G3.4 Given that the function $f$ satisfies $f(2+x)=f(2-x)$ for every real number $x$ and that $f(x)=0$ has exactly four distinct real roots. Find the sum of these four distinct real roots.
Let two of these distinct roots be $2+\alpha, 2+\beta$, where $\alpha \neq \beta$ and $\alpha, \beta \geq 0$.
$$
f(2+x)=f(2-x) \Leftrightarrow f(2+\alpha)=f(2-\alpha)=0 ; f(2+\beta)=f(2-\beta)=0
$$

If $\alpha=0$ and $\beta \neq 0 \Rightarrow$ there are only three real roots $2,2+\beta, 2-\beta$ contradiction, rejected.
$\therefore \alpha \neq 0$ and $\beta \neq 0 \Leftrightarrow$ The four roots are $2+\alpha, 2-\alpha, 2+\beta, 2-\beta$.
Sum of roots $=2+\alpha+2-\alpha+2+\beta+2-\beta=8$