Hi, I've been trying to solve this problem algebraically for a while, but I can't seem to figure it out. I don't THINK it's pre university math... but maybe I'm just missing something. How would you solve this step by step? Thanks:

x^3 + x = 40

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- Dec 10th 2009, 12:18 PMbandofhorses11Seemingly Easy??
Hi, I've been trying to solve this problem algebraically for a while, but I can't seem to figure it out. I don't THINK it's pre university math... but maybe I'm just missing something. How would you solve this step by step? Thanks:

x^3 + x = 40 - Dec 10th 2009, 06:18 PMtreetheta
- Dec 11th 2009, 05:40 AMHallsofIvy
Yes, it is, but we can't expect people taking basic "algebra" to have any idea what "abstract algebra " is, so-

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so, this implies and

By the "rational root theorem" any rational root must be an integer divisor of 40. By "Descartes' rule of signs", this equation has no negative roots and , , while any positive integer divisor of 40 larger than 2 already has cube larger than 40. Therefore there are no rational roots and you would probably have to resort to "Cardano's cubic formula" http://en.wikipedia.org/wiki/Cubic_function - which very complex. There is not going to be any simple solution to this equation.

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And,