Have been dealing with this equation for a while, but yet able to get a simple expression.
I have already done numerical solution and seems to be working just fine, but I would like to get a closed expression.
The problem is to find $x$ form the following:
$\frac{2LxB}{2Lx+\sqrt{b^2+x^2}} + 2b\sqrt{\frac{2LxB}{2Lx+\sqrt{b^2+x^2}}} = Ax^2$
What I tried:
$y^2 = \frac{2LxB}{2Lx+\sqrt{b^2+x^2}}$ So the equation above resumes to:
$y^2 + 2by = Ax^2$
Then found the solution for $y$:
$y = \sqrt{b^2+Ax^2} -b$.
Then, trying equating the above, with the definition of $y$:
$\frac{2LxB}{2Lx+\sqrt{b^2+x^2}} = (\sqrt{b^2+Ax^2} -b)^2$
However, the problem is the term inside the square root, since $A$ is present in only one of them.
Any trick or technique you may know to solve this?
Thanks folks!
The problem is that, beside the trivial $x=0$, you face a quintic polynomial $$-64 b^6 B^2 L^2+16 B^2 L^2 \left(L^2 \left(B-4 b^2\right)^2-4 (A+1) b^4\right)x-$$ $$8 \left(A B L^2 \left(A b^2 \left(B-4 b^2\right)+8 B \left(L^2 \left(B-4 b^2\right)+b^2\right)\right)\right)x^2+$$ $$A^2 \left(A^2 b^4+8 B L^2 \left(2 (A+2) b^2-B\right)+32 B L^4 \left(3 B-4 b^2\right)\right)x^3-$$ $$2 \left(A^3 \left(4 L^2-1\right) \left(A b^2+8 B L^2\right)\right)x^4+A^4 \left(1-4 L^2\right)^2 x^5=0$$
So, no hope for a closed form solution.