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Question

(C) Given the quadratic equation $x^2-4x+8=0$. Using the quadratic formula,the roots are $\alpha, \beta = \frac{4 \pm \sqrt{16-32}}{2} = \frac{4 \pm 4

Vedclass · Accepted Answer

$2^{3n+1} \cos \frac{n\pi}{2}$

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(C) Given the quadratic equation $x^2-4x+8=0$. Using the quadratic formula,the roots are $\alpha, \beta = \frac{4 \pm \sqrt{16-32}}{2} = \frac{4 \pm 4i}{2} = 2 \pm 2i$. Converting to polar form: $\alpha, \beta = 2\sqrt{2}(\frac{1}{\sqrt{2}} \pm i\frac{1}{\sqrt{2}}) = 2\sqrt{2}(\cos \frac{\pi}{4} \pm i\sin \frac{\pi}{4})$. Using De Moivre's Theorem,$\alpha^{2n} = (2\sqrt{2})^{2n}(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4})^{2n} = (2^{3/2})^{2n}(\cos \frac{n\pi}{2} + i\sin \frac{n\pi}{2}) = 2^{3n}(\cos \frac{n\pi}{2} + i\sin \frac{n\pi}{2})$. Similarly,$\beta^{2n} = 2^{3n}(\cos \frac{n\pi}{2} - i\sin \frac{n\pi}{2})$. Adding these,$\alpha^{2n} + \beta^{2n} = 2^{3n}(\cos \frac{n\pi}{2} + i\sin \frac{n\pi}{2} + \cos \frac{n\pi}{2} - i\sin \frac{n\pi}{2}) = 2^{3n} \cdot 2 \cos \frac{n\pi}{2} = 2^{3n+1} \cos \frac{n\pi}{2}$.

If $\alpha, \beta$ are the roots of the equation $x^2-4x+8=0$,then for any $n \in N$,$\alpha^{2n}+\beta^{2n}$ equals

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