If $1, \omega, \omega^2$ are the cube roots of unity, then the roots of the equation $8 z^3-12 z^2+6 z-28=0$…
If $1, \omega, \omega^2$ are the cube roots of unity, then the roots of the equation $8 z^3-12 z^2+6 z-28=0$ are
- $2,2 \omega, 3 \omega^2+1$
- $2, \frac{3 \omega+1}{2}, \frac{3 \omega^2+1}{2}$
- $2, \frac{1+3 \omega}{3}, \frac{1+3 \omega^2}{3}$
- $2, \frac{1-\omega}{2}, \frac{1-\omega^2}{2}$
Solution
$8 z^3-12 z^2+6 z-28=0$ ...(i)
Since $z=2$ satisfies equation (i). Hence $z=2$ is one of solution equation (i).
Therefore,
$
\begin{aligned}
& \Rightarrow 8 z^2(z-2)+4 z(z-2)+14(z-2)=0 \\
& \Rightarrow \quad\left(8 z^2+4 z+14\right)(z-2)=0 \\
& \Rightarrow \quad\left(4 z^2+2 z+7\right)(z-2)=0 \\
& \Rightarrow \quad z=2 \text { or } 4 z^2+2 z+7=0 \\
& z=\frac{-2 \pm \sqrt{(2)^2-4 \times 4 \times 7}}{2 \times 4}
\end{aligned}
$
$z=\frac{-1 \pm 3 \sqrt{3} i}{4}$...(1)
Since we know that cube root of unity are, $1, \omega, \omega^2$. Where $\omega=\frac{-1+i \sqrt{3}}{2}$ and $\omega^2=\frac{-1-i \sqrt{3}}{2}$
Now $\omega=\frac{-1+i \sqrt{3}}{2}$
$
\begin{aligned}
& \Rightarrow \quad \frac{3 \omega+1}{2}=\frac{-1 \pm 3 \sqrt{3} i}{2} \\
& \Rightarrow \frac{3 \omega+1}{2}=z
\end{aligned}
(from (1))$
Similary, $\omega^2=\frac{-1-i \sqrt{3}}{2}$
$
\begin{aligned}
& \Rightarrow \frac{3 \omega^2+1}{2}=\frac{-1-3 \sqrt{3} i}{4} \Rightarrow \frac{3 \omega^2+1}{2}=z \\
& \Rightarrow \frac{3 \omega^2+1}{2}=z
\end{aligned}
$
Hence required roots are $2, \frac{3 \omega+1}{2}, \frac{3 \omega^2+1}{2}$. ie option (b) is correct
Asked in: AP EAMCET 2023 (19 May Shift 1)
Practice more Complex Number questions on Aicharya