Ans: D
$\begin{array}{rcl}
(x-a)(x-a-1) & = & (x-a) \\
(x-a)(x-a-1)-(x-a) & = & 0 \\
(x-a)(x-a-1-1) & = & 0 \\
(x-a)(x-a-2) & = & 0
\end{array}$
$\begin{array}{rcl}
(x-a)(x-a-1) & = & (x-a) \\
(x-a)(x-a-1)-(x-a) & = & 0 \\
(x-a)(x-a-1-1) & = & 0 \\
(x-a)(x-a-2) & = & 0
\end{array}$
$\therefore x-a=0 $ or $x-a-2=0$.
$\therefore x=a$ or $x=a+2$.
