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07 July 2009 @ 08:51 am
Don't ask me what, ask me how!  
So I has a maths placement test on Thursday. Yay. Taught myself quadratic equations, linear equations, inequalities... and uh... some other shit. Fuck me if I remember anything. I'm still just a little kid when it comes to math. Do this? Oh ok. Why? I don't know, you told me to, that's why.

But anyway, all I have to know is that x equals negative b plus or minus the square root of b squared minus 4ac over 2a. What that does is fuck all to me, but it works, and I can do it.

Simplifying complex rational fractions or some shit? Yeah, I'll get back to you on that.

Simplify (x^2+3x-2/x-2) - (3x-1/x+2)

Right... simplify... hrmm... ok...

(x^2+3x-2/x-2) - (3x-1/x2)

=zero.

There... much simpler!
 
 
( 8 comments — Post a new comment )
Tim[info]fb_sheex on July 7th, 2009 03:54 pm (UTC)
...x equals negative b plus or minus the square root of b squared minus 4ac [all] over 2a.

Fuck yeah!
Arturis[info]arturis on July 7th, 2009 05:36 pm (UTC)
When you have something that fits the form y = ax^2 + bx + c, it can instead be rewritten into the form y = n(x + m)(x + o) where n, m, and o are derived from a, b, and c. The value of doing this is that, through the vagaries of math, -m and -o are the two points where the equation intercepts the x-axis. Why you care about that usually depends on what the equation represents, which in basic math is usually nothing. But equations of that form crop up in a lot of places and can represent lots of real-world phenomena, and it's a generally useful skill in math to be able to find the x-axis intercepts of a parabola.

So -m and -o are x-intercepts because if y=0 (which is necessarily true if we're talking about something being on the x-axis) then either x + m = 0 or x + o = 0. So finding m and o is equivalent to setting y=0 and solving the original equation for x. You can try to do that yourself and you'll quickly get frustrated.

0 = ax^2 + bx + c
c = x(ax + b)
...

Anyway, four pages of clever algebra later leads you to the equation you memorized, the quadratic equation. If you want to find the x-intercepts of a parabola, which is something that winds up being way more common in math (and physics) than you might think, the quadratic equation is the generic answer that it would otherwise take a lot of effort to get on your own.
Lover of Horses[info]meus_ovatio on July 7th, 2009 10:15 pm (UTC)
I'm a man with a loaded gun and I don't know what I'm doing! Watch out everybody.

Haha no, not totally, but factoring isn't something I can teach on my own with internets help. Real basic stuff and FOIL is all old stuff that comes back, but yeah... all that work when there is just a tool for it... bah I say!
Arturis[info]arturis on July 7th, 2009 10:52 pm (UTC)
That's exactly my point. I'm someone who earned a math minor, studied differential equations and took at least one horrible class on rigorous proofs. I've tried to derive the quadratic equation and failed. It's a lot of tricky algebra that isn't worth anyone's time when you can just memorize the damn thing and plug in some variables. But having struggled with it, I appreciate how valuable a tool it really is to just be given.
Derek[info]dichotomy2 on July 8th, 2009 05:15 pm (UTC)
We all stand on the shoulders of giants.

I think I would've been a much better math student if it was explained to me at the time what some of the various practical uses for these things were. Having just done that above, you greatly increased my appreciation for trinomial algebra.
Derek[info]dichotomy2 on July 8th, 2009 05:17 pm (UTC)
In other words, I think you'd make an excellent math and/or science teacher, because unlike many teachers, you often bring out the practical side of things in your explanations in a way that engages your students.
Arturis[info]arturis on July 8th, 2009 07:35 pm (UTC)
Thanks!
All Cars Eat Gas[info]voxbaryton on July 7th, 2009 11:50 pm (UTC)
www.purplemath.com
 
 

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