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How to solve a nonlinear equation?

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CS on 20 Oct 2020
Commented: Matt J on 21 Oct 2020
I have an equation as follows
How can I solve for x?
Thanks for any help!


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Answers (1)

Matt J
Matt J on 20 Oct 2020
Edited: Matt J on 20 Oct 2020
[x,fval] = fzero( @(x) x^(8.5)+3*x.^2-3000,nthroot(3000,8.5))
x = 2.5629
fval = 4.5475e-13


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CS on 21 Oct 2020
How about the below one?
[x,fval] = fzero( @(x) ((1/(3.52*10.^(22)))*x^(8.14))+(1/207000)*x.^2-4.52,0)
CS on 21 Oct 2020
It gives the error
Exiting fzero: aborting search for an interval containing a sign change
because complex function value encountered during search.
(Function value at -0.0282843 is -4.52+3.0076e-36i.)
Check function or try again with a different starting value.
How to solve this?
Matt J
Matt J on 21 Oct 2020
[x,fval] = fzero( @(x) ((1/(3.52*10.^(22)))*abs(x)^(8.14))+(1/207000)*x.^2-4.52,0)
x = -656.6949
fval = -8.8818e-16

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