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

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Answer 1 · 2015-05-04T14:56:48

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

Answer 2 · 2015-05-04T18:06:17

In this way:

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

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

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

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