mcerd1
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| posted on 4/12/12 at 04:08 PM |
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OT - Maths / Excel help
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 !!
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mookaloid
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| posted on 4/12/12 at 04:56 PM |
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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 
"That thing you're thinking - it wont be that."
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Rod Ends
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 posted on 4/12/12 at 05:02 PM |
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try Wolfram
Alpha
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me!
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| posted on 4/12/12 at 06:05 PM |
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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)
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matt_gsxr
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| posted on 4/12/12 at 06:06 PM |
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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)
}}
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mark chandler
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| posted on 4/12/12 at 06:11 PM |
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That's made it much easier 
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jps
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| posted on 4/12/12 at 08:29 PM |
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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)
On a similar vein, surely if a and B are known values you just stick the whole formula into excel but with C referencing a cell - then keep changing
the value in the cell (trial and error) until you hit a solution...?
I know that's not the point of maths, but it'd give you an answer...
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perksy
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| posted on 4/12/12 at 08:32 PM |
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I'll get me coat....
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nick205
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| posted on 4/12/12 at 09:19 PM |
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That must have been one of the bits I missed at school 
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coyoteboy
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| posted on 4/12/12 at 09:39 PM |
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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.
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austin man
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| posted on 4/12/12 at 10:59 PM |
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surely the answer has to be 7
Life is like a bowl of fruit, funny how all the weird looking ones are left alone
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mcerd1
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| posted on 5/12/12 at 08:42 AM |
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cheers guys, wolframalpha.com now bookmarked
one down 12 more to go
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hughpinder
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| posted on 5/12/12 at 10:54 AM |
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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.
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mcerd1
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| posted on 5/12/12 at 11:35 AM |
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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]
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