Question

If $\alpha, \beta$ are the roots of $x^{2}-x+1=0$, then the quadratic equation whose roots are $\alpha^{2015}$, $\beta^{2015}$ is

• A. $x^{2}-x+1=0$
• B. $x^{2}+x+1=0$
• C. $x^{2}+x-1=0$
• D. $x^{2}-x-1=0$

Answer 1

Answer: (B)

We have, $x^{2}-x+1=0$
$\Rightarrow x=\omega, \omega^{2}$
Then, $\alpha=\omega$ and $\beta=\omega^{2}$
Now, $\alpha^{2015}=\omega^{2015}=\omega^{3 \times 671+2}=\omega^{2}$
$ \omega^{2} $ and $ \beta^{2015} =\left(\omega^{2}\right)^{2015}=\omega^{4030} $
$=\omega^{3 \times 1343}+1=\omega $
$\therefore \alpha^{2015}+\beta^{2015} =\omega^{2}+\omega=-1 $
and $\alpha^{2015} \cdot \beta^{2015}=\omega^{2} \cdot \omega=\omega^{3}=1$
$\therefore $ Equation whose roots are $\alpha^{2015}$ and $\beta^{2015}$ will be
$ x^{2} -(-1) x+1=0 $
$\Rightarrow x^{2}+x+1=0 $