If $$f(5+x)=f(5-x)$$ for every real x, and $$f(x)=0$$ has four distinct real roots, then the sum of these roots is

Let 'r' be the root of the function. It follows that f(r) = 0. We can represent this as $$f\left(r\right)=f\left\{5-\left(5-r\right)\right\}$$

Based on the relation: $$f\left(5-x\right)=f\left(5+x\right)$$; $$f\left(r\right)=f\left\{5-\left(5-r\right)\right\}=f\left\{5+\left(5-r\right)\right\}$$

$$\therefore\ f\left(r\right)=f\left(10-r\right)$$

Thus, every root 'r' is associated with another root '(10-r)' [these form a pair]. For even distinct roots, in this case four, let us assume the roots to be as follows: $$r_1,\ \left(10-r_1\right),\ r_2,\ \left(10-r_2\right)$$

The sum of these roots = $$r_1\ +\left(10-r_1\right)+\ r_2+\ \left(10-r_2\right)\ =\ 20$$

Hence, Option D is the correct answer.