Let alpha and beta be the roots of x^{2}+x+1= | vedclass

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

Question

(A) Given the quadratic equation $x^{2}+x+1=0$.Using the quadratic formula, the roots are $x = \frac{-1 \pm \sqrt{1^{2}-4(1)(1)}}{2} = \frac{-1 \pm \s

Vedclass · Accepted Answer

$2 \cos \left(\frac{2 n \pi}{3}\right)$

Vedclass · Answer

(A) Given the quadratic equation $x^{2}+x+1=0$.Using the quadratic formula, the roots are $x = \frac{-1 \pm \sqrt{1^{2}-4(1)(1)}}{2} = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm i\sqrt{3}}{2}$.Let $\alpha = \frac{-1+i\sqrt{3}}{2} = e^{i(2\pi/3)}$ and $\beta = \frac{-1-i\sqrt{3}}{2} = e^{-i(2\pi/3)}$.Then, $\alpha^{n}+\beta^{n} = (e^{i(2\pi/3)})^{n} + (e^{-i(2\pi/3)})^{n} = e^{i(2n\pi/3)} + e^{-i(2n\pi/3)}$.Using Euler's formula, $e^{i	heta} + e^{-i	heta} = 2\cos(	heta)$.Therefore, $\alpha^{n}+\beta^{n} = 2\cos\left(\frac{2n\pi}{3}ight)$.

Let $\alpha$ and $\beta$ be the roots of $x^{2}+x+1=0$. If $n$ is a positive integer, then $\alpha^{n}+\beta^{n}$ is

Similar Questions

If $\omega$ is a complex cube root of unity,then $\left(1-\omega+\omega^2\right)^6+\left(1-\omega^2+\omega\right)^6=$

If $\omega$ is a complex cube root of unity,then $(x+1)(x+\omega)(x-\omega-1)$ is equal to

If the cube roots of unity are $1, \omega, \omega^2$,then the roots of the equation $(x - 2)^3 + 27 = 0$ are

The roots of $(2 - 2i)^{1/3}$ are

If $n$ is an integer which leaves a remainder of $1$ when divided by $3$,then $(1+\sqrt{3}i)^n + (1-\sqrt{3}i)^n$ equals

Vedclass Test Series

Exam Paper Generator

Online Exam Module