2y^3+2y^2-12y+9<0
2006-11-29 02:07:20 · 5 answers · asked by Phill W 1 in Science & Mathematics ➔ Mathematics
I'm still struggling with this one.
+ + - + (2 or 0 pos sol'n)
- + + + (1 negative sol'n)
I would have thought it would have to be factorable somehow, but none of:
{ -9, -4.5, -3, -1.5, -1, -.5}
work. In fact, just trying numbers, it looks like the only place it crosses the x-axis is somewhere in the vicinity of -3.265.
I don't understand why none of those factors works, though. Since all the coefficients are integers, I would have expected one of them to work....?
Ugh, dude, I found an equation solver on the web.
Plug your equation in here:
http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=quickmath&s1=equations&s2=solve&s3=basic
The only real root is approximately -3.26215. So the answer would be y < -3.26215
Are you sure you copied the coefficients correctly?
2006-11-29 02:49:32 · answer #1 · answered by Jim Burnell 6 · 0⤊ 0⤋
I see an inequality. I don't see a question. You want to solve for y?
There is just one real zero at y = -3.262. For y > than that, the value of the polynomial is > 0, and for y < that, the value of the polynomial is < 0.
2006-11-29 12:06:50 · answer #2 · answered by Philo 7 · 0⤊ 0⤋
What's your question? Are you trying to find y or what?
2006-11-29 10:08:51 · answer #3 · answered by leaptad 6 · 0⤊ 0⤋
what are you looking for here? the y value?? Solve it for y??...factors??? Please be specific about the question.
2006-11-29 10:09:32 · answer #4 · answered by Ashi 2 · 0⤊ 0⤋
No you don't ;-)
2006-11-29 10:11:38 · answer #5 · answered by Dik voor Mekaar 1 · 0⤊ 0⤋