Question

If \( \alpha \) and \( \beta \) are the roots of \( x^{2} - 2x + 4 = 0 \), then the value of \( \alpha^{6} + \beta^{6} \) is

• A. 32
• B. 64
• C. 128
• D. 256

Hint:

Roots of $x^2 - x + 1$ are $-\omega, -\omega^2$. Use these for higher powers.

Answer 1

The Correct Option is C

Step 1: Solve Equation
For $x^2 - 2x + 4 = 0$, roots are $\frac{2 \pm \sqrt{4 - 16}}{2} = 1 \pm \sqrt{3}i$.
Step 2: Omega Relation
$\alpha = 1 + \sqrt{3}i = -2(\frac{-1-\sqrt{3}i}{2}) = -2\omega^2$. $\beta = 1 - \sqrt{3}i = -2(\frac{-1+\sqrt{3}i}{2}) = -2\omega$.
Step 3: Power Calculation
$\alpha^6 + \beta^6 = (-2\omega^2)^6 + (-2\omega)^6 = 64\omega^{12} + 64\omega^6$.
Step 4: Use $\omega^3 = 1$
Since $\omega^3 = 1$, then $\omega^{12} = 1$ and $\omega^6 = 1$. Value = $64(1) + 64(1) = 128$.
Final Answer: (c)