If alpha, beta are the roots of the equation | vedclass

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Question

(B) The given equation is $x^2 - 2\sqrt{3}x + 4 = 0$.Using the quadratic formula,the roots are $\alpha, \beta = \frac{2\sqrt{3} \pm \sqrt{12 - 16}}{2}

Vedclass · Accepted Answer

$2^{2025}$

Vedclass · Answer

(B) The given equation is $x^2 - 2\sqrt{3}x + 4 = 0$.Using the quadratic formula,the roots are $\alpha, \beta = \frac{2\sqrt{3} \pm \sqrt{12 - 16}}{2} = \sqrt{3} \pm i$.In polar form,$\alpha = 2(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6})$ and $\beta = 2(\cos \frac{-\pi}{6} + i \sin \frac{-\pi}{6})$.Then $\alpha^{2024} = 2^{2024}(\cos \frac{2024\pi}{6} + i \sin \frac{2024\pi}{6})$ and $\beta^{2024} = 2^{2024}(\cos \frac{-2024\pi}{6} + i \sin \frac{-2024\pi}{6})$.Note that $\frac{2024\pi}{6} = \frac{1012\pi}{3} = 337\pi + \frac{\pi}{3}$.$\alpha^{2024} - \beta^{2024} = 2^{2024} [2i \sin(\frac{2024\pi}{6})] = 2^{2024} [2i \sin(337\pi + \frac{\pi}{3})] = 2^{2024} [2i (-\sin \frac{\pi}{3})] = 2^{2024} [2i (-\frac{\sqrt{3}}{2})] = -i \sqrt{3} \cdot 2^{2024}$.Therefore,$\frac{2}{\sqrt{3}} |\alpha^{2024} - \beta^{2024}| = \frac{2}{\sqrt{3}} | -i \sqrt{3} \cdot 2^{2024} | = \frac{2}{\sqrt{3}} \cdot \sqrt{3} \cdot 2^{2024} = 2^{2025}$.

If $\alpha, \beta$ are the roots of the equation $x+\frac{4}{x}=2 \sqrt{3}$,then $\frac{2}{\sqrt{3}}\left|\alpha^{2024}-\beta^{2024}\right|=$

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