8.53 The function $f(x)$ satisfies $f(2+x)=f(2-x)$ for all real numbers $x$. If the equation $f(x)=0$ has exactly four distinct real roots, then the sum of these roots is
(A) 0.
(B) 2.
(C) 4.
(D) 6.
(E) 8.
(35th American High School Mathematics Examination, 1984)
[Solution] Given $f(2+x)=f(2-x)$, if $r_{1}=2+a$ is a root of $f(x)=0$, then $r_{2}=2-a$ must also be a root, and we can conclude that all different roots appear in pairs, with the sum of each pair of real roots being 4.

Since it has exactly four distinct roots, 2 must not be one of its roots (if $r_{1}=2+a=2$, then $a=0$, implying $r_{2}=2$, so $r_{1}=r_{2}$, and this only happens when $a=0$. This contradicts the condition that the function has exactly four distinct roots), thus the sum of these four roots is
8.
Therefore, the answer is $(E)$.