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

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Question

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

Vedclass · Accepted Answer

$-256$

Vedclass · Answer

(A) Given the quadratic equation $x^2 + 2x + 2 = 0$.Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $x = \frac{-2 \pm \sqrt{4 - 8}}{2} = -1 \pm i$.Let $\alpha = -1 + i$ and $\beta = -1 - i$.In polar form,$\alpha = \sqrt{2} e^{i(3\pi/4)}$ and $\beta = \sqrt{2} e^{-i(3\pi/4)}$.Then $\alpha^{15} + \beta^{15} = (\sqrt{2})^{15} [e^{i(45\pi/4)} + e^{-i(45\pi/4)}]$.Using Euler's formula,this is $2^{7} \cdot \sqrt{2} \cdot 2 \cos(45\pi/4)$.Since $45\pi/4 = 11\pi + \pi/4$,$\cos(45\pi/4) = \cos(11\pi + \pi/4) = -\cos(\pi/4) = -1/\sqrt{2}$.Thus,$\alpha^{15} + \beta^{15} = 2^8 \cdot \sqrt{2} \cdot (-1/\sqrt{2}) = -2^8 = -256$.

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

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