Key Concepts and Formulas

• Quadratic Equations and Complex Conjugate Roots: If a quadratic equation $ax^2 + bx + c = 0$ with real coefficients has complex roots, they are complex conjugates of each other, i.e., $z = \alpha + i\beta$ and $\bar{z} = \alpha - i\beta$.
• Vieta's Formulas: For the quadratic equation $ax^2 + bx + c = 0$, the sum of the roots is $-b/a$ and the product of the roots is $c/a$.
• Modulus of a Complex Number: For a complex number $z = x + iy$, the modulus is $|z| = \sqrt{x^2 + y^2}$.

Step 1: Define the Roots and Apply Vieta's Formulas

We are given the quadratic equation $x^2 + bx + 45 = 0$. Since the roots are complex conjugates, let them be $z = \alpha + i\beta$ and $\bar{z} = \alpha - i\beta$, where $\alpha, \beta \in \mathbb{R}$ and $\beta \neq 0$. We apply Vieta's formulas to relate the roots to the coefficients of the quadratic equation.

• Sum of roots: $z + \bar{z} = (\alpha + i\beta) + (\alpha - i\beta) = 2\alpha$. From the equation, the sum of the roots is $-b/1 = -b$. Therefore, $2\alpha = -b \quad (1)$
• Product of roots: $z \cdot \bar{z} = (\alpha + i\beta)(\alpha - i\beta) = \alpha^2 + \beta^2$. From the equation, the product of the roots is $45/1 = 45$. Therefore, $\alpha^2 + \beta^2 = 45 \quad (2)$

Step 2: Apply the Modulus Condition

We are given $|z + 1| = 2\sqrt{10}$. Substitute $z = \alpha + i\beta$ into this equation and simplify. $|(\alpha + i\beta) + 1| = 2\sqrt{10}$ $|(\alpha + 1) + i\beta| = 2\sqrt{10}$ Now, using the definition of the modulus: $\sqrt{(\alpha + 1)^2 + \beta^2} = 2\sqrt{10}$ Square both sides to eliminate the square root: $(\alpha + 1)^2 + \beta^2 = (2\sqrt{10})^2$ $(\alpha + 1)^2 + \beta^2 = 40 \quad (3)$

Step 3: Solve for $\alpha$

Expand equation (3) and substitute equation (2) to solve for $\alpha$. $\alpha^2 + 2\alpha + 1 + \beta^2 = 40$ $(\alpha^2 + \beta^2) + 2\alpha + 1 = 40$ Substitute $\alpha^2 + \beta^2 = 45$ from equation (2): $45 + 2\alpha + 1 = 40$ $2\alpha = 40 - 46$ $2\alpha = -6$ $\alpha = -3$

Step 4: Solve for $b$

Substitute the value of $\alpha$ into equation (1) to find $b$. $2\alpha = -b$ $2(-3) = -b$ $-6 = -b$ $b = 6$

Step 5: Verify the Options

Substitute $b = 6$ into each of the given options to determine which one is satisfied.

• (A) $b^2 - b = 42$ $6^2 - 6 = 36 - 6 = 30 \neq 42$
• (B) $b^2 + b = 12$ $6^2 + 6 = 36 + 6 = 42 \neq 12$
• (C) $b^2 + b = 72$ $6^2 + 6 = 36 + 6 = 42 \neq 72$
• (D) $b^2 - b = 30$ $6^2 - 6 = 36 - 6 = 30$

Option (D) is satisfied by $b=6$. However, the stated "Correct Answer" is (A), which is incorrect. There appears to be an error in the problem statement or provided answer. Since we were instructed to arrive at the provided answer, let's find the b that would make option (A) correct. If $b^2 - b = 42$, then $b^2 - b - 42 = 0$, which factors to $(b-7)(b+6) = 0$. Thus, $b=7$ or $b=-6$.

If $b=7$, then $2\alpha = -7$, so $\alpha = -3.5$. Then $\alpha^2 + \beta^2 = 45$, so $(-3.5)^2 + \beta^2 = 45$, which means $12.25 + \beta^2 = 45$, so $\beta^2 = 32.75$. Also, $(\alpha+1)^2 + \beta^2 = 40$, so $(-3.5+1)^2 + \beta^2 = 40$, which means $(-2.5)^2 + \beta^2 = 40$, so $6.25 + \beta^2 = 40$, and $\beta^2 = 33.75$. Since the $\beta^2$ values are different, $b=7$ is not a solution.

If $b=-6$, then $2\alpha = 6$, so $\alpha = 3$. Then $\alpha^2 + \beta^2 = 45$, so $3^2 + \beta^2 = 45$, which means $9 + \beta^2 = 45$, so $\beta^2 = 36$. Also, $(\alpha+1)^2 + \beta^2 = 40$, so $(3+1)^2 + \beta^2 = 40$, which means $16 + \beta^2 = 40$, and $\beta^2 = 24$. Since the $\beta^2$ values are different, $b=-6$ is not a solution.

Therefore, the problem is flawed.

Common Mistakes & Tips

• Double-check calculations and substitutions to avoid algebraic errors.
• When dealing with complex numbers, remember the definition of the modulus and its properties.
• Always verify the solution by substituting the obtained value back into the original equations.

Summary

We systematically applied Vieta's formulas and the modulus condition to solve for the unknown coefficient $b$. We found that $b=6$, which corresponds to option (D). However, the provided correct answer is (A), which is not satisfied by our solution. There seems to be an error in the problem statement or the provided answer. The problem cannot be solved with the provided answer.

Final Answer The problem is flawed. If we proceed with our solution, we find $b = 6$, which satisfies the condition $b^2 - b = 30$, which corresponds to option (D).