Functional equation: $ f(x^2+x+3)+2f(x^2-3x+5)= 6x^2 -10 x + 17$

Question

$f: \mathbb R \to \mathbb R, f(x^2+x+3)+2f(x^2-3x+5)= 6x^2 -10 x + 17 \forall x \in \mathbb R $, then find the function $f(x)$

I have:

$f(15/4)= 16/3$ (both quadratics intersect at $x=1/2$)

$f(3)= 3$ (by substituting $x=0$ and $x=1$ and then solving the simultaneous equations obtained)

$f(5)= 7$

I am not getting anything fruitful from these.

Could anyone provide me a hint on how to proceed?

Edit:

Someone had commented (now it's deleted) that we can assume $f(x) = ax +b$, how can we do that? I got the right answer using that. Is it always okay to assume that way? When is it a reliable assumption?

Answer 1

Replace $x$ by $1-x$,

\begin{align*} f((1-x)^2+(1-x)+3)+2f((1-x)^2−3(1-x)+5)&=6(1-x)^2−10(1-x)+17\\ f(x^2-3x+5)+2f(x^2+x+3)&=6x^2-2x+13\\ \end{align*}

Solving with $f(x^2+x+3)+2f(x^2-3x+5)= 6x^2 -10 x + 17$, we have $f(x^2+x+3)=2x^2+2x+3$.

$f(x)=2x-3$.

Answer 2

If $x^2-3x+5=0$, $$f(4x-2)+2f(0)=8x-13 $$

If $x^2+x+3=0$ $$f(0)+2f(-4x+2)=-16x-1$$

$$\rightarrow f(4x-2)+f(-4x+2)+\frac52f(0)=\frac{-27}2$$

put $x=\frac12$ $$\frac92f(0)=-27/2$$ $\rightarrow f(0)=-3$

Then $$f(4x-2)=8x-7$$ Put $4x-2=y$ $$\rightarrow f(y)=8(\frac12+\frac y4)-7=2y-3$$