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JEE Mains · Maths · STD 11 - 4.1 complex nubers

यदि समीकरण \(x ^{2}+ bx +45=0,( b \in R )\) के संयुग्मी सम्मिश्र मूल हैं, जो \(|z+1|=2 \sqrt{10}\) को संतुष्ट करते हैं, तो

  1. A \(b^{2}-b=42\)
  2. B \(b^{2}+b=12\)
  3. C \(b^{2}+b=72\)
  4. D \(b^{2}-b=30\)
Verified Solution

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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