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

$\omega$ is an imaginary cube root of unity. If $(1 + \omega^2)^m = (1 + \omega^4)^m$,then the least positive integral value of $m$ is

If $x^{2}+x+1=0$,then the value of $(x+\frac{1}{x})^{4}+(x^{2}+\frac{1}{x^{2}})^{4}+(x^{3}+\frac{1}{x^{3}})^{4}+\dots+(x^{25}+\frac{1}{x^{25}})^{4}$ is:

If $\alpha \neq 1$ is any $n^{th}$ root of unity,then $S = 1 + 3\alpha + 5\alpha^2 + \dots$ up to $n$ terms,is equal to

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 ........ .

$\frac{(\sin \frac{\pi}{8} + i \cos \frac{\pi}{8})^8}{(\sin \frac{\pi}{8} - i \cos \frac{\pi}{8})^8}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module