x^3 + px^2 + qx + r = 0
p = -3
q = -2
r = -1
3 solutions, one real, and a complex conjugate pair
x1 = 3.62736508471183
x2 = -0.31368254235592 + i0.421052806987577
x3 = -0.31368254235592 - i0.421052806987577
See calculation:
http://www.geocities.com/derekcowley/public/cubic.xls
You will need the "Analysis ToolPak" and "Analysis ToolPak -VBA" add-ins in Excel to properly see this file.
based on cubic depression:
http://www.karlscalculus.org/cubic.html
2006-06-08 17:00:26
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answer #1
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answered by none2perdy 4
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The only possible rational factors are 1 and -1.
If you plug both of them in for x, you will see that neither is a solution.
Using my graphing calculator, I found that
x = 3.6273651 is the only real root.
2006-06-08 15:58:49
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answer #2
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answered by MsMath 7
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Using my texas voyage 200 I found the number is 3.62737, wich gives a result of 0.000077, if I had more decimals, it would approach more to zero.
2006-06-08 21:04:14
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answer #3
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answered by Anonymous
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x=-0.43016
I did a quick search on it. I may be wrong (Well, PC may be) but I'll try it myself. If you can't answer it, just write this answer down.
Or you might wanna listen to MathGirl )), for the name tells you, that she's the one to listen to (no joke). Hope She's right. Still, trying to find it..
2006-06-08 16:01:33
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answer #4
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answered by nikolka1992 3
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Try http://www.quickmath.com/ It's great for those stubborn problems you don't have time to mess with.
2006-06-08 16:05:58
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answer #5
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answered by anonymous 7
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do ur own homework kid
2006-06-08 15:55:56
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answer #6
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answered by Anonymous
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