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

<ol> <li><p><strong>Find the roots of the quadratic equation:</strong></p> <p>The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. For the quadratic equation $x^2 - \sqrt{2}x + 2 = 0$, we have $a = 1, b = -\sqrt{2}, c = 2$. Plugging these into the quadratic formula gives: <br/><br/>$$x = \frac{\sqrt{2} \pm \sqrt{(-\sqrt{2})^2 - 4(1)(2)}}{2(1)} = \frac{\sqrt{2} \pm \sqrt{2 - 8}}{2} = \frac{\sqrt{2} \pm \sqrt{6}i}{2}.$$ <br/><br/>So, we have two roots, $\alpha$ and $\beta$, which are: <br/><br/>$\alpha = \frac{\sqrt{2} + \sqrt{6}i}{2},$ <br/><br/>$\beta = \frac{\sqrt{2} - \sqrt{6}i}{2}.$</p> </li> <li><p><strong>Express the roots in exponential form:</strong></p> <p>We can express complex numbers in the form $re^{i\theta}$. For $\alpha$ and $\beta$, we find the magnitude $r = \sqrt{2}$ and the arguments $\theta = \frac{\pi}{3}, -\frac{\pi}{3}$ respectively. So, we have: <br/><br/>$\alpha = \sqrt{2}e^{i\frac{\pi}{3}},$ <br/><br/>$\beta = \sqrt{2}e^{-i\frac{\pi}{3}}.$</p> </li> <li><p><strong>Calculate the 14th power of the roots:</strong></p> <p>To find $\alpha^{14}$ and $\beta^{14}$, we use the property of exponents which says that $(a^m)^n = a^{mn}$. So, we have: <br/><br/>$$\alpha^{14} = (\sqrt{2}e^{i\frac{\pi}{3}})^{14} = 2^7e^{i\frac{14\pi}{3}} = 128e^{i\frac{2\pi}{3}},$$ <br/><br/>$$\beta^{14} = (\sqrt{2}e^{-i\frac{\pi}{3}})^{14} = 2^7e^{-i\frac{14\pi}{3}} = 128e^{-i\frac{2\pi}{3}}.$$</p> </li> <li><p><strong>Add the 14th powers of the roots:</strong></p> <p>We want to find the real part of $\alpha^{14} + \beta^{14}$. To do this, we use the property that $e^{ix} = \cos(x) + i\sin(x)$. We have: <br/><br/>$$\alpha^{14} + \beta^{14} = 128e^{i\frac{2\pi}{3}} + 128e^{-i\frac{2\pi}{3}} = 128(2)\cos\left(\frac{2\pi}{3}\right) = -128.$$</p> </li> </ol>