JEE Mains · Maths · STD 11 - 4.1 complex nubers
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 \(b^{2}-b=30\)
Answer & Solution
Correct Answer
(D) \(b^{2}-b=30\)
Step-by-step Solution
Detailed explanation
Assuming \(z\) is a root of the given equation, \(z=\frac{-b \pm i \sqrt{180-b^{2}}}{2}\) so, \(\left(1-\frac{b}{2}\right)^{2}+\frac{180-b^{2}}{4}=40\) \(\Rightarrow-4 b+184=160 \Rightarrow b=6\)
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