Let alpha and beta be two roots of the equati | vedclass

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(A) The given equation is $x^2 + 2x + 2 = 0$.Completing the square,we get $(x+1)^2 + 1 = 0$,so $(x+1)^2 = -1$.Thus,$x+1 = \pm i$,which gives $x = -1 \

Vedclass · Accepted Answer

$-256$

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(A) The given equation is $x^2 + 2x + 2 = 0$.Completing the square,we get $(x+1)^2 + 1 = 0$,so $(x+1)^2 = -1$.Thus,$x+1 = \pm i$,which gives $x = -1 \pm i$.In polar form,$x = \sqrt{2} \left( \cos \frac{3\pi}{4} \pm i \sin \frac{3\pi}{4} ight) = \sqrt{2} e^{\pm i(3\pi/4)}$.Using De Moivre's Theorem,$\alpha^{15} + \beta^{15} = (\sqrt{2})^{15} e^{i(45\pi/4)} + (\sqrt{2})^{15} e^{-i(45\pi/4)}$.$\alpha^{15} + \beta^{15} = 2^{7} \sqrt{2} 	imes 2 \cos \left( \frac{45\pi}{4} ight)$.Since $\frac{45\pi}{4} = 11\pi + \frac{\pi}{4}$,$\cos \left( \frac{45\pi}{4} ight) = \cos \left( 11\pi + \frac{\pi}{4} ight) = -\cos \left( \frac{\pi}{4} ight) = -\frac{1}{\sqrt{2}}$.Therefore,$\alpha^{15} + \beta^{15} = 2^8 \sqrt{2} 	imes \left( -\frac{1}{\sqrt{2}} ight) = -2^8 = -256$.

Let $\alpha$ and $\beta$ be two roots of the equation $x^2 + 2x + 2 = 0$. Then,the value of $\alpha^{15} + \beta^{15}$ is equal to:

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