# If f(x) has a property that f(2-x) = f(2+x) for all x and f(x) has exactly 4 real zeros, how do you find their sum?

Jul 24, 2016

8

#### Explanation:

Since $f \left(2 - x\right) = f \left(2 + x\right)$ for all $x$, we must have $f \left(t\right) = f \left(4 - t\right)$ for all $t$.

Let $a$ be one of the real roots, then $f \left(a\right) = f \left(4 - a\right) = 0$, so that $4 - a$ must be another real zero. (This works even if $a = 2$, in which case $4 - a = 2$ is to be counted as another of the real roots,and 2 must be a double root). If $b$ is a root different from both $a$ and $4 - a$, then $4 - b$ must be the fourth zero (it is easy to check that this is different from either $a$ or $4 - a$).

So, each pair of zeroes add up to 4, and the sum of all four is 8.