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
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
$\displaystyle x(x^2 +1) = 40$
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,
$\displaystyle x = -sqrt(39)$
Yes, it is, but we can't expect people taking basic "algebra" to have any idea what "abstract algebra " is, so-
NO, it does NOT! In fact, that is impossible. If x= 40 and $\displaystyle x^2+ 1= 40$, them $\displaystyle x(x^2+1)= 40(40)= 1600$, 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.$\displaystyle x(x^2 +1) = 40$
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.
$\displaystyle x^2 + 1 = 40 $
$\displaystyle x^2 = 40-1$
$\displaystyle x = +sqrt(39)$ And,
$\displaystyle x = -sqrt(39)$