Let , be the roots of the equation x^2-x+2=0 with Im( )Im( ).

Question

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

Answer: 13

Accepted Answer

Both roots satisfy $x^2=x-2$. Reducing powers of $\alpha$ (and similarly $\beta$) using this relation:
$$\alpha^2=\alpha-2,\quad \alpha^3=\alpha\cdot\alpha^2=\alpha(\alpha-2)=\alpha^2-2\alpha=-\alpha-2,$$
$$\alpha^4=\alpha\cdot\alpha^3=-\alpha^2-2\alpha=-3\alpha+2.$$

Similarly $\beta^4=-3\beta+2$.

$$\alpha^6=\alpha^2\cdot\alpha^4=(\alpha-2)(-3\alpha+2)=-3\alpha^2+8\alpha-4=-3(\alpha-2)+8\alpha-4=5\alpha+2.$$

Now:
$$\alpha^6+\alpha^4+\beta^4-5\alpha^2 = (5\alpha+2)+(-3\alpha+2)+(-3\beta+2)-5(\alpha-2)$$
$$= 5\alpha+2-3\alpha+2-3\beta+2-5\alpha+10 = -3\alpha-3\beta+16 = -3(\alpha+\beta)+16.$$

Since $\alpha+\beta=1$ (sum of roots of $x^2-x+2=0$):
$= -3(1)+16=13.$

Solutions · 1