$2 \cdot 49$ The equations $x^{2}+a x+1=0$ and $x^{2}-x-a=0$ have a common real solution. The number of possible values for the coefficient $a$ is
(A) 0.
(B) 1.
(C) 2.
(D) 3.
(E) infinitely many.
(28th American High School Mathematics Examination, 1977)
［Solution］Subtract the second equation from the first equation given:
$$
a x + x + (1 + a) = 0,
$$

which simplifies to $(a+1)(x+1)=0$. Solving this, we get $a=-1$ or $x=-1$.
If $a=-1$, then the two given equations are identical and have no real solutions (but have two complex solutions);
When $x=-1$, we get $a=2$, and the given equations have one common real solution.
Therefore, the answer is $(B)$.