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

I'm pretty sure this is the wrong section, but there is no there moving it so.. i'll help solve

$\displaystyle x(x^2 +1) = 40$

so, this implies$\displaystyle x = 40$ and $\displaystyle (x^2 + 1) = 40$

$\displaystyle x^2 + 1 = 40 $

$\displaystyle x^2 = 40-1$

$\displaystyle x = +sqrt(39)$ And,

$\displaystyle x = -sqrt(39)$ - 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-

Quote:

$\displaystyle x(x^2 +1) = 40$

so, this implies$\displaystyle x = 40$ and $\displaystyle (x^2 + 1) = 40$

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 $\displaystyle 1^3+ 1= 2\ne 40$, $\displaystyle 2^3+ 2= 10\ne 40$, 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.

Quote:

$\displaystyle x^2 + 1 = 40 $

$\displaystyle x^2 = 40-1$

$\displaystyle x = +sqrt(39)$ And,

$\displaystyle x = -sqrt(39)$