Find root of equation $\frac{2bnx}{a\sqrt{x^2-b^2}}+\frac{2bx}{a}+\frac{2ax}{b}-2m=0$

Question

How I can find $x$ for this equation

$$\frac{2bnx}{a\sqrt{x^2-b^2}}+\frac{2bx}{a}+\frac{2ax}{b}-2m=0$$ where $a, b, m, n$ are constant numbers.

Thanks.

Answer 1

I just wolframed it.
It will be a quartic equation once you remove the root, and it cannot be factorized by anything (so far).
So, the equation must be solved by the formula for quartic equations.
The link is below:

https://www.wolframalpha.com/input/?i=solve+bnx%2F(a*sqrt(x%5E2-b%5E2))%2Bbx%2Fa%2Bax%2Fb-m%3D0

Answer 2

There are too parameters here and the answer depends on these all, but we can find a restriction for answer regard to a condition. With $x=b\cosh u$ we simplify $$\frac{2bnx}{a\sqrt{x^2-b^2}}+\frac{2bx}{a}+\frac{2ax}{b}-2m=0$$ to $$(a^2+b^2)\sinh u\cosh u=am\sinh u-bn\cosh u$$ by Cauchy-Schwarz inequality $$(am\sinh u-bn\cosh u)^2\leq(a^2+b^2)(m^2+n^2)(\sinh^2u+\cosh^2u)$$ or $$\frac{1}{\sinh^2u}+\frac{1}{\cosh^2u}\geq\frac{a^2+b^2}{m^2+n^2}$$ maybe this condition close us to an answer.