If the equation,
Question:

If the equation, $x^{2}+b x+45=0(b \in R)$ has conjugate complex roots and they satisfy $|z+1|=2 \sqrt{10}$, then:

1. (1) $b^{2}-b=30$

2. (2) $b^{2}+b=72$

3. (3) $b^{2}-b=42$

4. (4) $b^{2}+b=12$

Correct Option: 1

Solution:

Let $z=\alpha \pm$ i $\beta$ be the complex roots of the equation

So, sum of roots $=2 \alpha=-b$ and

Product of roots $=\alpha^{2}+\beta^{2}=45$

$(\alpha+1)^{2}+\beta^{2}=40$

Given, $|z+1|=2 \sqrt{10}$

$\Rightarrow \quad(\alpha+1)^{2}-\alpha^{2}=-5$ $\left[\because \beta^{2}=45-\alpha^{2}\right]$

$\Rightarrow \quad 2 \alpha+1=-5 \Rightarrow 2 \alpha=-6$

Hence, $b=6$ and $b^{2}-b=30$