4-1.Complex numbers
hard

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

A

$b^{2}-b=42$

B

$b^{2}+b=12$

C

$b^{2}+b=72$

D

$b^{2}-b=30$

(JEE MAIN-2020)

Solution

Assuming $z$ is a root of the given equation,

$z=\frac{-b \pm i \sqrt{180-b^{2}}}{2}$

so, $\left(1-\frac{b}{2}\right)^{2}+\frac{180-b^{2}}{4}=40$

$\Rightarrow-4 b+184=160 \Rightarrow b=6$

Standard 11
Mathematics

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