How do you factor #x^2 +1 - x#?

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Answer 1 · 2016-08-26T15:48:32

#x=1/2+-[1/2sqrt(3)]i#

given:#" "x^2-x+1#

The discriminant is: #b^2-4ac -> (-1)^2-4(1)(1) = -3#

AS the discriminant is negative there are no factors for #x^2-x+1=0 # for #x !in RR#. The graph does not cross the x-axis.

Tony B

#x=(-b+-(sqrt(b^2-4ac)))/(2a)" " ->" " (1+-sqrt((-1)^1-4(1)(1)))/(2(1))#

#x=1/2+-(sqrt(-3))/2#

#x=1/2+-[1/2sqrt(3)]i#