Hi,
how do you solve the following equation by completing the square:
x2 +6x+17=0
Thanks!
+6251 \(x^2 + 6x + 17 =0 \\ x^2 + 6x +9 -9 + 17 = 0\\ (x+3)^2+8=0 \\ (x+3)^2 = -8 \\ \mbox{and this clearly has no solution over the real numbers as any number squared is non-negative.}\)
\(\mbox{If we solve this over the complex numbers we get}\\ (x+3)^2=-8 = (-1)8 \\ (x+3) = \pm \sqrt{-1}\sqrt{8} = \pm \imath \sqrt{8} = \pm \imath 2\sqrt{2} \\ x = -3 \pm \imath 2\sqrt{2}\)
.+6251 \(x^2 + 6x + 17 =0 \\ x^2 + 6x +9 -9 + 17 = 0\\ (x+3)^2+8=0 \\ (x+3)^2 = -8 \\ \mbox{and this clearly has no solution over the real numbers as any number squared is non-negative.}\)
\(\mbox{If we solve this over the complex numbers we get}\\ (x+3)^2=-8 = (-1)8 \\ (x+3) = \pm \sqrt{-1}\sqrt{8} = \pm \imath \sqrt{8} = \pm \imath 2\sqrt{2} \\ x = -3 \pm \imath 2\sqrt{2}\)
