1. Key Concepts and Formulas

• Common Root: If $\alpha$ is a common root of two equations $f(x) = 0$ and $g(x) = 0$, then $f(\alpha) = 0$ and $g(\alpha) = 0$.
• Solving Quadratic Equations: Methods for solving quadratic equations, including factoring, completing the square, or using the quadratic formula.
• Modulus of a Complex Number: If $z = a + bi$ is a complex number, its modulus is $|z| = \sqrt{a^2 + b^2}$.

2. Step-by-Step Solution

Step 1: Eliminate $x^2$ to find $x$ in terms of $b$

We are given the equations: Equation (1): $x^2 + bx - 1 = 0$ Equation (2): $x^2 + x + b = 0$

Subtracting Equation (2) from Equation (1) will eliminate the $x^2$ term, allowing us to solve for $x$ in terms of $b$. $(x^2 + bx - 1) - (x^2 + x + b) = 0$ $x^2 + bx - 1 - x^2 - x - b = 0$ $(b - 1)x - (1 + b) = 0$ $(b - 1)x = b + 1$ $x = \frac{b + 1}{b - 1}$ This is valid as long as $b \neq 1$. We will address the case where $b = 1$ later.

Step 2: Check the case where $b=1$

If $b=1$, the equations become $x^2 + x - 1 = 0$ and $x^2 + x + 1 = 0$. Subtracting the equations yields $-2 = 0$, which is a contradiction. Thus, $b \neq 1$.

Step 3: Substitute $x$ into one of the original equations

Substitute the expression for $x$ into Equation (2): $x^2 + x + b = 0$ $\left(\frac{b + 1}{b - 1}\right)^2 + \left(\frac{b + 1}{b - 1}\right) + b = 0$ Multiply by $(b-1)^2$ to eliminate the fractions: $(b + 1)^2 + (b + 1)(b - 1) + b(b - 1)^2 = 0$

Step 4: Solve for $b$

Expand and simplify the equation: $(b^2 + 2b + 1) + (b^2 - 1) + b(b^2 - 2b + 1) = 0$ $b^2 + 2b + 1 + b^2 - 1 + b^3 - 2b^2 + b = 0$ $b^3 + (b^2 + b^2 - 2b^2) + (2b + b) + (1 - 1) = 0$ $b^3 + 3b = 0$ $b(b^2 + 3) = 0$ This gives us the solutions $b = 0$ or $b^2 = -3$, so $b = \pm i\sqrt{3}$.

Step 5: Check the condition $x \neq -1$

The problem states that the common root is different from $-1$. If $x = -1$, then substituting into Equation (2): $(-1)^2 + (-1) + b = 0$ $1 - 1 + b = 0$ $b = 0$ Since we are given that $x \neq -1$, we must reject $b = 0$. Therefore, the only valid solutions are $b = \pm i\sqrt{3}$.

Step 6: Calculate $|b|$

For $b = i\sqrt{3}$, $|b| = \sqrt{0^2 + (\sqrt{3})^2} = \sqrt{3}$. For $b = -i\sqrt{3}$, $|b| = \sqrt{0^2 + (-\sqrt{3})^2} = \sqrt{3}$. In both cases, $|b| = \sqrt{3}$.

3. Common Mistakes & Tips

• Forgetting Conditions: Always remember to check any conditions given in the problem, such as the common root not being equal to $-1$.
• Division by Zero: Be mindful of potential division by zero when manipulating equations. In this case, we needed to consider the case $b=1$.
• Extraneous Solutions: Always check for extraneous solutions, especially when dealing with square roots or rational expressions.

4. Summary

We found the common root in terms of $b$ by eliminating $x^2$ from the given quadratic equations. Substituting this expression back into one of the original equations allowed us to solve for $b$. After finding the possible values for $b$, we used the condition that the common root is not equal to $-1$ to eliminate $b=0$. Finally, we calculated the modulus of the remaining values of $b$, which resulted in $|b| = \sqrt{3}$.

5. Final Answer

The final answer is $\boxed{\sqrt{3}}$, which corresponds to option (D).