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Mathematics · Complex Numbers

JEE Main 2024 — 29 January, Shift 1 — Question 22

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

Answer: 13

Numerical answer — enter this value.

Step-by-step solution

α6+α4+β4−5α2=α4(α−2)+α4−5α2+(β−2)2=α5−α4−5α2+β2−4β+4=α3(α−2)−α4−5α2+β−2−4β+4=−2α3−5α2−3β+2=−2α(α−2)−5α2−3β+2=−7α2+4α−3β+2=−7(α−2)+4α−3β+2=−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} α6+α4+β4−5α2=13\boxed{\alpha^{6}+\alpha^{4}+\beta^{4}-5 \alpha^{2}=13}

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Exam
JEE Main 2024
Paper
29 January, Shift 1
Subject
Mathematics
Chapter
Complex Numbers
Topic
Introduction to Complex Numbers

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