The given equations are

$${x^2} + bx - 1 = 0$$

$${x^2} + x + b = 0$$ ....... (1)

Common root is $$(b - 1)x - 1 - b = 0$$.

$$ \Rightarrow x = {{b + 1} \over {b - 1}}$$

This value of x satisfies Eq. (1), we get

$${{{{(b + 1)}^2}} \over {{{(b - 1)}^2}}} + {{b + 1} \over {b - 1}} + b = 0$$

$$ \Rightarrow b = i\sqrt 3 ,\, - i\sqrt 3 ,\,0$$