II. For what value of $b$ do the equations $x^{2}-b x-2=0$ and $x^{2}-2 x-b(b-1)=0$ have the same roots? And find their common roots.
Let the common root of the two quadratic equations be $\alpha$, then
$$
\begin{array}{l}
\alpha^{2}+b \alpha-2=0, \\
\alpha^{2}-2 \alpha-b(b-1)=0 .
\end{array}
$$
Subtract (1) from (2): we get
$$
b \alpha-2 \alpha-b(b-1)+2=0 .
$$

After factoring, we get
$$
(b-2)(\alpha-b-1)=0 .
$$

When $b \neq 2$, then $\alpha-b-1=0$.
$$
\begin{array}{l}
\therefore \alpha=b+1 . \text { Substituting into (1) or (2) gives } \\
\quad b=1 . \\
\therefore \alpha=2 . \\
\text { When } b=2 \text {, (1) and (2) both become } \\
\alpha^{2}-2 \alpha-2=0 .
\end{array}
$$

When $b=2$, (1) and (2) both become

Solving it, we get
$$
\begin{array}{l}
\alpha_{1}=1+\sqrt{3}, \\
\alpha_{2}=1-\sqrt{3} .
\end{array}
$$

Answer: When $b=1$, the common root of the two equations is 2; when $b=2$, the common roots of the two equations are $1+\sqrt{3}$ and $1-\sqrt{3}$.