A nonlinear problem

**markel** · September 2nd 2012, 10:01 AM

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.

**Soroban** · September 2nd 2012, 12:26 PM

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.

**markel** · September 2nd 2012, 12:56 PM

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.

**Prove It** · September 2nd 2012, 07:50 PM

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.

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