Math Help - Seemingly Easy??

1. Seemingly 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

2. Originally Posted by bandofhorses11
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:

$x^3 + x = 40$

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

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

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

$x^2 + 1 = 40$
$x^2 = 40-1$
$x = +sqrt(39)$ And,
$x = -sqrt(39)$

3. Originally Posted by treetheta
I'm pretty sure this is the wrong section, but there is no there moving it so.. i'll help solve
Yes, it is, but we can't expect people taking basic "algebra" to have any idea what "abstract algebra " is, so-

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

so, this implies $x = 40$ and $(x^2 + 1) = 40$
NO, it does NOT! In fact, that is impossible. If x= 40 and $x^2+ 1= 40$, them $x(x^2+1)= 40(40)= 1600$, not 40. You may be thinking of the fact that if ab= 0 then either a=0 or (not "and") b= 0, but that is a special property of "0" and is not true for any other number.

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

$x^2 + 1 = 40$
$x^2 = 40-1$
$x = +sqrt(39)$ And,
$x = -sqrt(39)$