How do you solve # x^2 + x + 1=0# using completing the square?

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Answer 1 · 2015-06-13T18:32:10

This quadratic has no real roots, but:

#x^2+x+1 = (x+1/2)^2+3/4#

giving complex solutions: #x = -1/2 +-sqrt(3)/2i#

#x^2+x+1# is one of the factors of #x^3-1#

#x^3-1 = (x-1)(x^2+x+1) = (x-1)(x-omega)(x-omega^2)#

where #omega = -1/2+sqrt(3)/2i = cos((2pi)/3)+isin((2pi)/3)#

is called the primitive complex root of unity.

Pretending we don't know that,

#x^2+x+1 = x^2 + x + 1/4 - 1/4 + 1#

#=(x+1/2)^2 + 3/4#

So this is zero when #(x+1/2)^2 = -3/4#

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

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