Let alpha and beta be the roots of the equati | vedclass

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

Question

(D) The roots of the equation $x^2 + x + 1 = 0$ are the complex cube roots of unity,$\omega$ and $\omega^2$.Let $\alpha = \omega$ and $\beta = \omega^

Vedclass · Accepted Answer

$x^2 + x + 1 = 0$

Vedclass · Answer

(D) The roots of the equation $x^2 + x + 1 = 0$ are the complex cube roots of unity,$\omega$ and $\omega^2$.Let $\alpha = \omega$ and $\beta = \omega^2$.We need to find the equation whose roots are $\alpha^{19}$ and $\beta^7$.$\alpha^{19} = \omega^{19} = (\omega^3)^6 \cdot \omega = 1^6 \cdot \omega = \omega$.$\beta^7 = (\omega^2)^7 = \omega^{14} = (\omega^3)^4 \cdot \omega^2 = 1^4 \cdot \omega^2 = \omega^2$.The roots of the new equation are $\omega$ and $\omega^2$,which are the same as the roots of the original equation $x^2 + x + 1 = 0$.

Let $\alpha$ and $\beta$ be the roots of the equation $x^2 + x + 1 = 0$. The equation whose roots are $\alpha^{19}$ and $\beta^7$ is

Similar Questions

If $\alpha, \beta$ are the roots of the equation $x^2-2x+4=0$,then $\alpha^n+\beta^n = \ldots \cos \left(\frac{n\pi}{3}\right)$ for any $n \in N$.

If $\omega$ is a complex root of the equation $z^3 = 1$,then $\omega + \omega^{\left( \frac{1}{2} + \frac{3}{8} + \frac{9}{32} + \frac{27}{128} + \dots \right)}$ is equal to

The real part of $\frac{(\cos a+i \sin a)^6}{(\sin b+i \cos b)^8}$ is

If the roots of the equation $(z-4)^3=8 i$ are $a-2 i, b+i$,and $c+i$,then $\sqrt{a b c}=$

One of the values of $(\sqrt{3}-i)^{\frac{1}{6}}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module