Question #b9b3d

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Answer 1 · 2016-10-30T20:06:39

#p=-1/2+-i#

Standard form:#" "y=ax^2+bx+c" " =>" " x=(-b+-sqrt(b^2-4ac))/(2a)#

In this #" "a=1"; "b=1"; "c=4# giving

#p=(-1+-sqrt(1^2-4(1)(4)))/(2(1))#

#p=-1/2+-sqrt(-16)/2#

Which is the same as:

#p=-1/2+-sqrt(4^2xx(-1))/2#

#p=-1/2+-(cancel(2)sqrt(-1))/(cancel(2))#

But #sqrt(-1) = i#

#p=-1/2+-i#