Question

completing the square
x^2+x+1=0

Answer 1

i think ur question is x^2+2x+1=0

(x+1)(x+1) i think its like this !

Answer 2

not that

Answer 3

\[x^2+x=-1\]
\[x^2+1\cdot x+(\frac{1}{2})^2=-1+(\frac{1}{2})^2\]
\[(x+\frac{1}{2})^2=-1+\frac{1}{4}\]
\[(x+\frac{1}{2})^2=\frac{-3}{4}\]
\[x+\frac{1}{2}=\pm \sqrt{\frac{-3}{4}}\]
\[x=-\frac{1}{2} \pm \sqrt{\frac{-3}{4}}\]
\[x=-\frac{-1}{2} \pm i \frac{\sqrt{3}}{2}\]

Answer 4

the steps:
x^2 + x + 1/4 = -1 + 1/4
(x+1/2)^2 = -3/4
x + 1/2 = (sqrt of -3 )over 2
x = [(i sqrt of 3) - 1]/2