The common roots of the equations z^3+2z^2+2z | vedclass

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

Question

(B) We have,$z^3+2z^2+2z+1=0$.Factoring the expression: $(z^3+1)+2z(z+1)=0$.$(z+1)(z^2-z+1)+2z(z+1)=0$.$(z+1)(z^2-z+1+2z)=0$.$(z+1)(z^2+z+1)=0$.The ro

Vedclass · Accepted Answer

$z^4+z^2+1=0$

Vedclass · Answer

(B) We have,$z^3+2z^2+2z+1=0$.Factoring the expression: $(z^3+1)+2z(z+1)=0$.$(z+1)(z^2-z+1)+2z(z+1)=0$.$(z+1)(z^2-z+1+2z)=0$.$(z+1)(z^2+z+1)=0$.The roots are $z=-1$ and the roots of $z^2+z+1=0$,which are $\omega$ and $\omega^2$.Now,test these roots in $z^{2018}+z^{2017}+1=0$:For $z=-1$: $(-1)^{2018}+(-1)^{2017}+1 = 1-1+1 = 1 
eq 0$.For $z=\omega$: $\omega^{2018}+\omega^{2017}+1 = (\omega^3)^{672} \cdot \omega^2 + (\omega^3)^{672} \cdot \omega + 1 = \omega^2+\omega+1 = 0$.For $z=\omega^2$: $(\omega^2)^{2018}+(\omega^2)^{2017}+1 = \omega^{4036}+\omega^{4034}+1 = \omega+\omega^2+1 = 0$.The common roots are $\omega$ and $\omega^2$.Checking the options for $z=\omega$ and $z=\omega^2$:For $z^4+z^2+1=0$:If $z=\omega$,$\omega^4+\omega^2+1 = \omega+\omega^2+1 = 0$.If $z=\omega^2$,$(\omega^2)^4+(\omega^2)^2+1 = \omega^8+\omega^4+1 = \omega^2+\omega+1 = 0$.Thus,the common roots satisfy $z^4+z^2+1=0$.

The common roots of the equations $z^3+2z^2+2z+1=0$ and $z^{2018}+z^{2017}+1=0$ satisfy the equation

Similar Questions

Let $\alpha$ and $\beta$ be the roots of $x^2 - \sqrt{2}x + 1 = 0$. Then the value of $\alpha^{50} + \beta^{50}$ is:

The sum of the least positive arguments of the distinct cube roots of the complex number $(1-i \sqrt{3})$ is

If $\omega$ is an imaginary cube root of unity,then the value of $\sin \left[ (\omega^{10} + \omega^{23})\pi - \frac{\pi}{4} \right]$ is

If $\alpha, \beta$ are the roots of the equation $x^2-2x+4=0$,then $\alpha^n+\beta^n = \ldots \cos \left(\frac{n\pi}{3}\right)$ for any $n \in N$.

If $1, \omega, \omega^2$ are the cube roots of unity and $(x+y)(x \omega+y \omega^2)(x \omega^2+y \omega)=f(x, y)$,then $f(2, 3)=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module