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}-4 a c}}{2 a}\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 $\beta=-\omega^{2}$
$\therefore \alpha^{2013}+\beta^{2013}=-(\omega)^{2013}+\left(-\omega^{2}\right)^{2013}$
$=-\omega^{2013}-\omega^{4026}$
$=-\left(\omega^{3}\right)^{671}-\left(\omega^{3}\right)^{1342}$
$=-(1)^{671}-(1)^{1342} \quad\left(\because \omega^{3}=1\right)$
$=-1-1=-2$