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

Step 1: Find the roots of the quadratic equation.

\[ x^2 - x + 2 = 0 \] \[ x = \frac{1 \pm \sqrt{1 - 8}}{2} = \frac{1 \pm i\sqrt{7}}{2} \]

Hence, \[ \alpha = \frac{1 + i\sqrt{7}}{2}, \quad \beta = \frac{1 - i\sqrt{7}}{2} \] Since \( \text{Im}(\alpha) > \text{Im}(\beta) \), we take: \[ \alpha = \frac{1 + i\sqrt{7}}{2} \]

Step 2: Use relations between roots.

From the equation \( x^2 - x + 2 = 0 \): \[ \alpha + \beta = 1, \quad \alpha\beta = 2 \]

Step 3: Express higher powers of \( \alpha \) and \( \beta \).

Since \( \alpha \) is a root: \[ \alpha^2 = \alpha - 2 \] We will use this recurrence relation: \[ \alpha^2 = \alpha - 2 \]

Step 4: Compute successive powers.

\[ \alpha^3 = \alpha \cdot \alpha^2 = \alpha(\alpha - 2) = \alpha^2 - 2\alpha \] Substitute \( \alpha^2 = \alpha - 2 \): \[ \alpha^3 = (\alpha - 2) - 2\alpha = -\alpha - 2 \]

\[ \alpha^4 = \alpha \cdot \alpha^3 = \alpha(-\alpha - 2) = -\alpha^2 - 2\alpha \] Substitute \( \alpha^2 = \alpha - 2 \): \[ \alpha^4 = -(\alpha - 2) - 2\alpha = -3\alpha + 2 \]

\[ \alpha^5 = \alpha \cdot \alpha^4 = \alpha(-3\alpha + 2) = -3\alpha^2 + 2\alpha \] Substitute \( \alpha^2 = \alpha - 2 \): \[ \alpha^5 = -3(\alpha - 2) + 2\alpha = -3\alpha + 6 + 2\alpha = -\alpha + 6 \]

\[ \alpha^6 = \alpha \cdot \alpha^5 = \alpha(-\alpha + 6) = -\alpha^2 + 6\alpha \] Substitute \( \alpha^2 = \alpha - 2 \): \[ \alpha^6 = -(\alpha - 2) + 6\alpha = -\alpha + 2 + 6\alpha = 5\alpha + 2 \]

Step 5: Compute \( \beta^4 \) similarly.

Since \( \beta \) satisfies the same relation \( \beta^2 = \beta - 2 \): \[ \beta^3 = -\beta - 2 \] \[ \beta^4 = -3\beta + 2 \]

Step 6: Substitute in the given expression.

We need: \[ \alpha^6 + \alpha^4 + \beta^4 - 5\alpha^2 \] Substitute the derived values: \[ \alpha^6 = 5\alpha + 2, \quad \alpha^4 = -3\alpha + 2, \quad \beta^4 = -3\beta + 2, \quad \alpha^2 = \alpha - 2 \]

Now: \[ \alpha^6 + \alpha^4 + \beta^4 - 5\alpha^2 = (5\alpha + 2) + (-3\alpha + 2) + (-3\beta + 2) - 5(\alpha - 2) \] Simplify: \[ = (5\alpha - 3\alpha - 5\alpha) + (-3\beta) + (2 + 2 + 2 + 10) \] \[ = (-3\alpha - 3\beta) + 16 \] \]

Step 7: Use \( \alpha + \beta = 1 \)

\[ -3(\alpha + \beta) + 16 = -3(1) + 16 = 13 \]

Final Answer:

\[ \boxed{13} \]

Answer 2

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:
\(= \alpha^4(\alpha - 2) + \alpha^4 - 5\alpha^2 + (\beta - 2)^2\)
\(= \alpha^5 - \alpha^4 - 5\alpha^2 + \beta^2 - 4\beta + 4\)
\(= \alpha^3(\alpha - 2) - \alpha^4 - 5\alpha^2 + \beta - 2 - 4\beta + 4\)
\(= -2\alpha^3 - 5\alpha^2 - 3\beta + 2\)
\(= -2\alpha(\alpha - 2) - 5\alpha^2 - 3\beta + 2\)
\(= -7\alpha^2 + 4\alpha - 3\beta + 2\)
\(= -7(\alpha - 2) + 4\alpha - 3\beta + 2\)
\(= -3\alpha - 3\beta + 16\)
\(= -3(1) + 16\)
\(= 13\)