Q48 / 160 · AP EAMCET 2023 (19 May Shift 1) · Mathematics

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

  • A
    $2,2 \omega, 3 \omega^2+1$
  • Correct answer
    B
    $2, \frac{3 \omega+1}{2}, \frac{3 \omega^2+1}{2}$
  • C
    $2, \frac{1+3 \omega}{3}, \frac{1+3 \omega^2}{3}$
  • D
    $2, \frac{1-\omega}{2}, \frac{1-\omega^2}{2}$

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.

Hence, the correct answer is option: 'B'.

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