Discussion Forum

1)

If $\alpha$  and $\beta$  are the roots of the equation

$x^{2}-2x+4=0$  , then $\alpha^{12}+\beta^{12}$=

Answer:

Option C

Explanation:

 We have,

  $\alpha, \beta$  are the  roots of equation  $x^{2}-2x+4$=0

  $x^{2}-2x+4=0$

$x^{2}-2x+1= -3 \Rightarrow  (x-1)^{2}=-3$

 $x-1= \pm\sqrt{3}i\Rightarrow x=1\pm\sqrt{3}i$

 $x=-2\omega,-2\omega^{2}$

   $\left[\because \frac{-1+\sqrt{3}i}{2}=\omega , \frac{-1-\sqrt{3}i}{2}=\omega^{2}\right]$

$\therefore$   $\alpha =-2 \omega$ , $\beta = -2 \omega^{2}$

 $\alpha^{12}+\beta ^{12}= (-2 \omega) ^{12} +(-2 \omega^{2})^{12}$

 = $2^{12} (\omega^{12}+ \omega^{24})= 2^{12} (1+1)$   [ $\because \omega^{3}=1$]

 =$2^{13}$