The common roots of the equations x^{12} - 1 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given equations are $x^{12} - 1 = 0$ and $x^4 + x^2 + 1 = 0$. Factorizing the first equation: $x^{12} - 1 = (x^6 - 1)(x^6 + 1) = (x^2 - 1)(x^4 + x

Vedclass · Accepted Answer

$\pm \omega, \pm \omega^2$

Vedclass · Answer

(C) Given equations are $x^{12} - 1 = 0$ and $x^4 + x^2 + 1 = 0$. Factorizing the first equation: $x^{12} - 1 = (x^6 - 1)(x^6 + 1) = (x^2 - 1)(x^4 + x^2 + 1)(x^6 + 1)$. Since $x^4 + x^2 + 1$ is a factor of $x^{12} - 1$,all roots of $x^4 + x^2 + 1 = 0$ are common roots. Solving $x^4 + x^2 + 1 = 0$: Let $y = x^2$,then $y^2 + y + 1 = 0$. The roots are $y = \omega$ and $y = \omega^2$,where $\omega = \frac{-1 + i\sqrt{3}}{2}$. Thus,$x^2 = \omega$ or $x^2 = \omega^2$. Taking square roots,$x = \pm \sqrt{\omega} = \pm \omega^2$ and $x = \pm \sqrt{\omega^2} = \pm \omega$. Therefore,the common roots are $\pm \omega, \pm \omega^2$.

The common roots of the equations $x^{12} - 1 = 0$ and $x^4 + x^2 + 1 = 0$ are:

Similar Questions

If the equations $x^2 - 11x + a = 0$ and $x^2 - 14x + 2a = 0$ have a common factor and $a \neq 0$,then what is the common factor?

Let $p_1(x) = x^3 - 2020x^2 + b_1x + c_1$ and $p_2(x) = x^3 - 2021x^2 + b_2x + c_2$ be polynomials having two common roots $\alpha$ and $\beta$. Suppose there exist polynomials $q_1(x)$ and $q_2(x)$ such that $p_1(x)q_1(x) + p_2(x)q_2(x) = x^2 - 3x + 2$. Then the correct identity is

If the equations $x^2+ax+b=0$ and $x^2+bx+a=0$ $(a \neq b)$ have a common root,then $a+b$ is equal to

If $3x^2 - 7x + 2 = 0$ and $15x^2 - 11x + a = 0$ have a common root and $a$ is a positive real number,then the sum of the roots of the equation $15x^2 - ax + 7 = 0$ is:

If the two equations $x^2 - cx + d = 0$ and $x^2 - ax + b = 0$ have one common root and the second equation has equal roots,then $2(b + d) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module