Results 1 to 9 of 9

- December 18th 2009, 01:46 PM #1

- Joined
- Nov 2009
- Posts
- 87

- December 18th 2009, 01:48 PM #2

- Joined
- Jun 2009
- From
- Africa
- Posts
- 641

- December 18th 2009, 01:53 PM #3

- Joined
- Nov 2009
- Posts
- 87

- December 18th 2009, 01:55 PM #4

- Joined
- Jun 2009
- From
- Africa
- Posts
- 641

- December 18th 2009, 01:59 PM #5

- Joined
- Nov 2009
- Posts
- 87

- December 18th 2009, 02:33 PM #6

- Joined
- Nov 2009
- Posts
- 268

- December 19th 2009, 05:49 AM #7

- Joined
- Apr 2005
- Posts
- 18,075
- Thanks
- 2390

- December 19th 2009, 05:54 AM #8

- December 22nd 2009, 02:36 PM #9

- Joined
- Dec 2009
- Posts
- 3,120
- Thanks
- 1

The imaginary zeros are obtained from x^2+2x+15,

since this factor of the cubic does not itself have real factors,

it has complex factors since b^2<4ac.

x^2+2x+15 is zero for x = {-2+sqrt(4-60)}/2

and for x = {-2-sqrt(4-60)}/2,

which are x = -1+i(sqrt[14]) and x = -1-i(sqrt[14]).