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 ^4-5α ^2 is equal to

Given equation ${x}^{2}-x+2$x2−x+2 has roots \alpha &\beta,

So, $\alpha +\beta =1,\alpha \beta =2$α+β=1,αβ=2

$$\Rightarrow {\alpha }^{4}+{\beta }^{4}={({\alpha }^{2}+{\beta }^{2})}^{2}-2{\alpha }^{2}{\beta }^{2}$$⇒α4+β4=(α2+β2)2−2α2β2

$$\Rightarrow {\alpha }^{4}+{\beta }^{4}={[{(\alpha +\beta )}^{2}-2\alpha \beta ]}^{2}-2{\alpha }^{2}{\beta }^{2}$$⇒α4+β4=[(α+β)2−2αβ]2−2α2β2

$$\Rightarrow {\alpha }^{4}+{\beta }^{4}={(1-4)}^{2}-2\times 4$$⇒α4+β4=(1−4)2−2×4

$$\Rightarrow {\alpha }^{4}+{\beta }^{4}=1$$⇒α4+β4=1

It is given that, $\alpha$α is a root of ${x}^{2}-x+2$x2−x+2.

$$\Rightarrow {\alpha }^{2}-\alpha +2=0$$⇒α2−α+2=0

$$\Rightarrow {\alpha }^{2}=\alpha -2$$⇒α2=α−2

$$\Rightarrow {\alpha }^{4}={\alpha }^{2}+4-4\alpha$$⇒α4=α2+4−4α

$$\Rightarrow {\alpha }^{4}=\alpha -2+4-4\alpha$$⇒α4=α−2+4−4α

$$\Rightarrow {\alpha }^{4}=2-3\alpha$$⇒α4=2−3α

Now, ${\alpha }^{6}-5{\alpha }^{2}={\alpha }^{2}({\alpha }^{4}-5)$α6−5α2=α2(α4−5)

$$\Rightarrow {\alpha }^{6}-5{\alpha }^{2}=(\alpha -2)(2-3\alpha -5)$$⇒α6−5α2=(α−2)(2−3α−5)

$$\Rightarrow {\alpha }^{6}-5{\alpha }^{2}=(\alpha -2)(-3\alpha -3)$$⇒α6−5α2=(α−2)(−3α−3)

$$\Rightarrow {\alpha }^{6}-5{\alpha }^{2}=-3({\alpha }^{2}-\alpha -2)$$⇒α6−5α2=−3(α2−α−2)

$$\Rightarrow {\alpha }^{6}-5{\alpha }^{2}=-3(\alpha -2-\alpha -2)$$⇒α6−5α2=−3(α−2−α−2)

$$\Rightarrow {\alpha }^{6}-5{\alpha }^{2}=12$$⇒α6−5α2=12

$$\Rightarrow {\alpha }^{6}+{\alpha }^{4}+{\beta }^{4}-5{\alpha }^{2}=13$$⇒α6+α4+β4−5α2=13