How do you solve using the quadratic formula: # x^2+x+1=0#?

1 Answer
Nov 20, 2015

Answer:

See explanation...

Explanation:

#x^2+x+1# is of the form #ax^2+bx+c# with #a = b = c = 1#

This has zeros given by the quadratic formula:

#x = (-b+-sqrt(b^2-4ac))/(2a) = (-1+-sqrt(1^2-(4xx1xx1)))/(2xx1)#

#=(-1+-sqrt(1-4))/2 = (-1+-sqrt(-3))/2 = (-1+-sqrt(3)i)/2#

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

The number #omega = -1/2+sqrt(3)/2i = cos((2pi)/3)+sin((2pi)/3)i#

is called the primitive Complex cube root of unity and is used extensively when solving cubic equations.

#omega^2 = bar(omega) = -1/2-sqrt(3)/2i#

#omega^3 = 1#