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## #1 2006-05-10 16:47:05

Lukie
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### Urgent Parabola Help!~!

I need to find "a" when ax² + 0x + 185
Luke

## #2 2006-05-10 16:55:18

kempos
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### Re: Urgent Parabola Help!~!

when ax^2 + 0x + 185 what?

## #3 2006-05-10 17:05:43

Lukie
Guest

### Re: Urgent Parabola Help!~!

ax^2 + bx + c = 0 is the normal parabolic eqaution.

i need to find "a" when b = 0 and c = 185

## #4 2006-05-10 17:11:51

kempos
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### Re: Urgent Parabola Help!~!

sorry, i dont get it :-(

maybe i'm just to tired :-)

## #5 2006-05-11 22:04:15

MATHSKITZO
Member

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### Re: Urgent Parabola Help!~!

I don't get it either

Last edited by MATHSKITZO (2006-05-11 22:07:41)

'Math is not a means to arrive at truth, Math IS the only truth!'

## #6 2006-05-11 22:40:44

MathsIsFun

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### Re: Urgent Parabola Help!~!

ax² + 0x + 185 = 0
ax² + 185 = 0
ax² = -185
a = -185/x²

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

## #7 2006-05-12 12:04:26

George,Y
Super Member

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### Re: Urgent Parabola Help!~!

You should find another constraint to determine a.
3 parameters a, b and c, at least need 2 constraints to determine, according to my experience.

X'(y-Xβ)=0

## #8 2006-05-13 00:53:11

Ricky
Moderator

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### Re: Urgent Parabola Help!~!

Right.  a can be anything you want it.  For example:

4x² + 0x + 185 = 0
104x² + 0x + 185 = 0

Are both perfectly fine equations.  In fact, anything that you want to put in for 'a' works.  So you have an infinite amount of solutions.  Either this is the answer, or there is something about the question you are missing or not understanding.  Could you post the entire question here?

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."