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
$x^2-(-1)x+1=0$
$\Rightarrow x^2+x+1=0$