I'm in a bit of a hurry here and it would really help me out if I could find a way of solving this for the real solution of this horrible thing using excel:

a^2 *b^3 + a*(2+c)*b^2 + (1 + 2c)*b - c*(2 + a) = 0

solving for b
a & c are known varibles


help !!

There was a time when I could have done that stuff - but it was over 30 years ago and I can't remember any of it now

try Wolfram Alpha

What are a and c? Should be easy to do by plotting it against a range of values for b, where the line crosses 0 are your answers (it looks like a cubic so there could be 3 solutions)

Mathematica says:

{
{b = -((2 a + a c)/(3 a^2)) - (2^(1/3) (-a^2 + 2 a^2 c - a^2 c^2))/(
3 a^2 (2 a^3 + 21 a^3 c + 54 a^4 c + 27 a^5 c + 6 a^3 c^2 -
2 a^3 c^3 + Sqrt[
4 (-a^2 + 2 a^2 c - a^2 c^2)^3 + (2 a^3 + 21 a^3 c +
54 a^4 c + 27 a^5 c + 6 a^3 c^2 - 2 a^3 c^3)^2])^(
1/3)) + (2 a^3 + 21 a^3 c + 54 a^4 c + 27 a^5 c + 6 a^3 c^2 -
2 a^3 c^3 + Sqrt[
4 (-a^2 + 2 a^2 c - a^2 c^2)^3 + (2 a^3 + 21 a^3 c + 54 a^4 c +
27 a^5 c + 6 a^3 c^2 - 2 a^3 c^3)^2])^(1/3)/(
3 2^(1/3) a^2)},

{b = -((2 a + a c)/(
3 a^2)) + ((1 + I Sqrt[3]) (-a^2 + 2 a^2 c - a^2 c^2))/(3 2^(2/3)
a^2 (2 a^3 + 21 a^3 c + 54 a^4 c + 27 a^5 c + 6 a^3 c^2 -
2 a^3 c^3 + Sqrt[
4 (-a^2 + 2 a^2 c - a^2 c^2)^3 + (2 a^3 + 21 a^3 c +
54 a^4 c + 27 a^5 c + 6 a^3 c^2 - 2 a^3 c^3)^2])^(
1/3)) - (1/(
6 2^(1/3)
a^2))(1 - I Sqrt[3]) (2 a^3 + 21 a^3 c + 54 a^4 c + 27 a^5 c +
6 a^3 c^2 - 2 a^3 c^3 + Sqrt[
4 (-a^2 + 2 a^2 c - a^2 c^2)^3 + (2 a^3 + 21 a^3 c +
54 a^4 c + 27 a^5 c + 6 a^3 c^2 - 2 a^3 c^3)^2])^(
1/3)},

{b = -((2 a + a c)/(
3 a^2)) + ((1 - I Sqrt[3]) (-a^2 + 2 a^2 c - a^2 c^2))/(3 2^(2/3)
a^2 (2 a^3 + 21 a^3 c + 54 a^4 c + 27 a^5 c + 6 a^3 c^2 -
2 a^3 c^3 + Sqrt[
4 (-a^2 + 2 a^2 c - a^2 c^2)^3 + (2 a^3 + 21 a^3 c +
54 a^4 c + 27 a^5 c + 6 a^3 c^2 - 2 a^3 c^3)^2])^(
1/3)) - (1/(
6 2^(1/3)
a^2))(1 + I Sqrt[3]) (2 a^3 + 21 a^3 c + 54 a^4 c + 27 a^5 c +
6 a^3 c^2 - 2 a^3 c^3 + Sqrt[
4 (-a^2 + 2 a^2 c - a^2 c^2)^3 + (2 a^3 + 21 a^3 c +
54 a^4 c + 27 a^5 c + 6 a^3 c^2 - 2 a^3 c^3)^2])^(1/3)
}}

That's made it much easier

quote:

---

Originally posted by me!
What are a and c? Should be easy to do by plotting it against a range of values for b, where the line crosses 0 are your answers (it looks like a cubic so there could be 3 solutions)

---

I'll get me coat....

That must have been one of the bits I missed at school

Yeah that's unpleasant, just drop it into wolframalpha.com like this, obviously fill in A and B:

a^2 *b^3 + a*(2+c)*b^2 + (1 + 2c)*b - c*(2 + a) = 0 solve for b where a= and b=

Or leave a and b alone and copy the beast of a real solution into excel.

surely the answer has to be 7

cheers guys, wolframalpha.com now bookmarked

one down 12 more to go

Just for reference, excel has a feaure called 'solver'- press F1 and type solver - you have to install it as an option, but ut will already be on your system so no internet access reqired, but then it allows you to enter equations in a cell, specify where your data is and the constraints and it gives you an answer. - its good for stuff like working out the best mis of work for maximum profit etc. Last time I showed someone how to use it they picked it up pretty easily. I'm sure there are some good examples on the web. I'm pretty sure it doesn't do imaginary numbers though, so pretty limited for advanced stuff.

never had much luck with excel's solver (needs to work in the old 97-03 format, which is a bit more limited I think)


this is actually a small part of calculating the effect of short circuits on high voltage overhead lines - basically catenary functions mixed with magnetic fields and a fair bit of damped harmonic motion

[Edited on 5/12/2012 by mcerd1]