My goal is to pull out all the $s$ from under the square root.
$$\sqrt{(a^2-4d)s^2+(4bd-4ac)s+4c^2}$$
where $a,b,c,d$ are complex numbers and $s$ is a variable.
I tried solving for $s$, but that only gave me $s_1=\frac{2\left(ac+\sqrt{d\left(db^2-2abc+4c^2\right)}-db\right)}{a^2-4d},\:s_2=\frac{2\left(ac-db-\sqrt{d\left(db^2-2abc+4c^2\right)}\right)}{a^2-4d};\quad \:a^2-4d\ne \:0 $
and i realized that i cannot take the squareroot of $(s-s_1)(s-s_2)$ if $s_1$ and $s_2$ are not equal.
Is it possible to multiply out at least by $s^2$ in this root?
On the same note is there a general way to check whether a parametric expression has a semi-rational form like:
$$(s\cdot \sqrt{\mathscr{\xi}})$$
?
Where $\xi$ is the irreducible part of the expression(preferably made up of only constants).
I think you're asking for some complex $k,l$ such that
$$ks+l=\sqrt{(a^2-4d)s^2+(4bd-4ac)s+4c^2}.$$ Unfortunately, this would make $a,b,c,d$ depend upon $k,l$ and two of their own within the set, or, by excluding $k,l$, have a single dependency. However, we may proceed anyway.
$$k^2s^2+2kls+l^2= (a^2-4d)s^2+4(bd-ac)s+4c^2$$ $$k=\sqrt{a^2-4d}$$ and $$2\sqrt{a^2-4d}\cdot ls+l^2=4(bd-ac)s+4c^2$$ $$l=\dfrac{2(bd-ac)}{\sqrt{a^2-4d}}$$ leaving the condition $$(bd-ac)^2c^2=a^2-4d.$$