ALL the time. The additional part of this strategy is to
dream optimistically about how you can make the thing you are dealing with
"nice" after doing stuff to it. In the case of completing the square, you are
hoping you can just straight up take a square root. That would be easy. Turns
out if you try enough stuff, you can.

------
bikenaga
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.

------
knzhou
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.

~~~
fyp
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")

------
fyp
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
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
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!

------
czrnb
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)

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

~~~
pfedak
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.

~~~
knolax
Oh wow I just read their about page and I stand corrected. I'm not sure if the
fact that they're a (supposedly) hundred year old newspaper makes me feel any
better about the quality of their articles though.

------
mkl
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
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
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.

------
thanhhaimai
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
> 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.

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

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

------
montalbano
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...](https://www.goodreads.com/book/show/19161684-a-friendly-approach-to-
complex-analysis)

------
ninjinxo
/r/math discussion:
[https://www.reddit.com/r/math/comments/dzbmbu/a_new_way_to_s...](https://www.reddit.com/r/math/comments/dzbmbu/a_new_way_to_solve_quadratic_equations_poshen_loh/)

Original source:
[https://www.poshenloh.com/quadratic](https://www.poshenloh.com/quadratic)

------
dooglius
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
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
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.

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

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

------
tel
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.

------
LatteLazy
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
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.

------
bograt
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
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.)

------
ivan_ah
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...](https://en.wikipedia.org/wiki/Completing_the_square#Integration)
or worked example here [https://www.khanacademy.org/math/ap-calculus-ab/ab-
integrati...](https://www.khanacademy.org/math/ap-calculus-ab/ab-integration-
new/ab-6-10/v/integration-using-completing-the-square-arctan)

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
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.

------
jimhefferon
I don't know that it was ever hard.

~~~
mikorym
>> 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
> 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".

------
edflsafoiewq
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

------
Tiruneh
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_...](https://www.researchgate.net/publication/337829551_A_simple_formula_for_solving_quadratic_equations_using_function_evaluation?fbclid=IwAR1yoUtuTQAc8Mpc-
BRME5vdKkBZ4QbXZj741lGhUdvc7s7R3rAugZE6VGg)

------
tomp
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. :)

------
ajarmst
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
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

------
6gvONxR4sf7o
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
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
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.

~~~
sykick
Well for one thing the simple substitution you mention is a hard concept. They
certainly will have a hard time with f(x-1) and introducing a new variable is
a mental block at this stage. At my college we don’t introduce functions until
the next course. I suggest your viewpoint is clouded by the fact that you know
this stuff so well that you no longer remember what the pain points are for
students learning it for the first time.

After learning to solve simple quadratic equations and then the quadratic
formula we typically introduce variable substitutions to solve things like x^4
+ 5x^2 - 6 = 0.

~~~
6gvONxR4sf7o
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 experience inapplicable.

To be clear, you're saying that if you tell a student

    
    
        y = x^2 + b*x + c
    

and you tell them

    
    
        x = z + 2
    

they aren't yet equipped to learn to combine those into

    
    
        y = (z+2)^2 + b*(z+2) + c.
    

Is that right? Why do you teach quadratic equations at this stage? I'd
consider those to be much more advanced than simple "replace x with (z+2)
everywhere you see it" plus some "alice's house to bob's house to carol's
house" problems along a single axis for the concepts.

It seems like you're saying they're taught quadratic equations before they're
equipped to poke around with them, which seems to be setting up for the black
box/memorize-the-formula version of math.

~~~
sykick
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 and
difference of squares. This reinforces those concepts and shouldn't be too
great of a leap at this stage. Also, there is a nice geometry behind the (b/2
+ z)(b/2 - z) idea. The approach is nice precisely because it isn't a
memorization approach. It's an approach that says, "Hey, let's analyze what
factoring trinomials is all about and what the relationship between b, c, r,
and s are.". It says, we know that r and s have to have the property that r+s
is b and rs is c. We know this because of our analysis of multiplying
binomials and our experience with factoring trinomials. We are using patterns
and pattern recognition and from this we are constructing solutions to an
equation that we can't solve by isolating the x like we did with linear
equations. This to me, is true mathematics and it's nice to show students this
exploration. To show how mathematicians think and approach problems. It's not
a black box. It builds upon previous ideas and uses them to solve problems we
weren't able to before.

Note that I'll be using this method in elementary algebra from now on but will
not be using it to prove the quadratic formula. Indeed I will not even tell
them what the quadratic formula is. I will save that for the next class.

~~~
jessaustin
'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...

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

------
spicymaki
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)

------
nestorD
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

------
eyeundersand
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!

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

------
aphextim
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.

------
shikoba
(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

------
prvc
Surely this is a prank?

