Solving Equation for c: Analytical or Numerical Method? | Possible Integer A

[i13m]

Hi, all

Homework Statement

After I differentiate a mathematical model, I got the following equation.

I wonder whether or not it is possible to solve c with the following equation

$$0=c^3\,\left(\ifrac{3\,\left(A^2+c\,A\right)^{\frac{4}{3}}}{c^{\frac{8}{3}}}-\ifrac{24\,\left(A^2+c\,A\right)^{\frac{2}{3}}}{c^{\frac{4}{3}}}+20\right)+c\,\left(-\ifrac{12\,A^2\,\left(A^2+c\,A\right)^{\frac{2}{3}}}{c^{\frac{4}{3}}}-\ifrac{4\,A^2\,\left(A^2+c\,A\right)^{\frac{1}{3}}}{c^{\frac{2}{3}}}\right)+c^5\,\left(\ifrac{72\,\left(A^2+c\,A\right)^{\frac{2}{3}}}{c^{\frac{4}{3}}}-44\right)+c^2\,\left(9\,A-\ifrac{2\,A\,\left(A^2+c\,A\right)^{\frac{1}{3}}}{c^{\frac{2}{3}}}\right)+7\,A^3-40\,c^4\,A$$

where A will be a possible integer

Can it be done analytically, or the numerical method is a possible solution.

Homework Equations

The Attempt at a Solution

Thanks

 

[gb7nash]

Unless stuff magically cancels out when you expand everything, no. However, there's two ways of going about it:

1) If you're left with a polynomial in c with rational coefficients after expanding everything, you could try the rational roots test to find the roots.

2) You could always try an approximation technique: bisection, Newton, Newton–Raphson

Of course both of these options assume that you know what A is. If you don't, I don't know how you would solve for the roots.

 

[Mentallic]

You don't need to know the value of A, other than it's a constant. It's similar to being able to solve for the roots of a general quadratic, without knowing their coefficients. The result will be in terms of its coefficients.

 

[epenguin]

We can't even see it; after each large bracket start another line if you want anyone to look at it please.

 

[i13m]

Thanks for all replies.

I hope the equation can now be broke into two lines.

$$0=c^3\,\left(\ifrac{3\,\left(A^2+c\,A\right)^{\fra c{4}{3}}}{c^{\frac{8}{3}}}-\ifrac{24\,\left(A^2+c\,A\right)^{\frac{2}{3}}}{c^ {\frac{4}{3}}}+20\right)+c\,\left(-\ifrac{12\,A^2\,\left(A^2+c\,A\right)^{\frac{2}{3} }}{c^{\frac{4}{3}}}-\ifrac{4\,A^2\,\left(A^2+c\,A\right)^{\frac{1}{3}} }{c^{\frac{2}{3}}}\right)+\newline c^5\,\left(\ifrac{72\,\l eft(A^2+c\,A\right)^{\frac{2}{3}}}{c^{\frac{4}{3}} }-44\right)+c^2\,\left(9\,A-\ifrac{2\,A\,\left(A^2+c\,A\right)^{\frac{1}{3}}}{ c^{\frac{2}{3}}}\right)+7\,A^3-40\,c^4\,A$$

I have tried and failed at expending this whole equation.

I will have a look at some approximation techniques.

Regards

 

[Mentallic]

Here's the equation:

$$0=c^3\,\left(\ifrac{3\,\left(A^2+c\,A\right)^{\fra c{4}{3}}}{c^{\frac{8}{3}}}-\ifrac{24\,\left(A^2+c\,A\right)^{\frac{2}{3}}}{c^ {\frac{4}{3}}}+20\right)+c\,\left(-\ifrac{12\,A^2\,\left(A^2+c\,A\right)^{\frac{2}{3} }}{c^{\frac{4}{3}}}-\ifrac{4\,A^2\,\left(A^2+c\,A\right)^{\frac{1}{3}} }{c^{\frac{2}{3}}}\right)$$

$$+ c^5\,\left(\ifrac{72\,\l eft(A^2+c\,A\right)^{\frac{2}{3}}}{c^{\frac{4}{3}} }-44\right)+c^2\,\left(9\,A-\ifrac{2\,A\,\left(A^2+c\,A\right)^{\frac{1}{3}}}{ c^{\frac{2}{3}}}\right)+7\,A^3-40\,c^4\,A$$

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