General solution for $x$ of $C = 100/(1+aX) + 100/(1+bX)+ \cdots + 100/(1+zX)$

Question

Please can someone help me find a general solution for $X$:

$$ C = \frac{100}{(1+aX)} + \frac{100}{(1+bX)}+ \cdots + \frac{100}{(1+zX)} $$

UPDATE

Its not ideal but if we make $C = 350$ would this help?

Answer 1

There is no general solution in radicals for $X$.

For ease of notation define $\frac{C}{100}=D$ and $a=a_1, b=a_2, \ldots , z=a_{26}$, so that $D=\sum\limits_{i=1}^{26}{\frac{1}{1+a_i X}}$.

Clearing denominators gives $D \prod\limits_{i=1}^{26}{(1+a_i X)}$ $=\sum\limits_{i=1}^{26}{\prod\limits_{1 \leq j \leq 26, j \not = i}{(1+a_j X)}}$

This reduces to $D \sum\limits_{i=0}^{26}{s_i(a_1, a_2, \ldots , a_{26})X^i}=\sum\limits_{i=0}^{26}{(26-i)s_i(a_1, a_2, \ldots , a_{26})X^i}$, where $s_i$ is the ith elementary symmetric polynomial in $a_1, a_2, \ldots , a_{26}$, so $\sum\limits_{i=0}^{26}{(D-(26-i))s_i(a_1, a_2, \ldots , a_{26}) X^i}=0$.

Note: Above, the following result was implicitly used: $\sum\limits_{j=1}^{n}{s_i(a_j, 1 \leq k \leq n, k \not =j)}=(n-i)s_i(a_1, a_2, \ldots , a_n)$.

Now, $X$ is a root of some general polynomial with degree $26$. It is well known (Abel's impossibility theorem) that for a general polynomial with degree $\geq 5$, there is no general algebraic solution. Thus there is no general algebraic solution for $X$.

For example, when we have 3 fractions, i.e. $D=\frac{1}{1+aX}+\frac{1}{1+bX}+\frac{1}{1+cX}$, this becomes $DabcX^3+(D-1)(ab+ac+bc)X^2+(D-2)(a+b+c)X+(D-3)=0$. It is possible to get the general solution in terms of radicals for up to 4 fractions.