Answer 1

how does \[r^2+2 r+\lambda = 0\] come out to be \[-1\pm \sqrt{1-\lambda}\] I got \[-1\pm \sqrt{4-4\lambda} = 1\pm2\sqrt{1-\lambda}\]

Answer 2

*edit- the last equation should be -1 plus/minus ...

Answer 3

\[r^2 + 2r + 1 - 1 + \lambda= 0 \]
\[(r + 1)^2 = 1 - \lambda\]
\[r + 1 = \pm\sqrt{1 - \lambda}\]
\[r = -1 \pm \sqrt{1 - \lambda}\]

Answer 4

wow, forgot to divide by 2 for both real AND imaginary parts using quadratic eq.. thanks!

Answer 5

No Problem :)