Common Root Condition for Two Quadratic Equations

To find the value of $b$ for which the given quadratic equations have a common root, let $\alpha$ be the common root of the two equations:
$x^2 + bx - 1 = 0 \quad \text{--- (1)}$
$x^2 + x + b = 0 \quad \text{--- (2)}$

Since $\alpha$ satisfies both equations, we have:
$\alpha^2 + b\alpha - 1 = 0 \quad \text{--- (3)}$
$\alpha^2 + \alpha + b = 0 \quad \text{--- (4)}$

Subtracting equation (4) from equation (3) to eliminate $\alpha^2$:
$(\alpha^2 + b\alpha - 1) - (\alpha^2 + \alpha + b) = 0$
$(b - 1)\alpha - (b + 1) = 0$

Assuming $b \neq 1$ (if $b = 1$, the subtraction gives $-2 = 0$, which is impossible), we can solve for $\alpha$:
$\alpha = \frac{b + 1}{b - 1}$

Now, substitute this expression for $\alpha$ back into equation (4):
$\left(\frac{b + 1}{b - 1}\right)^2 + \left(\frac{b + 1}{b - 1}\right) + b = 0$

Multiplying the entire equation by $(b - 1)^2$:
$(b + 1)^2 + (b + 1)(b - 1) + b(b - 1)^2 = 0$

Expanding each term:
$(b^2 + 2b + 1) + (b^2 - 1) + b(b^2 - 2b + 1) = 0$
$2b^2 + 2b + b^3 - 2b^2 + b = 0$
$b^3 + 3b = 0$
$b(b^2 + 3) = 0$

This gives two cases:

1. $b = 0$
2. $b^2 + 3 = 0 \implies b = \pm i\sqrt{3}$

Comparing with the given options, $b = -i\sqrt{3}$ is present in option (B).

Hence, the correct option is (B).