If x + frac{1}{x} = sqrt{3}, then x = | vedclass

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

Question

(D) Given $x + \frac{1}{x} = \sqrt{3}$.Multiplying by $x$,we get $x^2 - \sqrt{3}x + 1 = 0$.Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4

Vedclass · Accepted Answer

$\cos \frac{\pi}{6} + i \sin \frac{\pi}{6}$

Vedclass · Answer

(D) Given $x + \frac{1}{x} = \sqrt{3}$.Multiplying by $x$,we get $x^2 - \sqrt{3}x + 1 = 0$.Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we have:$x = \frac{\sqrt{3} \pm \sqrt{(\sqrt{3})^2 - 4(1)(1)}}{2(1)}$$x = \frac{\sqrt{3} \pm \sqrt{3 - 4}}{2} = \frac{\sqrt{3} \pm \sqrt{-1}}{2} = \frac{\sqrt{3} \pm i}{2}$.Thus,$x = \frac{\sqrt{3}}{2} \pm \frac{i}{2}$.Taking the positive sign,$x = \frac{\sqrt{3}}{2} + i \frac{1}{2} = \cos \frac{\pi}{6} + i \sin \frac{\pi}{6}$.

If $x + \frac{1}{x} = \sqrt{3},$ then $x =$

Similar Questions

The modulus of the complex number $\frac{(1+i)^2(1+3 i)}{(2-6 i)(2-2 i)}$ is

The conjugate of the complex number $\frac{2 - 3i}{4 - i}$ is

Sum of the moduli of the complex roots of the equation $(x^2+\frac{1}{x^2})-5(x+\frac{1}{x})+6=0$ is

The complex number $\frac{1+2i}{1-i}$ lies in

The modulus of the square root of the conjugate of $-7+24 \sqrt{-1}$ is ....

Vedclass Test Series

Exam Paper Generator

Online Exam Module