[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.

[Anon84]

It's phrased in a funny way, but this: ""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" is pretty obvious.

If x+y=B then the average(x+y) = (x+y)/2 = B/2

B/2 is then the number in between x and y so you can represent x and y as B/2 + z and B/2 - z (where z is just half the distance between x and y, or |x-y|/2)