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:
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 $b^{2}-b=30$