If alpha, beta are the roots of the equation | vedclass

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

Question

(B) The given quadratic equation is $x^2-2x+4=0$. Using the quadratic formula,the roots are $\alpha, \beta = \frac{2 \pm \sqrt{4-16}}{2} = \frac{2 \pm

Vedclass · Accepted Answer

$2^{n+1}$

Vedclass · Answer

(B) The given quadratic equation is $x^2-2x+4=0$. Using the quadratic formula,the roots are $\alpha, \beta = \frac{2 \pm \sqrt{4-16}}{2} = \frac{2 \pm 2\sqrt{3}i}{2} = 1 \pm i\sqrt{3}$. Expressing the roots in polar form: $\alpha = 2\left(\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}ight)$ and $\beta = 2\left(\cos \left(-\frac{\pi}{3}ight) + i \sin \left(-\frac{\pi}{3}ight)ight)$. Using De Moivre's Theorem,$\alpha^n + \beta^n = 2^n \left(\cos \frac{n\pi}{3} + i \sin \frac{n\pi}{3}ight) + 2^n \left(\cos \left(-\frac{n\pi}{3}ight) + i \sin \left(-\frac{n\pi}{3}ight)ight)$. Since $\cos(-	heta) = \cos 	heta$ and $\sin(-	heta) = -\sin 	heta$,we get: $\alpha^n + \beta^n = 2^n \left(2 \cos \frac{n\pi}{3}ight) = 2^{n+1} \cos \left(\frac{n\pi}{3}ight)$.

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 $n \in N$.

Similar Questions

If $1, \omega, \omega^2$ are the cube roots of unity,then $\frac{1}{1+2 \omega}+\frac{1}{2+\omega}-\frac{1}{1+\omega}=$

If $\alpha$ and $\beta$ are imaginary cube roots of unity,then the value of $\alpha^4 + \beta^{28} + \frac{1}{\alpha \beta}$ is

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

The value of $i^{1/3}$ is

One of the values of $(-32 i)^{\frac{2}{5}}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module