Consider the equation below
$$ z -1 - \left( \frac{1}{\sqrt{z^2+z + 1}} - \frac{2}{\sqrt{z^2 + 4}} \right) = 0. $$
The function on the lhs is a four-valued function. To get its zeros, we can convert it into a 12th order polynomial equation by squaring.
Then we will get 12 zeros.
The question is then, are all these zeros indeed zeros of the original equation? More precisely, for each of them, is it always possible to make the function on the lhs vanish by taking the signs of the square roots correctly?
Is there any general theory for such kind of problem?
Yes. Let $u = \sqrt {z^2+z+1}$ and $v = \sqrt {z^2+4}$, so that we have $u^2 = z^2+z+1$ and $v^2=z^2+4$.
Your equation $z-1-1/u+2/v = 0$ is equivalent to $(z-1)uv-v+2u = 0$.
Now if you want to get an equation that only talks about $z$ from this, you have to multiply $(z-1)uv-v+2u$ with its conjugates.
In terms of Galois theory, $\Bbb C(z)[u,v]$ is a Galois extension of $\Bbb C(z)$, its Galois group is isomorphic to $(\Bbb Z/2\Bbb Z)^2$ and is generated by $(u,v) \mapsto (-u,v)$ and $(u,v) \mapsto (u,-v)$.
Taking the norm of $(z-1)uv-v+2u$ will give you the expression
$((z-1)uv-v+2u)(-(z-1)uv-v-2u)(-(z-1)uv+v+2u)((z-1)uv+v-2u)$
This is invariant when you replace $u$ with $-u$ or $v$ with $-v$, so when you expand it and simplify, it is a polynomial expression in $z,u^2,v^2$.
Now you can replace $u^2,v^2$ with the corresponding polynomials in $z$, and you get that this product is actually a degree $12$ polynomial in $z$ (probably the same one that you've got).
Now, if $z$ is a root of the degree $12$ polynomial, and if you pick for $u$ and $v$ any of their two corresponding values and plug everything into the product, since the product evaluates to $0$, you must have that at least one of its factors also evaluates to zero. This means that there must be a choice for $u$ and $v$ that makes the original expression evaluate to $0$.