Can this equation be numerically solved using online resources or Matlab code?

[ColdFusion85]

I need to use a program to numerically solve the following equation:

$$exp(-2.222)=\frac{(\frac{2x}{2+2x})^2(\frac{x}{2+2x})}{(\frac{2-x}{2+2x})^2}$$

I tried using Matlab to solve it but I kept on getting an error message. I'm not a programmer or familiar with these numerical solvers at all. Is there some place online I can solve this equation, or can someone post the Matlab code to solve it? The code is not part of the homework, but I arrived at the above equation and need to solve it in order to finalize my answer, so if anyone could help me solve it that would be great. Thanks.

 

[Avodyne]

Well, you can simplify the right-hand side quite a bit. Ultimately you can turn it into a cubic equation for x.

 

[D H]

First, simplify the expression. Then reformulate it in the form $f(x)=0$.

You now have something amenable to the zero finding functions.

$x=2$ and $x=-1$ are obviously problematic. You want to find a pair of $x$ values $x_1$ and $x_2$ such that neither of the poles is between $x_1$ and $x_2$ and such that $f(x_1)$ and $f(x_2)$ have opposite signs. There is a zero between these points.

 

[D H]

Well, you can simplify the right-hand side quite a bit. Ultimately you can turn it into a cubic equation for x.

Not quite that much. You get a rational polynomial, a simple cubic divided by a quadratic.

EDIT: I worked to much today.

Ultimately, you're right. Simplifying the RHS leads to $p(x)/q(x)-c=0$ where $p(x)$is cubic and $q(x)$ is quadratic. Multiplying by $q(x)$ yields $p(x)-cq(x)=0$, a cubic.

 

[ColdFusion85]

For the RHS I expanded everything out and got this:

$$\frac{4x^3(4x^2+8x+4)}{(8x^3+24x^2+24x+8)(x^2-4x+4)}$$

Is this correct?

 

[D H]

Don't expand! Get rid of the common factor instead. Look at the numerator and denominator.

 

[ColdFusion85]

Oh, ok so one gets:

$$\frac{4x^3}{(2-x)^2(2+2x)}$$?

 

[D H]

Yes. You can also get rid of a factor of two. And oops, I said cubic/quadratic. What you have is a ratio of two cubics, which is correct. Now multiply through by the denominator to get a cubic equation (valid so long as x is not 2 or -1).