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

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Answer 1 · 2015-06-13T18:32:10

This quadratic has no real roots, but:

$x^{2} + x + 1 = {(x + \frac{1}{2})}^{2} + \frac{3}{4}$ $x^2+x+1 = (x+1/2)^2+3/4$

giving complex solutions: $x = - \frac{1}{2} \pm \frac{\sqrt{3}}{2} i$ $x = -1/2 +-sqrt(3)/2i$

$x^{2} + x + 1$ $x^2+x+1$ is one of the factors of $x^{3} - 1$ $x^3-1$

$x^{3} - 1 = (x - 1) (x^{2} + x + 1) = (x - 1) (x - ω) (x - ω^{2})$ $x^3-1 = (x-1)(x^2+x+1) = (x-1)(x-omega)(x-omega^2)$

where $ω = - \frac{1}{2} + \frac{\sqrt{3}}{2} i = cos (\frac{2 π}{3}) + i sin (\frac{2 π}{3})$ $omega = -1/2+sqrt(3)/2i = cos((2pi)/3)+isin((2pi)/3)$

is called the primitive complex root of unity.

Pretending we don't know that,

$x^{2} + x + 1 = x^{2} + x + \frac{1}{4} - \frac{1}{4} + 1$ $x^2+x+1 = x^2 + x + 1/4 - 1/4 + 1$

$= {(x + \frac{1}{2})}^{2} + \frac{3}{4}$ $=(x+1/2)^2 + 3/4$

So this is zero when ${(x + \frac{1}{2})}^{2} = - \frac{3}{4}$ $(x+1/2)^2 = -3/4$

Hence $x + \frac{1}{2} = \pm \sqrt{- \frac{3}{4}} = \pm \frac{\sqrt{3}}{2} i$ $x+1/2 = +-sqrt(-3/4) = +-sqrt(3)/2i$

So $x = - \frac{1}{2} \pm \frac{\sqrt{3}}{2} i$ $x =-1/2+-sqrt(3)/2i$

How do you solve x2+x+1=0 x^2 + x + 1=0 using completing the square?