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

Given equation

$ x^2 - \sqrt{2}x + 2 = 0 $

Let roots be $ \alpha, \beta $

Then

$ \alpha + \beta = \sqrt{2} $

$ \alpha \beta = 2 $

We use recurrence:

$ S_n = \alpha^n + \beta^n $

Formula:

$ S_n = (\alpha + \beta) S_{n-1} - (\alpha \beta) S_{n-2} $

Initial values:

$ S_0 = 2 $

$ S_1 = \sqrt{2} $

Compute stepwise:

$ S_2 = \sqrt{2} \cdot \sqrt{2} - 2 \cdot 2 = 2 - 4 = -2 $

$ S_3 = \sqrt{2}(-2) - 2(\sqrt{2}) = -2\sqrt{2} - 2\sqrt{2} = -4\sqrt{2} $

$ S_4 = \sqrt{2}(-4\sqrt{2}) - 2(-2) = -8 + 4 = -4 $

Pattern:

$ S_2 = -2 $

$ S_4 = -4 $

$ S_6 = -8 $

$ S_8 = -16 $

$ S_{10} = -32 $

$ S_{12} = -64 $

$ S_{14} = -128 $

Final Answer:

$ \boxed{-128} $