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JEE Mains · Maths · STD 11 - 4.1 complex nubers

Let \(\alpha, \beta\) be the roots of the quadratic equation \(x^2+\sqrt{6} x+3=0\). Then \(\frac{\alpha^{23}+\beta^{23}+\alpha^{14}+\beta^{14}}{\alpha^{15}+\beta^{15}+\alpha^{10}+\beta^{10}}\) is equal to 

  1. A \(729\)
  2. B \(72\)
  3. C \(81\)
  4. D \(9\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(81\)

Step-by-step Solution

Detailed explanation

\(\alpha, \beta=\frac{-\sqrt{6} \pm \sqrt{6-12}}{2}=\frac{-\sqrt{6} \pm \sqrt{6} i }{2}\) \(=\sqrt{3} e ^{ \pm \frac{3 \pi i }{4}}\) Required expression…