6. Given four distinct real numbers $x_{1}, x_{2}, x_{3}, x_{4} \left(x_{1}<x_{2}, x_{3}<x_{4}\right)$. Also, $a$ is a real number, the function $y_{1}=$ $x^{2}-4 x+a$ intersects the $x$-axis at points $\left(x_{1}, 0\right)$ and $\left(x_{2}, 0\right)$, and the function $y_{2}=x^{2}+a x-4$ intersects the $x$-axis at points $\left(x_{3}, 0\right)$ and $\left(x_{4}, 0\right)$. If these four intersection points are labeled from left to right as $A, B, C, D$, and $A B=B C=C D$, then the value of $a$ is $(\quad)$.
(A) $a=-3$
(B) $a$ is less than 0
(C) $a=0$
(D) $a$ is greater than 0
6. C.
$$
\begin{array}{l}
x_{1}<x_{2}<x_{3}<x_{4}, x_{1}<x_{3}<x_{2}<x_{4}, x_{1}<x_{3}<x_{4}<x_{2}, \\
x_{3}<x_{4}<x_{1}<x_{2}, x_{3}<x_{1}<x_{4}<x_{2}, x_{3}<x_{1}<x_{2}<x_{4} .
\end{array}
$$

Among the above 6 cases, the 3rd and 6th cases cannot occur (otherwise, the axes of symmetry of the two functions would be the same, so $a=-4$. Thus, $x_{1}=x_{3}, x_{2}=x_{4}$, which contradicts the problem statement). In the other 4 cases, we have
$$
\left|x_{2}-x_{1}\right|=\left|x_{4}-x_{3}\right| .
$$

Therefore, $\sqrt{16-4 a}=\sqrt{a^{2}+16}$.
Solving this, we get $a=0$ or -4 (discard -4).
Upon verification, $a=0$ satisfies the problem statement.