If the equations $$x^2 + bx - 1 = 0$$ and $$x^2 + x + b = 0$$ have a common root different from $$-1$$, then $$|b|$$ is equal to:

Let $$\alpha$$ (where $$\alpha \neq -1$$) be the common root of $$x^2 + bx - 1 = 0$$ and $$x^2 + x + b = 0$$. Since $$\alpha$$ satisfies both equations, subtracting the first from the second gives $$(1 - b)\alpha + (b + 1) = 0$$. Solving for $$\alpha$$: $$(1 - b)\alpha = -(b + 1)$$, so $$\alpha = \frac{b + 1}{b - 1}$$ (valid when $$b \neq 1$$).

Substituting $$\alpha = \frac{b+1}{b-1}$$ into the second equation $$\alpha^2 + \alpha + b = 0$$ and multiplying through by $$(b-1)^2$$: $$(b+1)^2 + (b+1)(b-1) + b(b-1)^2 = 0$$. Expanding: $$(b^2 + 2b + 1) + (b^2 - 1) + (b^3 - 2b^2 + b) = 0$$, which simplifies to $$b^3 + 3b = 0$$, i.e., $$b(b^2 + 3) = 0$$.

If $$b = 0$$, then $$\alpha = (0+1)/(0-1) = -1$$, which is excluded. So $$b^2 + 3 = 0$$, giving $$b^2 = -3$$. Since $$b$$ is complex, $$|b|^2 = |{-3}| = 3$$, and therefore $$|b| = \sqrt{3}$$.