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

1 Answer
Aug 26, 2016

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

Explanation:

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#