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TS EAMCET · Maths · Quadratic Equation

If \(\alpha, \beta, \gamma, \delta\) are the roots of the equation \(x^4+x^2+1=0\) such that \(\alpha+\beta=-1, \gamma+\delta=1, \alpha^2=\beta\) and \(\gamma^2=-\delta\), then \(\alpha^{2023}+\beta^{2023}+\gamma^{2022}+\delta^{2022}=\)

  1. A 1
  2. B 0
  3. C \(1+3 \omega\)
  4. D \(\omega-2 \omega^2\)
Verified Solution

Answer & Solution

Correct Answer

(A) 1

Step-by-step Solution

Detailed explanation

\begin{aligned} & \text x^4+x^2+1=0 \\ & \left(x^2+x+1\right)\left(x^2-x+1\right)=0 \\ & x^2+x+1=0 \text { have roots } \omega, \omega^2 \\ & x^2-x+1=0 \text { have roots }-\omega,-\omega^2 \\ & \omega+\omega^2=-1 \Rightarrow \alpha=\omega, \beta=\omega^2 \\ & -\omega-\omega^2=1…