Solveeit Logo
Login

Question

Mathematics Question on complex numbers

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

Answer

We aim to compute:
α6+α4+β4−5α2\alpha^6 + \alpha^4 + \beta^4 - 5\alpha^2
Since α and β satisfyx2−x+2=0 x^2 - x + 2 = 0, we know:
α2=α−2,β2=β−2\alpha^2 = \alpha - 2, \quad \beta^2 = \beta - 2

Using these relations, we compute higher powers of α:
=α4(α−2)+α4−5α2+(β−2)2= \alpha^4(\alpha - 2) + \alpha^4 - 5\alpha^2 + (\beta - 2)^2
=α5−α4−5α2+β2−4β+4= \alpha^5 - \alpha^4 - 5\alpha^2 + \beta^2 - 4\beta + 4
=α3(α−2)−α4−5α2+β−2−4β+4= \alpha^3(\alpha - 2) - \alpha^4 - 5\alpha^2 + \beta - 2 - 4\beta + 4
=−2α3−5α2−3β+2= -2\alpha^3 - 5\alpha^2 - 3\beta + 2
=−2α(α−2)−5α2−3β+2= -2\alpha(\alpha - 2) - 5\alpha^2 - 3\beta + 2
=−7α2+4α−3β+2= -7\alpha^2 + 4\alpha - 3\beta + 2
=−7(α−2)+4α−3β+2= -7(\alpha - 2) + 4\alpha - 3\beta + 2
=−3α−3β+16= -3\alpha - 3\beta + 16
=−3(1)+16= -3(1) + 16
=13= 13