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#1
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y=f(x)
is it possible to solve: 0,5= x^2/(x-2)(3-0,5x)?
how can I do it? |
#2
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y=f(x)
Use algebra, not Excel:
x = (-b +/-SQRT(b^2-4*a*c))/2a Simplify your formula into the form ax^2 + bx + c = 0 You'll find there are no real roots..... HTH, Bernie "bbombel" wrote in message ... is it possible to solve: 0,5= x^2/(x-2)(3-0,5x)? how can I do it? |
#3
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y=f(x)
Tushar Mehta has a good add-in that solves quadratics,
Once you've done what Bernie suggested (rearrange into a quadratic) you can use the add-in to find the roots (yes it deals with imaginary numbers...) Check it out at http://www.tushar-mehta.com Under "Solve Polynomials" http://www.tushar-mehta.com/excel/so...als/index.html Dan E "bbombel" wrote in message ... is it possible to solve: 0,5= x^2/(x-2)(3-0,5x)? how can I do it? |
#4
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y=f(x)
Just to mention. Excel would consider the following
0.5 = x^2/(x-2)(3-0.5x) as 0.5 = ( (x^2) / (x - 2) ) * (3 - 0.5*x) with a real solution at 5.887850847558446 You may have meant to group the last two items as in the following = (x^2) / ( (x - 2) * (3 - 0.5*x) ) If this is the case, you would get two imaginary solutions as Bernie mentioned.. 0.05607457622077704 - 0.580119090916421*I 0.05607457622077704 + 0.580119090916421*I -- Dana DeLouis Using Windows XP & Office XP = = = = = = = = = = = = = = = = = "bbombel" wrote in message ... is it possible to solve: 0,5= x^2/(x-2)(3-0,5x)? how can I do it? |
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