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

May 4, 2015

This equation can't be solved by completing the square.
Since D = b^2 - 4ac = 1 - 4 = -3 < 0, there are complex roots.

May 4, 2015

In this way:

${x}^{2} + x + 1 = 0 \Rightarrow {x}^{2} + x + \frac{1}{4} - \frac{1}{4} + 1 = 0 \Rightarrow$

${\left(x + \frac{1}{2}\right)}^{2} = - 1 + \frac{1}{4} \Rightarrow {\left(x + \frac{1}{2}\right)}^{2} = - \frac{3}{4}$,

that is an impossible equation (no solutions) because a square can't be negative.