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if you can solve ax^2-40x+40=0 for a and x that would be great

- Thread starter dbn
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- #1

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if you can solve ax^2-40x+40=0 for a and x that would be great

- #2

AKG

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[tex]x = \frac{20 \pm 2\sqrt{10(10 - a)}}{a}[/tex]

- #3

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use this formula [tex]\frac{-b ^+_- \sqrt{b^2 - 4ac}}{2a}[/tex]where a=a, b=-40 and c=40if you can solve ax^2-40x+40=0 for a and x that would be great

Paden Roder

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Paden Roder

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i still don't get it can some one give me a walk through on how to use them with this problem/

- #6

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a= 0 x= 1

i just fount it

i just fount it

- #7

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*groan*

I bet there are millions of answers for this one.

I came up with a = 10, x = 2

I bet there are millions of answers for this one.

I came up with a = 10, x = 2

- #8

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:uhh:phreak said:I bet there are millions of answers for this one.

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.

- #9

arildno

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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:dbn said:i still don't get it can some one give me a walk through on how to use them with this problem/

[tex]ax^{2}-40x+40=0[/tex]

Rewrite this as:

[tex]a=40\frac{x-1}{x^{2}}[/tex]

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

- #10

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Unless it's an 'express x in terms of a' type question you can't solve it with only that information.

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