| Author | Comment |
Barrett
Administrator
Posted: 5 Feb 2007
16:15 GMT
Total Posts: 1676 | Can someone do me a favor and solve 0=Ax^2+Bx+C for X?
I remember my math teacher doing it for me my freshman year of high school but it's been like 4 years since I've taken any math classes... this is something I feel that I should know...
(Of course, I know the answer, I just want to see the work).
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-Barrett A |
threefingeredguy
Ghost
Posted: 18 Feb 2007
07:22 GMT
Total Posts: 1189 | x = (-b + sqrt(b^2 - 4ac))/(2a)
and x = (-b - sqrt(b^2 - 4ac))/(2a)
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Someone call for an exterminator? |
Barrett
Administrator
Posted: 22 Feb 2007
21:03 GMT
Total Posts: 1676 | alright, but what about the work?
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-Barrett A |
threefingeredguy
Ghost
Posted: 23 Feb 2007
09:57 GMT
Total Posts: 1189 | That basically is the work. Just follow the steps on a piece of paper.
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Someone call for an exterminator? |
Barrett
Administrator
Posted: 4 Mar 2007
09:50 GMT
Total Posts: 1676 | and... what are the steps?
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-Barrett A |
banjo2E
Wraith
Posted: 5 Mar 2007
13:31 GMT
Total Posts: 689 | The quadratic formula is:
X= -b ± sqrt(b^2 - 4ac)
____________________
2a
There really isn't any work.
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Barrett
Administrator
Posted: 5 Mar 2007
13:32 GMT
Total Posts: 1676 | um... yes... there is work... that formula is derived from the equation that I wrote up on top, 0=Ax^2+Bx+C
It's simply solved for x. There ARE several steps (well, there are enough steps that I can't do the work in my head).
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-Barrett A |
banjo2E
Wraith
Posted: 5 Mar 2007
13:40 GMT
Total Posts: 689 | Ohhh...now I get what you're trying to do: prove the Formula. Um...try Wikipedia. I can't do it off the top of my head and I'm on the computer illegal-ish.
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Barrett
Administrator
Posted: 5 Mar 2007
13:44 GMT
Total Posts: 1676 | good idea... found it... http://en.wikipedia.org/wiki/Quadratic_formula#Derivation
unfortunately I'll need to brush up on my math skills before I understand what the crap they're talking about.... it's been WAY too long.
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-Barrett A |
banjo2E
Wraith
Posted: 6 Mar 2007
16:48 GMT
Total Posts: 689 | My unedited first reaction to that link:
ZOMG o.0 WTF?!?!?
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Zachary940
Wraith
Posted: 7 Mar 2007
15:52 GMT
Total Posts: 714 | Barret that link is way too confussing. If I had the time I would show you but unfortinutly I'm at work. :)
All you need to do is solve for X. Start out with dividing every thing by a. At the end you should come up with the quadratic forumula equalling x.
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It is much easier to suggest solutions when you know nothing about the problem. |
