Let alpha and beta be the roots of x^2 - sqrt | vedclass

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

Question

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

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) Given the quadratic equation $x^2 - \sqrt{2}x + 1 = 0$.Using the quadratic formula,the roots are $x = \frac{\sqrt{2} \pm \sqrt{2 - 4}}{2} = \frac{\sqrt{2} \pm i\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \pm i\frac{1}{\sqrt{2}}$.These can be written in polar form as $\alpha = e^{i\pi/4}$ and $\beta = e^{-i\pi/4}$.Then,$\alpha^{50} + \beta^{50} = (e^{i\pi/4})^{50} + (e^{-i\pi/4})^{50} = e^{i25\pi/2} + e^{-i25\pi/2}$.Since $e^{i25\pi/2} = e^{i(12\pi + \pi/2)} = e^{i\pi/2} = i$ and $e^{-i25\pi/2} = -i$.Therefore,$\alpha^{50} + \beta^{50} = i + (-i) = 0$.

Let $\alpha$ and $\beta$ be the roots of $x^2 - \sqrt{2}x + 1 = 0$. Then the value of $\alpha^{50} + \beta^{50}$ is:

Similar Questions

If $\sin \theta + 2\sin \phi + 3\sin \psi = 0$ and $\cos \theta + 2\cos \phi + 3\cos \psi = 0$,then the value of $\cos 3\theta + 8\cos 3\phi + 27\cos 3\psi = $

If $n, K \in N$ such that $n \neq 3K$,then $(\sqrt{3}+i)^{2n} + (\sqrt{3}-i)^{2n} = $

If $2 \alpha = -1 - i \sqrt{3}$ and $2 \beta = -1 + i \sqrt{3}$,then $5 \alpha^4 + 5 \beta^4 + 7 \alpha^{-1} \beta^{-1}$ is equal to

Common roots of the equations $z^3 + 2z^2 + 2z + 1 = 0$ and $z^{1985} + z^{100} + 1 = 0$ are

If the complex cube roots of $(-i)$ are $\alpha, \beta, \gamma$,then $\alpha^2+\beta^2+\gamma^2=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module