‥ For the equation greater than $x$, $2 x^{2}+3 a x+a^{2}-a=0$ to have at least one root with a modulus equal to 1, the value of the real number $a$ ().
(A) does not exist
(B) has one
(C) has two
(D) has four
6. C.
$$
2 x^{2}+3 a x+a^{2}-a=0 .
$$
(1) When the roots are real numbers
$$
\Delta=9 a^{2}-8\left(a^{2}-a\right)=a(a+8) \geqslant 0,
$$

we have $a \geqslant 0$ or $a \leqslant \cdots 8$. Substituting $x=1$ into (1) yields $a^{2}+2 a+2=0$.
Since $a \in \mathbb{R}$, this equation has no solution. Substituting $x=-1$ into (1) yields
$$
a^{2}-4 a+2=0 \text {. }
$$

Solving for $a$ gives $a=2 \pm \sqrt{2}$, which meets the requirements within the range of (2).
(2) When the roots are complex numbers,
$$
\begin{array}{l}
\Delta=a(a+8)<0 . \\
\text { Hence }-8<a<0 .
\end{array}
$$

Solving (1) yields $x=\frac{-3 a \pm \sqrt{-a^{2}-8} i}{4}$.
$$
\begin{array}{l}
\because|x|=1 . \\
\therefore\left(-\frac{3 a}{4}\right)^{2}+\left( \pm \frac{\sqrt{-a^{2}-8 a}}{4}\right)^{2}=1,
\end{array}
$$

which simplifies to $a^{2}-a-2=0$.
Solving for $a$ gives $a=-1, a=2$.
$\because a=2$ is not within the range of (3),
$$
\therefore a=-1 \text {. }
$$

In summary, $a=2 \pm \sqrt{2}, a=-1$.