If the equation x 2 + b x + 45 = 0 ,   b ∈ R has conjugate complex roots and they satisfy z + 1 =…

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 $z=\overline{\alpha +i\beta }=\alpha -i\beta$.

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

So, the sum of roots of the given equation, $\left(\alpha +i\beta \right)+\left(\alpha -i\beta \right)=-\frac{b}{1}\Rightarrow 2\alpha =-b ......\left(\mathrm{i}\right)$.

Also, the product of roots of the given equation, $\left(\alpha +i\beta \right)\left(\alpha -i\beta \right)=\frac{45}{1}\Rightarrow {\alpha }^2+{\beta }^2=45 ......\left(\mathrm{ii}\right)$.

Now, given that $\left|z+1\right|=2\sqrt{10}$.

$\Rightarrow \left|\alpha \pm i\beta +1\right|=2\sqrt{10}$

$\Rightarrow \sqrt{{\left(\alpha +1\right)}^2+{\beta }^2}=2\sqrt{10}$

$\Rightarrow {\left(\alpha +1\right)}^2+{\beta }^2=40 ......\left(\mathrm{iii}\right)$

Subtracting equation $\left(\mathrm{ii}\right)$ from equation $\left(\mathrm{iii}\right)$, we get

${\left(\alpha +1\right)}^2-{\alpha }^2=-5$

$\Rightarrow \left(\alpha +1+\alpha \right)\left(\alpha +1-\alpha \right)=-5$ $\left[\because a^2-b^2=\left(a+b\right)\left(a-b\right)\right]$

$\Rightarrow 2\alpha +1=-5$

$\Rightarrow 2\alpha =-6$

Comparing from equation $\left(\mathrm{i}\right)$, we get

$b=6$

$\therefore b^2-b=6^2-6=36-6=30$.

Also, $b^2+b=6^2+6=36+6=42$.

Hence, option $\left(a\right)$ is correct.