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(D) Given the quadratic equation $x^2-\sqrt{2} x+2=0$.Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$,we get $x = \frac{\sqrt{2} \p

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$-128$

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(D) Given the quadratic equation $x^2-\sqrt{2} x+2=0$.Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$,we get $x = \frac{\sqrt{2} \pm \sqrt{2-8}}{2} = \frac{\sqrt{2} \pm \sqrt{-6}}{2} = \frac{\sqrt{2} \pm i\sqrt{6}}{2}$.We can write the roots in polar form: $\alpha = \sqrt{2} \left( \frac{1}{2} + i\frac{\sqrt{3}}{2} ight) = \sqrt{2} e^{i\pi/3}$ and $\beta = \sqrt{2} e^{-i\pi/3}$.Now,calculate $\alpha^{14} + \beta^{14}$:$\alpha^{14} = (\sqrt{2})^{14} e^{i14\pi/3} = 2^7 e^{i(4\pi + 2\pi/3)} = 128 e^{i2\pi/3}$.$\beta^{14} = (\sqrt{2})^{14} e^{-i14\pi/3} = 128 e^{-i2\pi/3}$.$\alpha^{14} + \beta^{14} = 128 (e^{i2\pi/3} + e^{-i2\pi/3}) = 128 	imes 2 \cos(2\pi/3)$.Since $\cos(2\pi/3) = -1/2$,we have $\alpha^{14} + \beta^{14} = 256 	imes (-1/2) = -128$.

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

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