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Let $\alpha, \beta$ be the roots of the equation $\mathrm{x}^{2}-\mathrm{x}+2=0$ with $\operatorname{Im}(\alpha)>\operatorname{Im}(\beta)$. Then $\alpha^{6}+\alpha^{4}+\beta^{4}-5 \alpha^{2}$ is equal to [JEE(Main) 29 Jan 2024 S I]

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Solution

$\alpha \& \beta$ are roots of $x^{2}-x+2=0 \Rightarrow \alpha+\beta=1, \alpha^{2}-\alpha+2=0, \alpha^{2}=\alpha-2$

\[ \begin{aligned} & \beta^{2}-\beta+2=0 \Rightarrow \beta^{2}=\beta-2 \Rightarrow \alpha^{6}+\alpha^{4}+\beta^{4}-5 \alpha^{2}=\alpha^{4}(\alpha-2)=\alpha^{4}(\alpha-2)+\alpha^{4}-5 \alpha^{2}+(\beta-2)^{2} \\ & =\alpha^{5}-\alpha^{4}-5 \alpha^{2}+\beta^{2}-4 \beta+4=\alpha^{3}(\alpha-2)-\alpha^{4}-5 \alpha^{2}+\beta-2-4 \beta+4=-2 \alpha^{3}-5 \alpha^{2}-3 \beta+2=-2 \alpha(\alpha-2)-5 \alpha^{2}-3 \beta+2 \\ & =-7 \alpha^{2}+4 \alpha-3 \beta+2=-7(\alpha-2)+4 \alpha-3 \beta+2=-3 \alpha-3 \beta+16=-3(\alpha+\beta)+16=13=-3(1)+16=13 \end{aligned} \]

[UPLIFTJEE]

Answer

13

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