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dbn
Jul12-04, 12:32 AM
if you can solve ax^2-40x+40=0 for a and x that would be great

AKG
Jul12-04, 01:05 AM
No. We have two unknowns (a and x) and one equation. At best, we can express a in terms of x, and vice versa, but no better. To express x in terms of a, simply apply the quadratic formula to the equation you have. You'll get:

x = \frac{20 \pm 2\sqrt{10(10 - a)}}{a}

PRodQuanta
Jul12-04, 01:18 AM
if you can solve ax^2-40x+40=0 for a and x that would be greatuse this formula \frac{-b ^+_- \sqrt{b^2 - 4ac}}{2a}where a=a, b=-40 and c=40

Paden Roder

PRodQuanta
Jul12-04, 01:19 AM
Well, sorry! I just stated the quadratic formula, as AKG said. This was the first time I have used LaTeX though, pretty cool, once you get the hang of it.

Paden Roder

dbn
Jul12-04, 02:40 AM
i still don't get it can some one give me a walk through on how to use them with this problem/

dbn
Jul12-04, 02:53 AM
a= 0 x= 1
i just fount it

phreak
Jul12-04, 03:32 AM
*groan*

I bet there are millions of answers for this one.

I came up with a = 10, x = 2

Zorodius
Jul12-04, 08:19 AM
I bet there are millions of answers for this one.
:uhh:

I bet there's more than that. In fact, I bet the set of all possible answers is uncountably infinite. I'd bet a lot of money on it, since, as AKG said, with one equation and two unknowns, this isn't going to go anywhere.

arildno
Jul12-04, 11:14 AM
i still don't get it can some one give me a walk through on how to use them with this problem/
Now, while the others have shown how to find x given a, here's how you may find the a-value for any choice of x-value:
ax^{2}-40x+40=0
Rewrite this as:
a=40\frac{x-1}{x^{2}}

Hence, for any non-zero choice of x, only a single value for a is allowed by the equation.

KnowledgeIsPower
Jul24-04, 05:12 AM
Is this the whole question?
Unless it's an 'express x in terms of a' type question you can't solve it with only that information.