If the equation x ^ 2 + bx + 45 = 0, b ∈ R has conjugate complex...

Let $z = \alpha + i\beta$ be one of the roots of the equation ${x}^{2} + bx + 45 = 0, b \in R$.

So, its other conjugate complex root will be $z = \overset{¯}{\alpha + i\beta} = \alpha - i\beta$.

We know that for a quadratic equation $A{x}^{2} + Bx + C = 0$, the sum and product of its roots are $- \dfrac{B}{A} \text{ and } \dfrac{C}{A}$ respectively.

So, the sum of roots of the given equation, \left(\alpha + i\beta\right) + \left(\alpha - i\beta\right) = - \dfrac{b}{1} ⇒ 2\alpha = - b . . . . . . \left(i\right)$.</p><p>Also, the product of roots of the given equation, \left(\alpha + i\beta\right)\left(\alpha - i\beta\right) = \dfrac{45}{1} ⇒ {\alpha}^{2} + {\beta}^{2} = 45 . . . . . . \left(ii\right)$.

Now, given that \left|z + 1\right| = 2\sqrt{10}$.</p><p>$⇒ \left|\alpha \pm i\beta + 1\right| = 2\sqrt{10}$</p><p>$⇒ \sqrt{{\left(\alpha + 1\right)}^{2} + {\beta}^{2}} = 2\sqrt{10}$</p><p>$⇒ {\left(\alpha + 1\right)}^{2} + {\beta}^{2} = 40 . . . . . . \left(iii\right)$</p><p>Subtracting equation \left(ii\right)$ from equation \left(iii\right)$, we get</p><p>${\left(\alpha + 1\right)}^{2} - {\alpha}^{2} = - 5$</p><p>$⇒ \left(\alpha + 1 + \alpha\right)\left(\alpha + 1 - \alpha\right) = - 5$\left[∵ {a}^{2} - {b}^{2} = \left(a + b\right)\left(a - b\right)\right]$

$⇒ 2\alpha + 1 = - 5$

$⇒ 2\alpha = - 6$

Comparing from equation \left(i\right)$, we get</p><p>$b = 6$</p><p>$∴ {b}^{2} - b = {6}^{2} - 6 = 36 - 6 = 30$.</p><p>Also,${b}^{2} + b = {6}^{2} + 6 = 36 + 6 = 42$.</p><p>Hence, option \left(a\right)$ is correct.