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 a root of the equation $ax^2 + bx + c = 0$ is the reciprocal of a root of the equation $a'x^2 + b'x + c' = 0$,then:

If the value of real number $a > 0$ for which $x^2 - 5ax + 1 = 0$ and $x^2 - ax - 5 = 0$ have a common real root is $\frac{3}{\sqrt{2\beta}}$,then $\beta$ is equal to

If the roots of the equations $x^2 + 3x + 2 = 0$ and $x^2 - x + \lambda = 0$ are in the same ratio,then what is the value of $\lambda$?

If the quadratic equations $3x^2 + ax + 1 = 0$ and $2x^2 + bx + 1 = 0$ have a common root,then what is the value of the expression $5ab - 2a^2 - 3b^2$?

If $a, b, c$ are in $A$.$P$. and if the equations $(b-c) x^2+(c-a) x+(a-b)=0$ and $2(c+a) x^2+(b+c) x=0$ have a common root, then

Vedclass Test Series

Exam Paper Generator

Online Exam Module