Question

Roots of the equation $x^4-2x^2+4=0$ forms a

• A. rectangle
• B. square
• C. rhombus
• D. paralleleogram

Answer 1

Answer: (A)

$x^4-2x^2+4=0$
$\therefore x^2=\frac{2\pm \sqrt{4-16}}{2}$
$=1\pm \sqrt{3}i=2\frac{1\pm \sqrt{3}i}{2}=2\left(\cos \frac{\pi }{3}\pm i\sin \frac{\pi }{3}\right)$
$\therefore x=\pm \sqrt{2}{\left(\cos \frac{\pi }{3}\pm i\sin \frac{\pi }{3}\right)}^{1/2}$
$=\pm \sqrt{2}\left(\cos \frac{\pi }{6}\pm i\sin \frac{\pi }{6}\right)$
$\therefore x=\sqrt{2}{e}^{i\pi /6},-\sqrt{2}{e}^{i\pi /6},\sqrt{2}{e}^{-i\pi /6},-\sqrt{2}{e}^{-i\pi /6}$
Clearly these points forms rectangle inscribed in circle of radius $\sqrt{2}$.