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

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