Question

If $\alpha, \beta$ are the roots of the quadratic equation $x^2 - 2x + 4 = 0$, then the value of $\alpha^n + \beta^n$ is:

• A. $2^{n+1} \cos\left(\frac{n\pi}{3}\right)$
• B. $2^{n+1} \sin\left(\frac{n\pi}{3}\right)$
• C. $2^n \cos\left(\frac{n\pi}{3}\right)$
• D. $2^n \sin\left(\frac{n\pi}{3}\right)$

Hint:

Test for $n = 1$: $\alpha^1 + \beta^1$ is the sum of the roots, which is $2$. Plugging $n = 1$ into option (A) gives $2^{1+1} \cos(\pi/3) = 4 \times 0.5 = 2$, validating the answer instantly.

Answer 1

The Correct Option is A

Step 1: Concept
Find the complex roots using the quadratic formula, express them in trigonometric polar form, and apply De Moivre's Theorem to find their powers.

Step 2: Meaning
The given quadratic equation is $x^2 - 2x + 4 = 0$.

Step 3: Analysis
Finding the roots: \[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(4)}}{2} = \frac{2 \pm \sqrt{4 - 16}}{2} = 1 \pm i\sqrt{3} \] Converting these roots into polar form: \[ \alpha = 1 + i\sqrt{3} = 2 \left( \frac{1}{2} + i \frac{\sqrt{3}}{2} \right) = 2 \left( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} \right) \] \[ \beta = 1 - i\sqrt{3} = 2 \left( \frac{1}{2} - i \frac{\sqrt{3}}{2} \right) = 2 \left( \cos \frac{\pi}{3} - i \sin \frac{\pi}{3} \right) \] Using De Moivre's Theorem for $n$-th powers: \[ \alpha^n = 2^n \left( \cos \frac{n\pi}{3} + i \sin \frac{n\pi}{3} \right) \] \[ \beta^n = 2^n \left( \cos \frac{n\pi}{3} - i \sin \frac{n\pi}{3} \right) \] Summing the powers: \[ \alpha^n + \beta^n = 2^n \left( 2 \cos \frac{n\pi}{3} \right) = 2^{n+1} \cos\left(\frac{n\pi}{3}\right) \]

Step 4: Conclusion
Thus, the sum of the $n$-th powers of the roots is $2^{n+1} \cos\left(\frac{n\pi}{3}\right)$. Final Answer: (A)