1. ## A nonlinear problem

I have been trying to find a solution to the following problem.

x(4/3) - x(-2/3) + c = 0 where -1 < x > 1

I think if I could solve,

s(-2) - s + c = 0,

I could use substitution to solve the first equation, but I have not had any luck with the second either.

Any help would be appreciated.

2. ## Re: A nonlinear problem

Hello, markel!

Is there a typo?
You have a positive exponent and a negative exponent.
Even if the signs were alike, that "c" makes for a messy solution.

$x^{\frac{4}{3}} -x^{-\frac{2}{3}} - c \:=\:0\;\;\;\text{ for }0 \,\le\,x\,<\,1$

I will wait for your clarification.

3. ## Re: A nonlinear problem

Dear Soroban,

Thanks for responding. I'm afraid the negative exponent is correct. The limit however should be -1 < x < 1. I have used an Excel spreadsheet to calculate some examples by trial and error and the formula seems to be working. The task of simplifying the formula, however, has eluded me.

4. ## Re: A nonlinear problem

Originally Posted by markel
I have been trying to find a solution to the following problem.

x(4/3) - x(-2/3) + c = 0 where -1 < x > 1

I think if I could solve,

s(-2) - s + c = 0,

I could use substitution to solve the first equation, but I have not had any luck with the second either.

Any help would be appreciated.
\displaystyle \begin{align*} x^{\frac{4}{3}} - x^{-\frac{2}{3}} + c &= 0 \\ x^{\frac{2}{3}}\left( x^{\frac{4}{3}} - x^{-\frac{2}{3}} + c \right) &= 0x^{\frac{2}{3}} \\ x^{\frac{6}{3}} - 1 + c\,x^{\frac{2}{3}} &= 0 \\ X^3 + c\,X - 1 &= 0 \textrm{ if } X = x^{\frac{2}{3}} \end{align*}

While there is a solution to this cubic, it is NOT pretty. I suggest you read the attachment.