Solve $\frac{x^2}{\left(a+\sqrt{a^2+x^2}\right)^2}+\frac{x^2(1+x^2)}{\left(a+\sqrt{a^2+x^2(1+x^2)} \right)^2}=1$ I have been trying to solve the following equation for $x$:
\begin{align}
\frac{x^2}{\left(a+\sqrt{a^2+x^2}\right)^2}+\frac{x^2(1+x^2)}{\left(a+\sqrt{a^2+x^2(1+x^2)} \right)^2}=1,
\end{align}
for some fixed $a \in (0,1/2]$. 
This equation came from an analysis of an electrical circuit. The solution is the current.
However, after trying to solve it by hand using the software it doesn't appear that there is a good way of finding a solution. 
My question:  Can we at least give a good estimate of the position of the positive zero?    For example, can show that the zero belong to the specific interval?
Here is an equaivalent polynomial:
\begin{align}
x^4 - 4a^4 - 4a^3(a^2 + x^2)^{1/2} + x^6 - 4a^3(x^2(x^2 + 1) + a^2)^{1/2} - 4a^2(x^2(x^2 + 1) + a^2)^{1/2}(a^2 + x^2)^{1/2}=0.
\end{align}
 A: Hints:
Let $x = a \tan (2 A)$ and $x\sqrt{1 +x^2} = a \tan (2 B)$, then, due to the $\tan$- half-angle theorem,  the equation in question transforms into 
$$
\tan^2 (A) + \tan^2(B) = 1
$$
This can be used for numerical analysis, e.g. (with fixed $a$) a Banach type fixed point iteration $x^{(n)} \to B \to A \to x^{(n+1)} \to  \dots $ 
Re-inserting $A$ and $B$  gives
$$
\tan^2 \left(\frac12 \arctan \frac{x}{a}\right) + \tan^2\left(\frac12 \arctan \frac{x\sqrt{1 +x^2}}{a}\right) = 1
$$
where one might check for further simplification.
EDIT (March 18,2018)
Matlab is able to plot the solutions by implicit plot:

A rough (underestimated) linearization is $x = 2.3 a$; near $a=0$ we have the pretty exact linearization $x \simeq 2.8 a$.
A: Your equation is simplified to:
$$\dfrac{t+\sqrt{1+t^2+\frac{a^4}{t^4}}}{2t} = (t+\sqrt{1+t^2})^2$$
with the substitution $a=xt.$ Conceivably, now you can explicitly find $a$ in terms of $t$ and then might succeed doing an asymptotic analysis, but I really doubt anything close to a closed form solution exists, despite the equation itself look like it should have a closed form solution. 
