[UPLIFTJEE]

Let $\alpha, \beta$ be the roots of the equation $x^{2}-\sqrt{2} x+2=0$. Then $\alpha^{14}+\beta^{14}$ is equal to [JEE (Main) 13 April 2023 SII]
(A) -128
(B) $-64 \sqrt{2}$
(C) -64
(D) $-128 \sqrt{2}$

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Solution

$\mathrm{x}^{2}-\sqrt{2} \mathrm{x}+2=0 \Rightarrow \mathrm{x}=\frac{\sqrt{2} \pm \sqrt{6} \mathrm{i}}{2} \Rightarrow \alpha=\frac{\sqrt{2}+\sqrt{6} \mathrm{i}}{2}=\sqrt{2} \mathrm{e}^{\frac{\mathrm{i} \pi}{3}}$

\[ \begin{aligned} & \beta=\sqrt{2} \mathrm{e}^{-\frac{\mathrm{i} \pi}{3}} \Rightarrow \alpha^{14}=2^{7} \mathrm{e}^{\frac{\mathrm{i} 14 \pi}{3}}=128\left[\mathrm{e}^{\frac{\mathrm{i} 2 \pi}{3}}\right] \Rightarrow \beta^{14}=128\left[\mathrm{e}^{-\frac{\mathrm{i} 2 \pi}{3}}\right] \\ & \alpha^{14}+\beta^{14}=128(2) \cos \left(\frac{2 \pi}{3}\right)=-128 \end{aligned} \]

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Answer

A

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