Question 22. Let \alpha,\beta be the roots of the equation x^{2}-x+2=0, with \operatorname{Im}(\alpha)>\operatorname{Im}(\beta).
Then the value of \alpha^{6}+\alpha^{4}+\beta^{4}-5\alpha^{2} is
Answer: 13
Explanation:
Consider the expression
\alpha^{6}+\alpha^{4}+\beta^{4}-5\alpha^{2}.
Rewrite it as =\alpha^{4}(\alpha-2)+\alpha^{4}-5\alpha^{2}+(\beta-2)^{2}.
This gives =\alpha^{5}-\alpha^{4}-5\alpha^{2}+\beta^{2}-4\beta+4.
Rewrite again as
=\alpha^{3}(\alpha-2)-\alpha^{4}-5\alpha^{2}+\beta-2-4\beta+4.
So, =-2\alpha^{3}-5\alpha^{2}-3\beta+2.
Now, =-2\alpha(\alpha-2)-5\alpha^{2}-3\beta+2.
This simplifies to =-7\alpha^{2}+4\alpha-3\beta+2.
Rewrite again as =-7(\alpha-2)+4\alpha-3\beta+2.
So, =-3\alpha-3\beta+16.
Using \alpha+\beta=1,
we get =-3(1)+16=13.
Hence, the required value is 13.