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Question: Answered & Verified by Expert
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 $\alpha^{2023}+\beta^{2023}+\gamma^{2022}+\delta^{2022}=$
MathematicsQuadratic EquationTS EAMCETTS EAMCET 2023 (13 May Shift 2)
Options:
  • A 1
  • B 0
  • C $1+3 \omega$
  • D $\omega-2 \omega^2$
Solution:
2982 Upvotes Verified Answer
The correct answer is: 1
$\begin{aligned} & \text x^4+x^2+1=0 \\ & \left(x^2+x+1\right)\left(x^2-x+1\right)=0 \\ & x^2+x+1=0 \text { have roots } \omega, \omega^2 \\ & x^2-x+1=0 \text { have roots }-\omega,-\omega^2 \\ & \omega+\omega^2=-1 \Rightarrow \alpha=\omega, \beta=\omega^2 \\ & -\omega-\omega^2=1 \Rightarrow \gamma=-\omega, \delta=-\omega^2 \\ & \therefore \quad \alpha^{2023}+\beta^{2023}+\gamma^{2022}+\delta^{2022} \\ & \quad=(\omega)^{2023}+\left(\omega^2\right)^{2023}+(-\omega)^{2022}+\left(-\omega^2\right)^{2022} \\ & \quad=\omega^{2023}+\omega^{4046}+\omega^{2022}+\omega^{4044} \\ & \quad=\omega+\omega^2+1+1=1 .\end{aligned}$

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