# Solving tricky functional equation resembling quadratic equation

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.

• This looks correct, you can see that $g(0)=0$ and so the second approach cancels to the first one. Apr 1 '19 at 8:46

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: 