Can anyone advise how this nonlinear equation with a single variable $x$ can be solved as a closed form? ${\left(\frac{x}{1-x}\right)}^2.{\left(\frac{x-C}{\left(1-x\right)-N}\right)}^2=H.\frac{2x-C}{2\left(1-x\right)-N}$ where $C$, $N$ and $H$ are real and positive values. I only know that $0<x<1$.
Please also guide me on what the best way to solve it computationally is.
I cannot speak to a closed form solution, but I can address the possibility of a numeric approach. To solve the equation, we will look for a root in $(0, 1)$ of the function $$f(x) = {\left(\frac{x}{1-x}\right)}^2 \cdot {\left(\frac{x-C}{\left(1-x\right)-N}\right)}^2 - H\cdot \frac{2x-C}{2\left(1-x\right)-N}.$$
First note that $f(x)\rightarrow \infty$ as $x\rightarrow 1,$ so you will need to bound the upper limit of the search interval for $x$ strictly below 1. Second, there may be more than one root of $f$ in the unit interval.
I did some quick numerical experiments, and it appears that $f$ can a bit volatile (very steep in places), making numeric methods a bit tricky. You could, for instance, try Newton's method, starting from some value inside the unit interval (maybe 0.5, maybe a different value), either using an explicit formula for the derivative $f^\prime(x)$ ($f$ is not too hard to differentiate) or using numerical approximations for the derivative. If Newton's method fails to converge to a root, you can retry with a different starting point. Depending on what your preferred programming language is, you may be able to find third party libraries that support Newton (also known as Newton-Raphson) or other root finding methods.
A lower tech, fairly easy approach to program is bisection search. Start by defining a sequence of grid points $x_0 = 0 < x_1 < x_2 < \dots < x_n < 1$ and evaluate $f(x_i)$ for each $i$. Find an index $i$ such that $f$ changes sign between $x_i$ and $x_{i+1}.$ Evaluate $f$ at the midpoint of the interval $[x_i, x_{i + 1}],$ replace the endpoint having the same sign for $f$ as the midpoint with the midpoint, and repeat until the width of the interval is small enough that you will accept the last midpoint as your estimate of the root of $f.$
Using Mathematica or Wolfram Engine, you could do something like this. Hope this helps.
In[11]:= NSolve[{((C - x)^2*x^2)/((-1 + x)^2*(-1 + N + x)^2) == (H*(C - 2*x))/(-2 + N + 2*x), x > 0, x < 1, N == 123, H
== 12, C == 321}]
Out[11]= {{C -> 321., H -> 12., N -> 123., x -> 0.681967}}
```
