A) \[1\,\,\pm \,\,i\]
B) \[-1\,\,\pm \,\,i\]
C) \[\omega \,\,or\,\,{{\omega }^{2}}\]
D) \[-\,\omega \,\,or\,\,-\,{{\omega }^{2}}\]
Correct Answer: A
Solution :
\[x=\sqrt{-2+2x}\] \[\Rightarrow {{x}^{2}}-2x+2=0\]
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