Question

If $\alpha$ and $\beta$ are the roots of $x^2-x+1=0$, then the value of ${\alpha }^{2013}+{\beta }^{2013}$ is equal to

• A. 2
• B. -2
• C. -1
• D. 1

Answer 1

Answer: (B)

Given equation is $x^2-x+1=0$
$x=\frac{1\pm \sqrt{1-4}}{2}\left(\because x=\frac{b\pm \sqrt{b^2-4ac}}{2a}\right)$
$=\frac{1\pm i\sqrt{3}}{2}=\frac{1+i\sqrt{3}}{2},\frac{1-i\sqrt{3}}{2}$
$\Rightarrow -x=\frac{-1+i\sqrt{3}}{2},\frac{-1-i\sqrt{3}}{2}$
$\Rightarrow +x=\omega ,-{\omega }^2$
Since, $(\alpha ,\beta )$ are the roots of given equation. Then, $\alpha =-\omega$ and $\beta =-{\omega }^2$
$\therefore {\alpha }^{2013}+{\beta }^{2013}=-(\omega {)}^{2013}+{\left(-{\omega }^2\right)}^{2013}$
$=-{\omega }^{2013}-{\omega }^{4026}$
$=-{\left({\omega }^3\right)}^{671}-{\left({\omega }^3\right)}^{1342}$
$=-(1{)}^{671}-(1{)}^{1342}\left(\because {\omega }^3=1\right)$
$=-1-1=-2$