is the real root of a higher order equation continuous with the parameters? I come across one equation that $a_{0}x^{2}+x(a_{1}\sqrt{f(x)}+a_{2})+a_{3}\sqrt{f(x)}+a_{4}=0$, in which $f(x)=b_{2}x^{2}+b_{1}x+b_{0}$ and $b_{1}^{2}-4b_{2}b_{0}\leq 0$ and $b_{2}>0$. I cannot find a way to solve this equation. Now I want to prove the real root (if any) of this equation is continuous with the parameters $a_{0}$, $a_{1}$, $a_{2}$, $a_{3}$, $a_{4}$, $b_{2}$, $b_{1}$ and $b_{0}$. Is it possible to prove this ? Please help, thank you. 
 A: The manipulations I am about to describe are valid unless $a_3 = a_1 = 0$, but if this is the case, the first equation is just a quadratic and the continuity of the location of the roots with respect to the coefficients follows in the same way.
Solve the first equation for $\sqrt{f}$.  Square both sides.  Substitute the quadratic for $f$.  Multiply both sides by $(a_3 + a_1 x)^2$.  Collect the results on one side of the equality.  The solutions of the original equation are a subset of the roots of this equation.  (Squaring both sides may have introduced additional/spurious roots.)  This is an equality between zero and a fourth degree polynomial.  Though solvable in principle, this is a problem instance that will produce magnificent intermediate expression swell (because, since the coefficients are arbitrary, we cannot restrict which Galois group is the automorphism group of the field extension containing the roots), so we don't want to do that.  You say you "want to prove the real root (if any) ..." which suggests you are not interested in the discontinuous behaviour of a real root appearing or disappearing depending on whether $a_4$ varies to include the roots associated with a turning point or not.
The answers to Continuity of the roots of a polynomial in terms of its coefficients address the fact that (assuming roots are not appearing/disappearing) they move continuously as the coefficients are varied.
