Question

The common roots of the equations $z^3+2z^2+2z+1=0$ and $z^{1985}+z^{100}+1=0$ are

• A. $\omega ,{\omega }^2$
• B. $\omega ,{\omega }^3$
• C. ${\omega }^2,{\omega }^3$
• D. None of these

Answer 1

The correct option is A

$\omega ,{\omega }^2$

Explanation for the correct option:

Step1. Find the root of the $z^3+2z^2+2z+1=0:$

Given $z^3+2z^2+2z+1=0.......\left(1\right)$

Put $z=-1$ in (1) .we get $=-1+2-2+1=0$

$z=-1$satisfy the equation (1) then it can be written as$(z+1)(z^2+z+1)=0$

so, the roots of (1) $=-1,\omega ,{\omega }^2$

Step 2. Find the common roots :

$z^{1985}+z^{100}+1=0.......\left(2\right)$

Put $z=-1$ in (1) .we get

$=-1+1+1\ne 0$

$z=-1$does not satisfy the equation (2)

Put $z=\omega$ in equation (2)

$$\begin{gathered} {\omega }^{1985}+{\omega }^{100}+1=0 \\ \Rightarrow {\omega }^2+\omega +1=0\mathbf{(}\mathbf{\because }{\mathit{\omega }}^{\mathbf{3}\mathbf{n}\mathbf{}}\mathbf{=}\mathbf{1}\mathbf{)} \end{gathered}$$

So $\omega$is the root of (2)

Put $z={\omega }^2$in equation (2)

$$\begin{gathered} {\omega }^{2870}+{\omega }^{200}+1=0 \\ \Rightarrow {\omega }^2+\omega +1=0\mathbf{(}\mathbf{\because }{\mathit{\omega }}^{\mathbf{3}\mathbf{n}\mathbf{}}\mathbf{=}\mathbf{1}\mathbf{)} \end{gathered}$$

So ${\omega }^2$is also the root of (2)

Therefore the common roots are $\omega ,{\omega }^2$.

Hence the correct option is A.