I'm looking to solve a fairly complex expression for $x$ which I will give below:
$$x^2-2b=c\left(x-\sqrt{x^2+a(x)}\right)-x\sqrt{x^2+a(x)}.$$
Here $b$ and $c$ are constants and $a(x)=2\sqrt{b-x^2}$. This expression is the result of solving a nonlinear system of 3 equations ($x$ is one of the unknowns in the system), and we know that $x>0$ and $b>0$. I have been unable to make any progress beyond this point (it already wasn't easy to get here). I would appreciate any tips on how to proceed in solving this equation for $x$. I know the result is quite complicated (but still tractable), so I would just like some assistance on what methods to try.
EDIT: I will provide the system this equation came from, in the event that there is a better way to generate an equation for $x$. The system is
$$-2x_2+\nu x_1+x_1^2+x_2^2=0$$ $$-x_3+\nu x_2+(x_1+x_3)x_2=\alpha\nu$$ $$\nu x_3+x_2^2+x_3^2=0$$
This should be approximately correct, since I had nondimensionalized at a later point in my derivations. The constants are then $b=\nu^2/4$ and $c=\alpha-1/2$ and
$$x=\sqrt{b-x_2^2}$$
It is desired to solve for $x_2$, since the final result I am looking for only depends on that unknown. I generated the equation for $x$ by solving the first and third equations for $x_3(x_2)$ and $x_1(x_2)$ and substituting into the second equation.
EDIT 2: I should also add there's a bit of physical insight that goes into dropping possible solutions from quadratics. Essentially, if $x_2=0$ we expect $x_1=x_3=0$ as well. This doesn't satisfy the system because I have assumed $x_2\neq0$ in my nondimensionalization ($\nu\to\infty$ for $x_2\to0$).
You can get a polynomial equation: $$x^2-2b=cx-(c+x)\sqrt{x^2+a(x)}\\x^2-cx-2b=-(c+x)\sqrt{x^2+a(x)}$$ Now you square both sides (make sure you check the solution because of this), so $$(x^2-cx-2b)^2=(c+x)^2(x^2+a(x))\\\frac{(x^2-cx-2b)^2}{(c+x)^2}=x^2+a(x)=x^2+\sqrt{b-x^2}$$ Move $x^2$ to the other side, then square it again: $$\left(\frac{(x^2-cx-2b)^2}{(c+x)^2}-x^2\right)^2=b-x^2$$ Multiply both sides with $(c+x)^4$ and you get a polymonial equation.