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Asked by  | 21st Sep, 2008, 08:01: AM

Expert Answer:

we know that roots of ax2 + bx +c=0 are

x =  ( -b±(b2 -4ac)  ) /2a

putting a=1,b=1 and c=1

we get x =  ( -1±3i ) /2  where i = -1

( (-1-3i)/2 )2  = 1/4 ( 1 -3 +23 i) = 1/4 (-2 +23i ) = 1/2 (-1 + 3i )

i.e. square of x1 is equal to x2

Answered by  | 21st Sep, 2008, 12:48: PM

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