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

The given quadratic equation is: \[ x^2 - \sqrt{2}x + 2 = 0 \] Let \( \alpha \) and \( \beta \) be the roots of this equation. The roots are given by: \[ x = \frac{\sqrt{2} \pm \sqrt{2 - 8}}{2} = \frac{\sqrt{2} \pm \sqrt{-6}}{2} \] \[ x = \frac{\sqrt{2} \pm i\sqrt{6}}{2} \] Thus, the roots are: \[ \alpha = -\sqrt{2} \omega, \quad \beta = -\sqrt{2} \omega^2 \] where \( \omega \) is a cube root of unity, i.e., \( \omega^3 = 1 \) and \( \omega^2 + \omega + 1 = 0 \). Now, we calculate \( \alpha^{14} + \beta^{14} \): \[ \alpha^{14} + \beta^{14} = 2^7 \left( \omega^{14} + \omega^{28} \right) \] Since \( \omega^3 = 1 \), we simplify the exponents: \[ \omega^{14} = \omega^2 \quad \text{and} \quad \omega^{28} = \omega \] Thus, \[ \alpha^{14} + \beta^{14} = 2^7 (\omega^2 + \omega) \] Using the identity \( \omega^2 + \omega = -1 \), we get: \[ \alpha^{14} + \beta^{14} = 2^7 (-1) = -128 \]