A) \[{{x}^{2}}-x-1=0\]
B) \[{{x}^{2}}-x+1=0\]
C) \[{{x}^{2}}+x-1=0\]
D) \[{{x}^{2}}+x+1=0\]
Correct Answer: D
Solution :
The roots of \[{{\operatorname{x}}^{2}} + x +1 = 0\] are \[\omega \] and \[{{\omega }^{2}}\] [see the cube roots of unity in complex numbers] Let \[\alpha = \omega , \beta = {{\omega }^{2}}\] Now \[{{\alpha }^{229}} = {{\omega }^{229}} = {{\omega }^{228}}\,.\,\,\omega = {{\left( {{\omega }^{3}} \right)}^{76}} .\,{{\omega }^{2}}\] \[= \omega = \alpha \left( \because \,{{\omega }^{3}} = 1 \right)\] \[{{\alpha }^{1004}}={{\omega }^{1002}}.\,{{\omega }^{2}}={{({{\omega }^{3}})}^{334}}.\,\,{{\omega }^{2}}={{\omega }^{2}}=\beta \] \[\therefore \] equation with roots \[{{\alpha }^{229}}\] and \[{{\alpha }^{1004}}\] is same as the equation with roots \[\alpha \] and \[\beta \] i.e. the original equation.
You need to login to perform this action.
You will be redirected in
3 sec
