If 1, omega, omega^2 are the cube roots of un | vedclass

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

Question

(B) Given equation: $8z^3 - 12z^2 + 6z - 28 = 0$ ...$(i)$By inspection,$z = 2$ is a root because $8(8) - 12(4) + 6(2) - 28 = 64 - 48 + 12 - 28 = 0$.Di

Vedclass · Accepted Answer

$2, \frac{3\omega + 1}{2}, \frac{3\omega^2 + 1}{2}$

Vedclass · Answer

(B) Given equation: $8z^3 - 12z^2 + 6z - 28 = 0$ ...$(i)$By inspection,$z = 2$ is a root because $8(8) - 12(4) + 6(2) - 28 = 64 - 48 + 12 - 28 = 0$.Dividing the equation by $(z - 2)$,we get:$8z^3 - 16z^2 + 4z^2 - 8z + 14z - 28 = 0$$8z^2(z - 2) + 4z(z - 2) + 14(z - 2) = 0$$(z - 2)(8z^2 + 4z + 14) = 0$$(z - 2)(4z^2 + 2z + 7) = 0$For $4z^2 + 2z + 7 = 0$,the roots are $z = \frac{-2 \pm \sqrt{4 - 112}}{8} = \frac{-2 \pm \sqrt{-108}}{8} = \frac{-2 \pm 6\sqrt{3}i}{8} = \frac{-1 \pm 3\sqrt{3}i}{4}$.We know $\omega = \frac{-1 + i\sqrt{3}}{2}$ and $\omega^2 = \frac{-1 - i\sqrt{3}}{2}$.Consider $\frac{3\omega + 1}{2} = \frac{3(\frac{-1 + i\sqrt{3}}{2}) + 1}{2} = \frac{-3 + 3i\sqrt{3} + 2}{4} = \frac{-1 + 3i\sqrt{3}}{4}$.Similarly,$\frac{3\omega^2 + 1}{2} = \frac{3(\frac{-1 - i\sqrt{3}}{2}) + 1}{2} = \frac{-3 - 3i\sqrt{3} + 2}{4} = \frac{-1 - 3i\sqrt{3}}{4}$.Thus,the roots are $2, \frac{3\omega + 1}{2}, \frac{3\omega^2 + 1}{2}$.

If $1, \omega, \omega^2$ are the cube roots of unity,then the roots of the equation $8z^3 - 12z^2 + 6z - 28 = 0$ are

Similar Questions

If $Z_{r} = \sin \frac{2 \pi r}{11} - i \cos \frac{2 \pi r}{11}$, then $\sum_{r=0}^{10} Z_{r}$ is equal to

If $1, \omega, \omega^2$ are the three cube roots of unity,then the value of $(a + b\omega + c\omega^2)^3 + (a + b\omega^2 + c\omega)^3$ is equal to,given that $a + b + c = 0$.

One of the values of $(\sqrt{3}-i)^{\frac{2}{5}}$ is

If $(1 + i\sqrt{3})^9 = a + ib$,then $b$ is equal to

If $\omega$ is a complex cube root of unity,then $\sum_{r=1}^9 r(r+1-\omega)(r+1-\omega^2) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module