ExamBro
ExamBro
JEE Mains · Maths · STD 11 - 4.2 Quadratic equations and inequations

यदि समीकरणों \(x^{2}+b x-1=0\) तथा \(x^{2}+x+b=0\) का \(-1\) से भिन्न एक सांझा मूल है, तो \(|b|\) बराबर है

  1. A \(2\)
  2. B \(3\)
  3. C \(\sqrt 3\)
  4. D \(\sqrt 2\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\sqrt 3\)

Step-by-step Solution

Detailed explanation

\(x^{2}+b x-1=0\) common root \(x^{2}+x+b=0\) \(x=\frac{b+1}{b-1}\) Put \(x=\frac{b+1}{b-1}\) in equation \(..........(2)\) \(\left(\frac{b+1}{b-1}\right)^{2}+\left(\frac{b+1}{b-1}\right)+b=0\) \((b+1)^{2}+(b+1)(b-1)+b(b-1)^{2}=0\)…
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app