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

If $n, K \in N$ such that $n \neq 3K$,then $(\sqrt{3}+i)^{2n} + (\sqrt{3}-i)^{2n} = $

If the least positive integer $n$ satisfying the equation $\left(\frac{\sqrt{3}+i}{\sqrt{3}-i}\right)^{n}=-1$ is $p$ and the least positive integer $m$ satisfying the equation $\left(\frac{1-\sqrt{3} i}{1+\sqrt{3} i}\right)^m=\operatorname{cis} \frac{2 \pi}{3}$ is $q$,then $\sqrt{p^2+q^2}=$

The product of the four values of $(1+i \sqrt{3})^{3/4}$ is

If $\alpha, \beta$ are the roots of $x^2-2x+4=0$, for $n \in N$, what is the value of $\alpha^n+\beta^n$?

Let $i=\sqrt{-1}$. If $\frac{(-1+i \sqrt{3})^{21}}{(1-i)^{24}}+\frac{(1+i \sqrt{3})^{21}}{(1+i)^{24}}=k$,and $n =[| k |]$ be the greatest integral part of $| k |$. Then $\sum_{ j =0}^{ n +5}( j +5)^{2}-\sum_{ j =0}^{ n +5}( j +5)$ is equal to ........ .

Vedclass Test Series

Exam Paper Generator

Online Exam Module