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

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

Question

(A) The given equation is $x^2 + 2x + 2 = 0$.Completing the square,we get $(x+1)^2 + 1 = 0$,so $(x+1)^2 = -1$.Thus,$x+1 = \pm i$,which gives $x = -1 \

Vedclass · Accepted Answer

$-256$

Vedclass · Answer

(A) The given equation is $x^2 + 2x + 2 = 0$.Completing the square,we get $(x+1)^2 + 1 = 0$,so $(x+1)^2 = -1$.Thus,$x+1 = \pm i$,which gives $x = -1 \pm i$.In polar form,$x = \sqrt{2} \left( \cos \frac{3\pi}{4} \pm i \sin \frac{3\pi}{4} ight) = \sqrt{2} e^{\pm i(3\pi/4)}$.Using De Moivre's Theorem,$\alpha^{15} + \beta^{15} = (\sqrt{2})^{15} e^{i(45\pi/4)} + (\sqrt{2})^{15} e^{-i(45\pi/4)}$.$\alpha^{15} + \beta^{15} = 2^{7} \sqrt{2} 	imes 2 \cos \left( \frac{45\pi}{4} ight)$.Since $\frac{45\pi}{4} = 11\pi + \frac{\pi}{4}$,$\cos \left( \frac{45\pi}{4} ight) = \cos \left( 11\pi + \frac{\pi}{4} ight) = -\cos \left( \frac{\pi}{4} ight) = -\frac{1}{\sqrt{2}}$.Therefore,$\alpha^{15} + \beta^{15} = 2^8 \sqrt{2} 	imes \left( -\frac{1}{\sqrt{2}} ight) = -2^8 = -256$.

Let $\alpha$ and $\beta$ be two roots of the equation $x^2 + 2x + 2 = 0$. Then,the value of $\alpha^{15} + \beta^{15}$ is equal to:

Similar Questions

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

If $\alpha, \beta$ are the roots of the equation $x+\frac{4}{x}=2 \sqrt{3}$,then $\frac{2}{\sqrt{3}}\left|\alpha^{2024}-\beta^{2024}\right|=$

The common roots of the equations $z^3+2z^2+2z+1=0$ and $z^{2018}+z^{2017}+1=0$ satisfy the equation

If $\alpha$ and $\beta$ are the roots of the equation $x^2-2x+4=0$,then $\alpha^9+\beta^9$ is equal to

The product of any two of the tenth roots of unity is

Vedclass Test Series

Exam Paper Generator

Online Exam Module