If alpha, beta are the roots of x^2-2x+4=0, f | vedclass

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(B) Given the quadratic equation $x^2-2x+4=0$. Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$, we get: $x = \frac{2 \pm \sqrt{4-16

Vedclass · Accepted Answer

$2^{n+1} \cos \left(\frac{n \pi}{3}\right)$

Vedclass · Answer

(B) Given the quadratic equation $x^2-2x+4=0$. Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$, we get: $x = \frac{2 \pm \sqrt{4-16}}{2} = \frac{2 \pm \sqrt{-12}}{2} = 1 \pm i\sqrt{3}$. Converting to polar form: $x = 2 \left( \frac{1}{2} \pm i\frac{\sqrt{3}}{2} ight) = 2 \left( \cos \frac{\pi}{3} \pm i \sin \frac{\pi}{3} ight)$. Let $\alpha = 2 e^{i\pi/3}$ and $\beta = 2 e^{-i\pi/3}$. Then $\alpha^n + \beta^n = (2 e^{i\pi/3})^n + (2 e^{-i\pi/3})^n = 2^n (e^{in\pi/3} + e^{-in\pi/3})$. Using Euler's formula $e^{i	heta} + e^{-i	heta} = 2 \cos 	heta$: $\alpha^n + \beta^n = 2^n (2 \cos \frac{n\pi}{3}) = 2^{n+1} \cos \frac{n\pi}{3}$. Thus, the correct option is $B$.

If $\alpha, \beta$ are the roots of $x^2-2x+4=0$, for $n \in N$, what is the value of $\alpha^n+\beta^n$?

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