We're all quadratics too

So let's do it as they do it in the quadratic formula.

Behold:

x = (-b ⁺/_- √b² - 4ac)/(2a)

Now Mathimoto's Complaint aims to cover a wide range of fans of math and today I'd like to reach out to the younger crowd. That precious younger crowd undoubtedly have seen the beauty of the quadric equation. But they and probably some older folks as well have never given an effort at deriving it. Well, I thought I'd give it a shot and show you the awesomeness of figuring out these formulas. Because Math rocks, it really does.

First start out with the generalized form of a basic quadratic equation (ie any equation with 1 variable (let's say x) and some instance of that variable raised to power 2 and possibly some instance of that variable raised to power 1. Okay, so it was harder to describe things rather than write it out, so let's do that)

Let a, b, and c be constants and x be a variable.

A basic general quadratic equation is:

ax² + bx + c = 0

Now it helps then to know one particular quadratic equation, that is what happens when you have

(x + b)² = 0

this can be expanded to

x² + 2bx + b² = 0

(Don't believe me, just use the distributive property of multiplication, ie,

(x+b)² = (x + b)(x + b) =

x (x + b) + b (x + b) = x² + bx + bx + b² =

x² + 2bx + b²

Ta-da)

Okay, now if you got a quadratic equation of the form

x² + 2bx + b² = 0

And you know

(x + b)² = x² + 2bx + b²

You can then say

(x + b)² = 0

and with an equation like that, the only time you have a number that can match the value of x (and still give you 0, ie, the solution of the equation) is

x = -b

Back to the general basic quadratic equation:

ax² + bx + c = 0

At this point we don't know how to find the solution value of x here, but since we know the solution for

(x + b)²

we can reconfigure our general equation to fit our particular equation. Just follow along.

If

ax² + bx + c = 0

then we can play around with this, our goal equation doesn't have a c, so let's just subtract it from both sides.

ax² + bx = -c

Well, our target equation doesn't have an &lsquot;a&rsquot; so let's get rid of the a by dividing it from both sides.

x² + ^bx/_a = ^-c/_a

Okay let's remember our target equation (or the essentials of it)

(x + b)² = x² + 2bx + b²

now let's make a little pretending. Let's say instead that the b in our target equation is really say some other letter, say d. Then:

(x + d)² = x² + 2dx + d²

And if the current state of our manipulation of the general equation is:

x² + ^bx/_a = ^-c/_a

we can get to our target a little easier if we say that

d = ^b/_2a

alright now let's take it up a notch by throwing in the new d, then we get

x² + 2dx = ^-c/_a

well now all we need is the d² and we can just add that to both sides, so:

x² + 2dx + d²= ^-c/_a + d²

Well, we can use our old target equation:

(x + d)² = x² + 2dx + d²

to simplify this:

(x + d)²= ^-c/_a + d²

Now remember, with a situation like this, the name of the game is find the x, and currently our x is trapped in a term that's raised to a power, so let's get rid of that with a little friend called the square root (but remember that with real square roots you have a positive root and a negative root, since the negative goes away from the squaring).

x + d = ⁺/_- √d² - ^c/_a

Now, now, now, we can FIND THE X (by subtracting d from each side)

x = -d ⁺/_- √d² - ^c/_a

now just one more step to define the x in the a, b, c constants we started out with, just reverse the d insertion with our

d = ^b/_2a

and we can get...

x = ^-b/_2a ⁺/_- √(^b/_2a)² - ^c/_a

so there we go, we've found the x, but it's kind of ugly so let's simplify things a little by doing some expansion and some common denominator and essentially simple algebra which I'm going to skim over a little:

x = (-b ⁺/_- √b² - 4ac)/(2a)

And there we go, we've got the quadratic formula! Yaaaah!!!! Behold it and be amazed!!!

x = (-b ⁺/_- √b² - 4ac)/(2a)

MATH RULES!!!! WOOOOO!!!!