Mathematics - Quadratic Equation Question with Solution | TestHub

Let the two roots of$x^2+2x+2=0$be$\alpha$and$\beta$

$\Rightarrow \alpha ,\beta =\frac{-2\pm \sqrt{4-8}}{2}=-1\pm i$

$\Rightarrow \alpha ,\beta =\sqrt{2}\left(\frac{-1}{\sqrt{2}}\pm i\frac{1}{\sqrt{2}}\right)$

$=\sqrt{2}\left(\mathrm{c}\mathrm{o}\mathrm{s}\frac{3\pi }{4}\pm i\mathrm{s}\mathrm{i}\mathrm{n}\frac{3\pi }{4}\right)$

$\Rightarrow {\alpha }^{15}+{\beta }^{15}={\left[\sqrt{2}\left(\mathrm{c}\mathrm{o}\mathrm{s}\frac{3\pi }{4}+i\mathrm{s}\mathrm{i}\mathrm{n}\frac{3\pi }{4}\right)\right]}^{15}+{\left[\sqrt{2}\left(\mathrm{c}\mathrm{o}\mathrm{s}\frac{ 3\pi }{4}-i\mathrm{s}\mathrm{i}\mathrm{n}\frac{3\pi }{4}\right)\right]}^{15}$

$\Rightarrow {\alpha }^{15}+{\beta }^{15}=2^{\frac{15}{2}}\left[\left(\mathrm{c}\mathrm{o}\mathrm{s}\frac{45\pi }{4}+i\mathrm{s}\mathrm{i}\mathrm{n}\frac{45\pi }{4}\right)+\left(\mathrm{c}\mathrm{o}\mathrm{s}\frac{45\pi }{4}-i\mathrm{s}\mathrm{i}\mathrm{n}\frac{45\pi }{4}\right)\right]$

$\Rightarrow {\alpha }^{15}+{\beta }^{15}={2.2}^{\frac{15}{2}}.\mathrm{c}\mathrm{o}\mathrm{s}\frac{45\pi }{4}=2^{\frac{17}{2}}.\mathrm{c}\mathrm{o}\mathrm{s}\frac{5\pi }{4}=-2^{\frac{17}{2}}.\frac{1}{\sqrt{2}}$

$\Rightarrow {\alpha }^{15}+{\beta }^{15}=-256$