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

यदि \(\alpha\) तथा \(\beta\) किसी \(k\) के लिए, समीकरण \(x^{2}-4 \sqrt{2} k x+2 e^{4 \ln k}-1=0\) के मूल हैं तथा \(\alpha^{2}+\beta^{2}=66\), है, तो \(\alpha^{3}+\beta^{3}\) बराबर है

  1. A \(248\sqrt 2 \)
  2. B \(280\sqrt 2 \)
  3. C \(-32\sqrt 2 \)
  4. D \(-280\sqrt 2 \)
Verified Solution

Answer & Solution

Correct Answer

(D) \(-280\sqrt 2 \)

Step-by-step Solution

Detailed explanation

\(x^{2}-4 \sqrt{2} k x+2 e^{4 \ln k}-1=0\) or, \(x^{2}-4 \sqrt{2} k x+2 k^{4}-1=0\) \(\alpha+\beta=4 \sqrt{2} k\) and \(\alpha . \beta=2 k^{4}-1\) Squaring both sides, we get \((\alpha+\beta)^{2}=(4 \sqrt{2} k)^{2}\) \(\Rightarrow \alpha^{2}+\beta^{2}+2 \alpha \beta=32 k^{2}\)…
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