Let alpha and beta be the roots of x^2 - sqrt | vedclass

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(A) Given the quadratic equation $x^2 - \sqrt{2}x + 1 = 0$.Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get:$x = \frac{\sqr

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(A) Given the quadratic equation $x^2 - \sqrt{2}x + 1 = 0$.Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get:$x = \frac{\sqrt{2} \pm \sqrt{2 - 4}}{2} = \frac{\sqrt{2} \pm \sqrt{-2}}{2} = \frac{\sqrt{2} \pm i\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \pm i\frac{1}{\sqrt{2}}$.These roots can be written in polar form as $\alpha = \cos(\frac{\pi}{4}) + i\sin(\frac{\pi}{4}) = e^{i\pi/4}$ and $\beta = \cos(-\frac{\pi}{4}) + i\sin(-\frac{\pi}{4}) = e^{-i\pi/4}$.Now,we need to find $\alpha^{50} + \beta^{50}$.$\alpha^{50} = (e^{i\pi/4})^{50} = e^{i50\pi/4} = e^{i25\pi/2} = e^{i(12\pi + \pi/2)} = e^{i\pi/2} = i$.$\beta^{50} = (e^{-i\pi/4})^{50} = e^{-i50\pi/4} = e^{-i25\pi/2} = e^{-i(12\pi + \pi/2)} = e^{-i\pi/2} = -i$.Therefore,$\alpha^{50} + \beta^{50} = i + (-i) = 0$.

Let $\alpha$ and $\beta$ be the roots of $x^2 - \sqrt{2}x + 1 = 0$. Then the value of $\alpha^{50} + \beta^{50}$ is:

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