jmc

Let $r$ be the common root, so $$\begin{array}{rl}{r}^2+ar+1& =0,\\ r^2-r-a& =0.\end{array}$$Subtracting these equations, we get $ar+r+a+1=0.$ This factors as $(r+1)(a+1)=0,$ so $r=-1$ or $a=-1.$

If $r=-1,$ then $1-a+1=0,$ so $a=2.$

If $a=-1,$ then $x^2-x+1=0,$ which has no real roots.

So the only possible value of $a$ is $\boxed{2}.$