# Roots of A Non-linear Equation

I have the following non-linear equation $$b_1\left(\frac{1}{f_1^2}-\frac{1}{(f_1-a_1)^2}\right)=b_2\left(\frac{1}{f_2^2}-\frac{1}{(f_2-a_2)^2}\right)$$ where $$f_1+f_2=A(\ \mbox{constant})$$ When I put $b_1=a_1^2,\ b_2=a_2^2$ in the above equation then I immediately get the equation $$\frac{1}{x_1^2}-\frac{1}{(x_1-1)^2}=\frac{1}{x_2^2}-\frac{1}{(x_2-1)^2}$$ where $\displaystyle x_1=\frac{f_1}{a_1},\ x_2=\frac{f_2}{a_2}$. From here I get $$\frac{1}{x_1^2}-\frac{1}{x_2^2}=\frac{1}{(x_1-1)^2}-\frac{1}{(x_2-1)^2}\\ \Rightarrow \left(\frac{(x_2-x_1)(x_2+x_1)}{x_1^2x_2^2}\right)=\frac{(x_2-x_1)(x_2+x_1-2)}{(x_1-1)^2(x_2-1)^2}$$ Hence $$x_2=x_1\\ \Rightarrow \frac{f_2}{a_2}=\frac{f_1}{a_1}$$ is an equation that I get from here and hence I can get an easy solution. But when I put $b_1=a_1,\ b_2=a_2$ (and this is the case, I need to solve), I can't find any such tractable equation. So, now the only way to get some insight about the dependence of $f_1,f_2$ on $a_1,a_2,a_3$ is to get insight about the roots of the equation (i.e. the equation in $f_1$). How should I proceed? Thank you.

Note: I can solve the equation to get numerical values of $f_1,f_2$, but I want to study its dependence on $a_1,a_2,A$.

• How do you arrive at $\frac{f_1}{a_1}=\frac{f_2}{a_2}$ from the first equation? – Antonio Vargas Aug 23 '13 at 21:51
• @AntonioVargas, I arrived at the result, when $b_1=a_1^2,b_2=a_2^2$, simply put $x_i=\frac{f_i}{a_i},\ i=1,2$, and you'll get a factor $\frac{1}{x_1}-\frac{1}{x_2}$ from the equation. – Samrat Mukhopadhyay Aug 24 '13 at 6:35
• Could you go into more detail? I'm really not seeing it. – Antonio Vargas Aug 24 '13 at 16:49
• @AntonioVargas, I have added the details. Hope this is clear now. – Samrat Mukhopadhyay Aug 25 '13 at 6:30

Multiplying your equation by its denominator yields a polynomial of degree 5 in $f_1$.