# Quadratic equation solution (it's actually a nice trick)

Po-shen loh's "A Simple Proof of the Quadratic Formula" was thrown at me few times last week. I read it through because I've explained solving of quadratic equation through formal logic some time ago. I don't remember if I've seen Po-shen's approach before, but it's actually a quite nice and a simple trick.

You've likely been asked to solve `x² + bx + c = 0` and you know it's got at most two roots. If you rewrite it in a different form you'll also see it:

``(x-s)(x-u) = x² - (s+u)x + s*u = 0``

The equation `(x-s)*(x-u) = 0` is satisfied if either `x-s` or `x-u` reaches `0`. That way you know there are two solutions `x=s` or `x=u`.

You know that you can solve this through factorization or by completing a square.

Po-shen Loh proposes that you can take the midpoint of the roots and use that as help. The midpoint of `s` and `u` is the `(s+u)/2`. Through polynomial factorization you know that `-b = (s+u)` and `c = s*u`, therefore you know that the midpoint of roots is `-b/2`.

Now you can introduce a variable `z` and represent `s` and `u` with their midpoint: `s = -b/2 + z`, `u = -b/2 - z`.

If you then substitute the midpoint-representations into the `c = s*u`, you get `c = -b²/4 - z²`, and you can use this equation to solve `z`. If you fiddle it around a bit you get the quadratic formula.

I guess you could think that the introduction of a "midpoint" is just a variation of the idea that the square root gives `|x|` and you get an another solution by negating its result.

I like this because it's simple and straightforward enough way to do it and resembles how you can directly solve linear equations `ax + b = 0`.

Presentation is a bit unusual and something I shy from because this is usually how you present something if you want to coinvince people without proper argumentation. But this time there was proper argumentation and it was just posh presentation. If this was pop-sci in the future I'd take it over the usual pop-sci!