Kriemhild Posted December 19, 2011 Share Posted December 19, 2011 Unfortunately, I got the numbers right, and it devolves into a quadratic equation that has imaginary numbers in it. So...yeah, I think that, even if I bothered to finish it, it's WAY over what the professor would want. Sorry. Um, imaginary numbers? In my attempt, I found all three solutions to be real... Are you sure you got imaginary numbers or did I make a mistake somewhere? Quote Link to comment Share on other sites More sharing options...
Aleph Posted December 20, 2011 Share Posted December 20, 2011 if it's quadratic then it's a feeble prablem regardless of imagination Quote Link to comment Share on other sites More sharing options...
Kngt_Of_Titania Posted December 20, 2011 Share Posted December 20, 2011 (edited) Um, imaginary numbers? In my attempt, I found all three solutions to be real... Are you sure you got imaginary numbers or did I make a mistake somewhere? I'll give it another shot, but I did get imaginary numbers for when I solved for s and t (not x or y) using that method. I know I checked my answer up to the "reduced" cubic, and it worked out when I plugged in -12.33333 for y (which would correspond to 4 for x, a known correct answer). So if I screwed up, it'd have to be when I converted it to a quadratic equation. EDIT: That being said, when I started pulling imaginary numbers, it was 1:30 in the morning, I had just gotten off a shift at 11:00, and I had to work a children's carnival at 8:00 in the morning, so I just said "Fuck it" and left it. Later on, I realized after doing it roughly in my head that, due to the way s and t are calculated, that the imaginary number would likely cancel out. Edited December 20, 2011 by Kngt_Of_Titania Quote Link to comment Share on other sites More sharing options...
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