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

$(27)^{1/3} = $

$\frac{(-1 + i\sqrt{3})^{15}}{(1 - i)^{20}} + \frac{(-1 - i\sqrt{3})^{15}}{(1 + i)^{20}}$ is equal to

If $\omega (\neq 1)$ is a cube root of unity and $(1 + \omega)^7 = A + B\omega$,then $A$ and $B$ are equal to:

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

For $n \in N$,$\left(\frac{1+\cos \theta+i \sin \theta}{1+\cos \theta-i \sin \theta}\right)^n=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module