Finding a root of a fractional equation I want to find a root of the following equation
\begin{equation}
\sum_{i=1}^m \frac{p_ia}{(1-xp_i)^2}=1
\end{equation}
where $x$ is variable with constraint $x>0$, $a$ and $p_i$ are some known positive number.
There might be many roots.
Is there any way to solve that equation?(or numerical method)
Any help would be appreciated!
 A: To simplify the problem, let us first change notations? Using $b_i=\frac a {p_i}$ and $c_i=\frac 1 {p_i}$, you want to find the zero's of function
$$f(x)=\sum_{i=1}^m \frac{b_i}{(x-c_i)^2}-1$$ which will show $2m$ roots among which some ( or none) would be real. Assuming, without any loss of generality, that the $c_i$'s are ordered $(c_1<c_2<\cdots<c_{m-1}<c_m)$ and different, there are potentially two roots between two asymptotes.
Suppose that your search for the roots between $c_k$ and $c_{k+1}$, amke the problem more continuous considering in this interval
$$g_k(x)=(x-c_k)^2(x-c_{k+1})^2 f(x)-(x-c_k)^2(x-c_{k+1})^2 $$ that is to say
$$g_k(x)=(x-c_k)^2(x-c_{k+1})^2\sum_{i=1}^{k-1} \frac{b_i}{(x-c_i)^2}+
b_k(x-c_{k+1})^2+$$
$$b_{k+1}(x-c_k)^2+(x-c_k)^2(x-c_{k+1})^2\sum_{i=k+2}^{m} \frac{b_i}{(x-c_i)^2}-(x-c_k)^2(x-c_{k+1})^2$$
At its boundaries
$$g_k(c_k)=b_k(c_{k+1}-c_{k})^2 \qquad \text{and} \qquad g_k(c_{k+1})=b_{k+1}(c_{k+1}-c_{k})^2$$ and you see that , in the range $c_k \leq x \leq c_{k+1}$, if  $b_{k}\,b_{k+1} <0 $,  function $g_k(x)$ looks like a straight line; otherwise  function $g_k(x)$ "looks more or less like a parabola".
If we are in the "nice" case, we can use as an estimate
$$x_0=\frac{b_{k+1}\,c_k- b_k\,c_{k+1}} {b_{k+1}-b_{k} }$$ and Newton method will converge like a charm.
To follow if you wish.
