Mathematics - Quadratic Equation Question with Solution | TestHub

Given,

$x^2-\sqrt{2}x+2=0$

$\Rightarrow x=\frac{\sqrt{2}\pm \sqrt{6}i}{2}$

$\Rightarrow x=\sqrt{2}\left(\frac{1\pm \sqrt{3}i}{2}\right)$

$\Rightarrow x=\sqrt{2}\left(\frac{1}{2}\pm \frac{\sqrt{3}i}{2}\right)$

$\Rightarrow x=\sqrt{2}\left(\cos \frac{\mathrm{\pi }}{3}\pm i\sin \frac{\mathrm{\pi }}{3}\right)$

So, $\alpha =\sqrt{2}\left(\cos \frac{\mathrm{\pi }}{3}+i\sin \frac{\mathrm{\pi }}{3}\right)=\sqrt{2}{e}^{\frac{\mathrm{i\pi }}{3}}$

And $\beta =\sqrt{2}\left(\cos \frac{\mathrm{\pi }}{3}-i\sin \frac{\mathrm{\pi }}{3}\right)=\sqrt{2}{e}^{\frac{-\mathrm{i\pi }}{3}}$

Now ${\alpha }^{14}+{\beta }^{14}={\left(\sqrt{2}{e}^{\frac{\mathrm{i\pi }}{3}}\right)}^{14}+{\left(\sqrt{2}{e}^{\frac{-\mathrm{i\pi }}{3}}\right)}^{14}$

$\Rightarrow {\alpha }^{14}+{\beta }^{14}={\left(\sqrt{2}\right)}^{14}\left(e^{\frac{i14\pi }{3}}+e^{\frac{-i14\pi }{3}}\right)$

$\Rightarrow {\alpha }^{14}+{\beta }^{14}={\left(\sqrt{2}\right)}^{14}\left(\cos \frac{14\mathrm{\pi }}{3}+i\sin \frac{14\mathrm{\pi }}{3}+\cos \frac{14\mathrm{\pi }}{3}-i\sin \frac{14\mathrm{\pi }}{3}\right)$

$\Rightarrow {\alpha }^{14}+{\beta }^{14}=2^7\left(2\cos \frac{14\mathrm{\pi }}{3}\right)$

$\Rightarrow {\alpha }^{14}+{\beta }^{14}=2^8\left(\cos \left(4\mathrm{\pi }+\frac{2\mathrm{\pi }}{3}\right)\right)$

$\Rightarrow {\alpha }^{14}+{\beta }^{14}=2^8\left(\cos \frac{2\mathrm{\pi }}{3}\right)$

$\Rightarrow {\alpha }^{14}+{\beta }^{14}=-2^8\times \frac{1}{2}=-128$