I have :
$$2z \frac{\sqrt{pq/n + z^2/4n^2}}{1+z^2/n} = w$$
and I want to demonstrate that $$n = \frac{2x^2pq - z^2w^2 ~\pm~ \sqrt{4z^4pq(pq - w^2) + w^2z^4}}{w^2}.$$
Can someone give me a detailed solution?
I tried to solve it and at the end I got $$ \frac{pq}{n} + \frac{z^2(1-w^2)}{4n^2} = \frac{w^2}{4z^2} + \frac{w^2}{2}, $$ but I am not sure if this is correct and I am even lost to find the exact solution for $n$.
Any help will be very appreciated.
If you multiply your last equation by $4n^2$ you get a degree 2 polynomial equation. $$ ax^2+bx+c=0. $$ The generic solution $$ n={-b\pm\sqrt{b^2-4ac}\over 2a} $$ looks a lot like what you are looking for...