Solving tricky functional equation resembling quadratic equation

Question

I have the following functional equation in hand, I can easily solve it for the case $(a, b ,c)=(1, 1,0)$ which gives $f(x)$ to be $x^2+x$.

$\begin{aligned}{g(x)=a\left[f(x)\right]^2+bf(x)+c, \text{where } g(x)=x^4+2x^3+2x^2+x}\end{aligned}$

I also tried setting $x=0$ that gives $af^2(0)+bf(0)+c=0$ which gives $f(0)=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$.

Also I've tried, using the quadratic formula to solve for $f(x)$ as follows:

$$a\left[f(x)\right]^2+bf(x)+\left(c-g(x)\right)=0\implies f(x)=\dfrac{-b\pm\sqrt{b^2-4a\left(c-g(x)\right)}}{2a}$$

Do both the functions solve the functional equation, if not, how can I proceed? Also the solution I found for the special case does not seem to be the same when we put in those values of $a, b, c$ in the general solution. Any hints are appreciated.

Answer 1

For every $x$,

$$a[f(x)]^2+bf(x)+c-g(x)=0$$

is an ordinary quadratic equation and

$$f(x)=\frac{-b\pm\sqrt{b^2-4ac+4ag(x)}}{2a}$$ indeed holds.

There are real roots when

$$b^2-4ac+4ag(x)\ge0,$$ a quartic inequation.

If there are no constraints on $f$, except that it must be a function, then for any $x$ such that there are distinct real roots, you can choose either of them. So there are infinitely, non-countably many solutions.

If $f$ is restricted to be continuous, then the choice of the sign must be consistent across the intervals of the domain.

---

For the case $(1,1,0)$, the roots are

$$\frac{-1\pm\sqrt{1+4(x^4+2x^3+2x^2+x)}}{2}=\frac{-1\pm(2x^2+2x+1)}2=x^2+x,-x^2-x-1.$$

A solution, among many many many others: