Question

The common roots of the equation $x^{12}-1=0,x^4+x^2+1=0$ are

• A. $\omega ,{\omega }^2$
• B. $-\omega ,-{\omega }^2$
• C. $\pm \omega ,\pm {\omega }^2$
• D. $\pm {\omega }^2,\pm {\omega }^5$

Answer 1

The correct option is C $\pm \omega ,\pm {\omega }^2$

Roots of $x^{12}=1$ are of form
$x=e^{\frac{2k\pi }{12}\cdot \left(i\right)}k=0,1,\cdots ,11$
i.e.,$1,e^{i\left(\frac{2\pi }{12}\right)},e^{i\left(\frac{4\pi }{12}\right)},e^{i\left(\frac{6\pi }{12}\right)},\cdots \cdots e^{i\left(\frac{20\pi }{12}\right)}$
Multiply and divide $x^4+x^2+1$ by $x^2-1$
$\frac{(x^2-1)(x^4+x^2+1)}{x^2-1}=0$
$\Rightarrow x^6-1=0$
$1,-1$ will not be the root due to multiplication of $x^2-1$
Hence by applying the formula roots will be of form
$x=e^{i\left(2k\right)\frac{\pi }{6}}k=0,1,\cdots ,5$
i.e., $e^{i\left(\frac{\pi }{3}\right)},e^{i\left(\frac{2\pi }{3}\right)},e^{i\left(\frac{4\pi }{3}\right)},e^{i\left(\frac{5\pi }{3}\right)}$ after neglecting $\pm 1$
Hence common roots will be
$e^{i\left(\frac{\pi }{3}\right)},e^{i\left(\frac{2\pi }{3}\right)},e^{i\left(\frac{4\pi }{3}\right)},e^{i\left(\frac{5\pi }{3}\right)}$
$\because e^{i\left(\frac{\pi }{3}\right)}=-{\omega }^2,e^{i\left(\frac{2\pi }{3}\right)}=\omega ,e^{i\left(\frac{4\pi }{3}\right)}={\omega }^2,e^{i\left(\frac{5\pi }{3}\right)}=-\omega$

Hence common roots will be
$\therefore \pm \omega ,\pm {\omega }^2$