Question

Let \( \alpha, \beta \) be the roots of the equation \( x^2 - \sqrt{2}x + 2 = 0 \), then \( \alpha^{14} + \beta^{14} \) is equal to:

• A. -256
• B. -128
• C. \(-128\sqrt2\)
• D. \(-256\sqrt2\)

Answer 1

The Correct Option is B

To solve for \( \alpha^{14} + \beta^{14} \), where \( \alpha \) and \( \beta \) are the roots of the quadratic equation \( x^2 - \sqrt{2}x + 2 = 0 \), follow these steps:

1. Identify the roots of the quadratic equation using Vieta's formulas. Based on the equation \( x^2 - \sqrt{2}x + 2 = 0 \):
   • The sum of the roots \( \alpha + \beta = \sqrt{2} \).
   • The product of the roots \( \alpha \beta = 2 \).
2. To calculate \( \alpha^{14} + \beta^{14} \), use the recurrence relation derived from powers of roots:
   • We start with \( \alpha^n + \beta^n \) which can be expressed using the recurrence: \[ \alpha^n + \beta^n = (\alpha^{n-1} + \beta^{n-1})\sqrt{2} - 2(\alpha^{n-2} + \beta^{n-2}) \]
3. Calculate initial conditions:
   • \(\alpha^0 + \beta^0 = 2\)
   • \(\alpha^1 + \beta^1 = \sqrt{2}\)
   • \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = (\sqrt{2})^2 - 2 \cdot 2 = 2 - 4 = -2\)
4. Using the recurrence relation, calculate successive terms up to \( \alpha^{14} + \beta^{14} \):
   • \(\alpha^3 + \beta^3 = (\alpha^2 + \beta^2)\sqrt{2} - 2\cdot(\alpha^1 + \beta^1) = -2\sqrt{2} - 2\sqrt{2} = -4\sqrt{2}\)
   • Continuing this process until \( \alpha^{14} + \beta^{14} \), employing the periodicity of the sequence formed by the powers of roots.
5. Finally, utilize the pattern observed for powers modulo a certain period (e.g., mod 6) to compute \( \alpha^{14} + \beta^{14} \) efficiently.
6. Calculating further, we find that \( \alpha^{14} + \beta^{14} = -128 \). This result is consistent with our solution's choice.

The correct answer is therefore -128.

This step-by-step calculation demonstrates how using algebraic identities and relations can simplify the computation without directly calculating each power individually.