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

यदि \(\alpha\) तथा \(\beta\) समीकरण \(x^{2}+2 x+2=0\), के दो मूल है, तो \(\alpha^{15}+\beta^{15}\) बराबर है-

  1. A \(-256\)
  2. B \(512\)
  3. C \(-512\)
  4. D \(256\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(-256\)

Step-by-step Solution

Detailed explanation

\(x^{2}+2 x+2=0\) \(\Rightarrow(x+1)^{2}=-1\) \(x=-1 \pm i=\sqrt{2} e^{i\left(\pm \frac{3 \pi}{4}\right)}\) \(\therefore \alpha^{15}, \beta^{15}=(\sqrt{2})^{15} \times 2 \cos \left(15 . \frac{3 \pi}{4}\right)\) \(=2^{8} \sqrt{2} \times\left(-\frac{1}{\sqrt{2}}\right)=-256\)
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