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1) 2 to the power x=1-x² solve for x. 2)|x²+x+1|=3 where x€R ,find x
2)-|x2 + x+ 1| = 3now when we open mod, there are two cases-1st Case-x2 + x+ 1 = 3x2 + x –2 = 0x = – 2, x = 12nd Case-x2 + x+ 1 = –3x2 + x+ 4 = 0in this equation D hence roots are imaginary.
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