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TS EAMCET · Maths · Quadratic Equation

If \(\alpha, \beta, \gamma, \delta\) are the roots of the equation \(x^4+x^2+1=0\), then \(\frac{\alpha^3+\beta^3+\gamma^3+\delta^3}{\alpha^6+\beta^6+\gamma^6+\delta^6}=\)

  1. A 0
  2. B 1
  3. C -1
  4. D \(\frac{1}{2}\)
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Answer & Solution

Correct Answer

(A) 0

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Detailed explanation

Given \(\alpha, \beta, \gamma, \delta\) are roots of \(x^4+x^2+1=0\) \(\therefore \quad x^2=\frac{-1 \pm \sqrt{1-4}}{2}=\frac{-1 \pm \sqrt{3} i}{2}\) \(\therefore \quad x= \pm \sqrt{\frac{1+\sqrt{3} i}{2}}, x= \pm \sqrt{\frac{-1-\sqrt{3} i}{2}}\) Let…