How to solve a nonlinear equation?

I have an equation as follows

x^(8.5)+3*x^(2)=3000

How can I solve for x?

Thanks for any help!

Answer 1

[Matt J]

[x,fval] = fzero( @(x) x^(8.5)+3*x.^2-3000,nthroot(3000,8.5))

x = 2.5629

fval = 4.5475e-13

[CS]

Thanks!

What is the function of nthroot(3000,9)?

[Matt J]

You're welcome. Please Accept-click if you are satisfied with the answer.

nthroot(3000,9) is the initial guess provided to fzero. With such a large exponent, we can expect the left hand side of the equation to be approximately x^9.

[CS]

And what is fval indicating?

[Matt J]

It is the value of the function at the point found by fzero. You use it to see if the point is approximately a root. Equivalently, you could do,

fun=@(x) x^(8.5)+3*x.^2-3000;

x = fzero( fun,nthroot(3000,8.5)),

x = 2.5629

fval=fun(x)

fval = 4.5475e-13

[CS]

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]

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]

[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