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

The correct option is A $x^2-x+1=0$

$x^2-x+1=0$

$\Rightarrow x=\frac{1\pm \sqrt{3}i}{2}=-\omega ,-{\omega }^2$, where $\omega ,{\omega }^2$ are the cube roots of unity

$\Rightarrow \alpha =-\omega ,\beta =-{\omega }^2$

So ${\alpha }^{2015}=(-\omega {)}^{2015}=-(\omega {)}^{2013+2}=-(\omega {)}^{671\times 3}\cdot {\omega }^2=-{\omega }^2$

and ${\beta }^{2015}=(-{\omega }^2{)}^{2015}=-(\omega {)}^{4030}=-(\omega {)}^{4029+1}=-(\omega {)}^{1343\times 3}\cdot \omega =-\omega$

Clearly the roots are same, that of given quadratic equation

Hence the quadratic equation will be same, which is $x^2-x+1=0$