Question

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

Answer 1

Correct Answer: 13

We aim to compute:
\(\alpha^6 + \alpha^4 + \beta^4 - 5\alpha^2\)
Since α and β satisfy\( x^2 - x + 2 = 0\), we know:
\(\alpha^2 = \alpha - 2, \quad \beta^2 = \beta - 2\)

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