Ans: B
$\begin{array}{rcl}
x^2 + ax +a & = & 1 \\
x^2 + ax + (a-1) & = & 0
\end{array}$
$\begin{array}{rcl}
x^2 + ax +a & = & 1 \\
x^2 + ax + (a-1) & = & 0
\end{array}$
Since the equation above has equal roots, then we have
$\begin{array}{rcl}
\Delta & = & 0 \\
a^2 – 4(1)(a-1) & = & 0 \\
a^2 -4a + 4 & = & 0 \\
(a-2)^2 & = & 0 \\
a & = & 2
\end{array}$
