Given that, $ f(5+x) = f(5-x) ; \forall x \in R \quad \longrightarrow (1) $
And, $ f(x) = 0 $
Let $ 5 + x = k $
$ \Rightarrow \boxed{x = k – 5} $
Now, from equation $(1),$
$ f(k) = f(5-k+5) $
$ \Rightarrow \boxed{ f(k) = f(10-k)} $
If $f(k)=0,$ then $f(10-k)=0$
Let $\alpha,\beta$ be two roots of $f(k)=0,$ then $10- \alpha, 10 – \beta$ are also be roots of $f(k)=0.$
$\therefore$ The sum of roots $ = \alpha + \beta + 10 – \alpha + 10 – \beta = 20 $
$\textbf{Second Method:}$ Given that, $f(5+x) = f(5-x) ; \forall x \in R \quad \longrightarrow (1) $
And, $ f(x) = 0 $
Put, $ 5 – x = x $ in equation $(1)$
$ f( 5 + 5 – x) = f(x) $
$ \Rightarrow \boxed{ f(x) = f( 10-x )} $
Let $\alpha,\beta$ be two roots of $f(x)=0,$ then $10- \alpha, 10 – \beta$ are also be roots of $f(x)=0.$
$\therefore$ The sum of roots $ = \alpha + \beta + 10 – \alpha + 10 – \beta = 20 $
Correct Answer$: \text{D}$