Related Formula
Since α$\alpha$ is a root of x² - x + 2 = 0$x^2 - x + 2 = 0$, it must satisfy the equation exactly:
α² - α + 2 = 0 ⇒ α² = α - 2$$\alpha^2 - \alpha + 2 = 0 \Rightarrow \alpha^2 = \alpha - 2$$
This is an essential root reduction property allowing polynomials of high degrees to be collapsed linearly.
Core Logic
We need to evaluate the expression E = α⁶ + α⁴ + β⁴ - 5α²$E = \alpha^6 + \alpha^4 + \beta^4 - 5\alpha^2$.
Use the substitution α² = α - 2$\alpha^2 = \alpha - 2$ to iteratively depress the powers of α$\alpha$.
α⁴ = (α²)² = (α - 2)² = α² - 4α + 4$$\alpha^4 = (\alpha^2)^2 = (\alpha - 2)^2 = \alpha^2 - 4\alpha + 4$$
Substitute α² = α - 2$\alpha^2 = \alpha - 2$ again into the result:
α⁴ = (α - 2) - 4α + 4 = -3α + 2$$\alpha^4 = (\alpha - 2) - 4\alpha + 4 = -3\alpha + 2$$
Now, generate α⁶$\alpha^6$ using α⁴$\alpha^4$:
α⁶ = α⁴ · α² = (-3α + 2)(α - 2) = -3α² + 6α + 2α - 4 = -3α² + 8α - 4$$\alpha^6 = \alpha^4 \cdot \alpha^2 = (-3\alpha + 2)(\alpha - 2) = -3\alpha^2 + 6\alpha + 2\alpha - 4 = -3\alpha^2 + 8\alpha - 4$$
Substitute α² = α - 2$\alpha^2 = \alpha - 2$ into the result again:
α⁶ = -3(α - 2) + 8α - 4 = -3α + 6 + 8α - 4 = 5α + 2$$\alpha^6 = -3(\alpha - 2) + 8\alpha - 4 = -3\alpha + 6 + 8\alpha - 4 = 5\alpha + 2$$
Step 1: Simplify the Full Expression
The symmetry of the roots dictates that β⁴$\beta^4$ behaves identically to α⁴$\alpha^4$. Thus:
β⁴ = -3β + 2$$\beta^4 = -3\beta + 2$$
Substitute all depressed linear forms back into E = α⁶ + α⁴ + β⁴ - 5α²$E = \alpha^6 + \alpha^4 + \beta^4 - 5\alpha^2$:
E = (5α + 2) + (-3α + 2) + (-3β + 2) - 5(α - 2)$$E = (5\alpha + 2) + (-3\alpha + 2) + (-3\beta + 2) - 5(\alpha - 2)$$
E = 5α - 3α - 5α - 3β + 2 + 2 + 2 + 10$$E = 5\alpha - 3\alpha - 5\alpha - 3\beta + 2 + 2 + 2 + 10$$
E = -3α - 3β + 16$$E = -3\alpha - 3\beta + 16$$
E = -3(α + β) + 16$$E = -3(\alpha + \beta) + 16$$
Step 2: Apply Sum of Roots
From the original quadratic equation x² - x + 2 = 0$x^2 - x + 2 = 0$, the sum of roots is:
α + β = -(-1)/(1) = 1$$\alpha + \beta = -\frac{-1}{1} = 1$$
Substitute this back:
E = -3(1) + 16 = 13$$E = -3(1) + 16 = 13$$
(Note: The condition Im(α) gt Im(β)$Im(\alpha) \gt Im(\beta)$ was a distractor since the expression simplified perfectly symmetrically into α + β$\alpha + \beta$ without needing the individual complex values of the roots).
Pattern Recognition
Never compute De Moivre polar forms for high root powers unless the quadratic has roots like ω$\omega$ or i$i$. Always use the characteristic quadratic relation α² = pα + q$\alpha^2 = p\alpha + q$ to rapidly step down degrees until everything is strictly linear.
Chapter Mix
Class 11 Mathematics: Complex Numbers and Quadratic Equations
