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Guru
Mar 15th, 2007, 05:45 AM
When solving an equation in the form

http://mathworld.wolfram.com/images/equations/QuadraticFormula/equation1.gif

using:

http://mathworld.wolfram.com/images/equations/QuadraticFormula/equation2.gif

we get two answers. Agreed?

Now, what I want is a way of creating an equation in the form

http://mathworld.wolfram.com/images/equations/QuadraticFormula/equation1.gif
from two values of x

eg x=14.468 , x=52.583

Any suggestions?

VBAhack
Mar 15th, 2007, 11:25 AM
You have 2 equations in 3 unknowns, so there are potentially infinitely many solutions. If you are willing to choose c arbitrarily, then you can find a solution by straightforward substitution because the problem reduces to 2 equations in 2 unknowns.

a = c/(x1*x2)

b = -c/x2 - a*x2 = -c(1/x2 + 1/x1)

:)

Logophobic
Mar 15th, 2007, 04:30 PM
eg x=14.468, x=52.583
(x - 14.468)(x - 52.583) = 0
x2 - 67.051*x + 760.770844 = 0

If you want 'a' to be anything other than 1, simply multiply through by whatever value you want 'a' to be.

Guru
Mar 16th, 2007, 04:02 AM
:thumb:

That's perfect!

Thanks