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QJEEM24Ma1853: JEE Main 2024 Mathematics - Class 11 JEEM24Ma1853 Solution: Let be the roots of the equation with . Then is equal to…

Let α,β\alpha, \beta be the roots of the equation x2−x+2=0x^2-x+2=0 with \operatorname{Im}(\alpha)>\operatorname{Im}(\beta) . Then α6+α4+β4−5α2\alpha^6+\alpha^4+\beta^4-5 \alpha^2 is equal to ___________.

Step-by-Step Verified Solution

amp;α6+α4+β4−5α2amp;=α4(α−2)+α4−5α2+(β−2)2amp;=α5−α4−5α2+β2−4β+4amp;=α3(α−2)−α4−5α2+β−2−4β+4amp;=−2α3−5α2−3β+2amp;=−2α(α−2)−5α2−3β+2amp;=−7α2+4α−3β+2amp;=−7(α−2)+4α−3β+2amp;=−3α−3β+16=−3(1)+16=13\begin{aligned}& \alpha^6+\alpha^4+\beta^4-5 \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(1)+16=13\end{aligned}

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