Question

Someone asked me to do this earlier. I figured I would post it....

Answer 1

\(ax^2+bx+c=0\\
ax^2+bx+c+\frac{b^2}{a2^2}-\frac{b^2}{a2^2}=0
\\ax^2+bx+\frac{b^2}{a2^2}+c-\frac{b^2}{a2^2}=0
\\ax^2+\frac{b}{2}x+\frac{b}{2}x+\frac{b^2}{2^2a}+c-\frac{b^2}{2^2a}=0\\
ax^2+\frac{bx}{2}+\frac{bx}{2}+\frac{b^2}{2^2a}+c-\frac{b^2}{2^2a}=0\\
a(x^2+\frac{bx}{2a}+\frac{bx}{2a}+\frac{b^2}{2^2a^2})+c-\frac{b^2}{2^2a}=0\\
a(x+\frac{b}{2a})^2+c-\frac{b^2}{2^2a}=0\\
a(x+\frac{b}{2a})^2=\frac{b^2}{2^2a}-c\\
a(x+\frac{b}{2a})^2=\frac{b^2}{2^2a}-\frac{c2^2a}{2^2a} \\
a(x+\frac{b}{2a})^2=\frac{b^2-4ac}{2^2a}\\
(x+\frac{b}{2a})^2=\frac{b^2-4ac}{2^2a^2}\\
(x+\frac{b}{2a})=\pm \sqrt{\frac{b^2-4ac}{2^2a^2}}\\
(x+\frac{b}{2a})=\pm \sqrt{\frac{b^2-4ac}{2^2a^2}}\\
(x+\frac{b}{2a})=\pm \frac{\sqrt{b^2-4ac}}{2a}\\
x=\pm \frac{\sqrt{b^2-4ac}}{2a}-\frac{b}{2a}\\
x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\





\)

Answer 2

This seems complicated, I don't remember this taking quite so many steps... #_#

Answer 3

I figure I would try and take a very obvious rout.

Answer 4

Yeah I've done this before to, and I didn't have that many steps either I don't think :P

Answer 5

Take my medal zz!

Answer 6

good because it was not for anyone of you lol

Answer 7

I assume you want to derive the Quadratic Equation?

Answer 8

I thought I did....

Answer 9

just adding 0

Answer 10

Okay, well anyway, here is how I derive it:
http://www.1728.org/quadr2.htm
and as iambatman said it does not require that many steps.

Answer 11

ffs this is not a competition. Go outside this is not for you if you know how to do it...

Answer 12

this place is silly. I am not trying to show you what I can do @wolf1728 I

it is for people to learn. The quickest rout is not the best always.

Answer 13

Hey man, wolf didn't mean any harm, I think your way is great :), and seems very fun actually.

Answer 14

If you read the steps you posted there is no justification for WHY they do it....I decided not to go through that rout and not treat it like magic.

Answer 15

@Jhannybean is doing phd in completing the square, she might also enjoy this derivation :)

few other methods to derive the same
http://www.pballew.net/quadsol.pdf

Answer 16

Thanks iambatman. Yes, I do not mean any harm whatsoever.
And zzr0ck3r if this is a place for people to learn, learning requires at least 1 person in a discussion to actually know how something is done doesn't it?
Well, to use your vernacular, I guess I will go outside.

Answer 17

I post something at with in 10 seconds there are 3 different people telling me there are better and faster ways to do it. This will not help others learn what I wanted to show. If I wanted to google how to derive something I am sure I could. There is a reason I do it like I did. Did you actually try and debate anything besides it could be shorter? You tried to get no discussion, you asked no why? you did nothing but say it could be shorter. This is, in my opinion, the reason that people don't learn.

Imagine someone comes in here to read what I posted, well this other dude said there is a shorter way..... and some other dude says yeah I did that to once..... I don't know man I find it all really silly and this is the reason I don't come here that much anymore.