Talk:Roots of a quadratic function: Difference between revisions
Content added Content deleted
No edit summary |
|||
| Line 1: | Line 1: | ||
I clearly see come of programs (Ada, and some marked as "translation of Ada") use |
|||
(b^2-4ac)/2*a, |
|||
there correct way is |
|||
(b^2-4ac)/(2*a). |
|||
I wonder why noone spotted this earlier! |
|||
<cite>Forsythe, Michael Malcolm and Cleve Mole suggest to try it on a=1, b=-10<sup>5</sup>, c=1</cite>, but Ada sample code uses -10e5, which is indeed -1e6 (-10<sup>6</sup>), if I was not wrong since I knew the "e" notation... --[[User:ShinTakezou|ShinTakezou]] 21:24, 22 June 2009 (UTC) |
<cite>Forsythe, Michael Malcolm and Cleve Mole suggest to try it on a=1, b=-10<sup>5</sup>, c=1</cite>, but Ada sample code uses -10e5, which is indeed -1e6 (-10<sup>6</sup>), if I was not wrong since I knew the "e" notation... --[[User:ShinTakezou|ShinTakezou]] 21:24, 22 June 2009 (UTC) |
||
Revision as of 10:56, 16 September 2010
I clearly see come of programs (Ada, and some marked as "translation of Ada") use (b^2-4ac)/2*a, there correct way is (b^2-4ac)/(2*a). I wonder why noone spotted this earlier!
Forsythe, Michael Malcolm and Cleve Mole suggest to try it on a=1, b=-105, c=1, but Ada sample code uses -10e5, which is indeed -1e6 (-106), if I was not wrong since I knew the "e" notation... --ShinTakezou 21:24, 22 June 2009 (UTC)
J example
Dumontier, hope you don't mind me replacing your example code. I understood that you were trying to illustrate the generality of p. however your example used a quadratic, that had already been shown above. If you were trying to illustrate some other point I apologise! --Tikkanz 23:14, 14 October 2009 (UTC)