Question

can anyone solve x^2+x+1=0---quadratic equation by completeting the square

Answer 1

We have
\[x^2+x+1=0\]
We know that \[(x+a)^2=x^2+2ax+a^2\]
here coefficient of x=1
so
\[2a=1\]
\[a=\frac{1}{2}\]
so we need to add and subtract \(\frac{1}{2^2}\) to complete square
\[x^2+x+\frac{1}{2^2}-\frac{1}{2^2}+1=0\]
we get
\[(x+\frac{1}{2})^2 +(\frac{\sqrt{3}}{2})^2=0\]
now we get
\[(x+\frac{1}{2})^2 =-(\frac{\sqrt{3}}{2})^2\]
we know \(\sqrt{-1}=i\)
so we get
\[(x+\frac{1}{2})=\pm i\frac{\sqrt{3}}{2}\]
we get
\[x=\frac{-1}{2}\pm i\frac{\sqrt{3}}{2}\]
or
\[x=\frac{-1\pm i\sqrt{3}}{2}\]

Answer 2

wow. thanks for be so detailed.

Answer 3

Welcome, did you understand?

Answer 4

I understand the formula of needing to find the (b) of 2ab but I got lost in the ^3/2 part

Answer 5

once i rework the problem I should be able to understand your answer

Answer 6

Yeah we get
\[-\frac{1}{4}+1= \frac{3}{4}= {(\frac{\sqrt 3}{2})}^2\]

Answer 7

o i c now. that makes sense!