Let $\alpha, \beta$ be the roots of the equation $x^{2}-x+2=0$ with $\operatorname{Im}(\alpha)>\operatorname{Im}(\beta)$. Then $\alpha^{6}+\alpha^{4}+\beta^{4}-5 \alpha^{2}$ is equal to

JEE Main · 2024 · 29 Jan 2024 · Shift 1 · Complex Numbers

Let

α, β

be the roots of the equation

x^{2} - x + 2 = 0

with

Im (α) > Im (β)

. Then

α^{6} + α^{4} + β^{4} - 5 α^{2}

is equal to

Answer: 13

α, β

are roots of

x^{2} - x + 2 = 0

, so

α + β = 1

α β = 2

, and each root satisfies

x^{2} = x - 2

α^{3} = α \cdot α^{2} = α (α - 2) = α^{2} - 2 α = (α - 2) - 2 α = - α - 2

α^{4} = α (- α - 2) = - α^{2} - 2 α = - (α - 2) - 2 α = - 3 α + 2

α^{6} = (α^{3})^{2} = (- α - 2)^{2} = α^{2} + 4 α + 4 = (α - 2) + 4 α + 4 = 5 α + 2

Since

β

satisfies the same relation,

β^{4} = - 3 β + 2

as well.

α^{6} + α^{4} + β^{4} - 5 α^{2} = (5 α + 2) + (- 3 α + 2) + (- 3 β + 2) - 5 (α - 2)

= - 3 α - 3 β + 16 = - 3 (α + β) + 16 = - 3 (1) + 16 = 13

Key confirmed: 13.

JEE Main · 2024 · 29 Jan 2024 · Shift 1 · Integer answer