1. real values for x

Hi, I have this part of a question that gets me:
I did all the previous steps and got to an equation that must =0:

x^2(-x+12)=0

I know that making x=12 is one value, but for the x^2 part, do i just ignore it?
In my answer book it says x=12 and nothing else, so I'm presuming its something to do with real values of x (even though I don't really know what a 'real' value of x is).

Thanks

2. For $\displaystyle x \neq 0$, divide both sides by $\displaystyle x^2$.

$\displaystyle \frac{x^2(12-x)}{x^2} = \frac{0}{x^2}$

$\displaystyle 12 - x = 0$

$\displaystyle x = 12$.

We found $\displaystyle x=12$, which makes the equation true, but don't forget that we didn't test $\displaystyle x$ for $\displaystyle 0$. If we do,
$\displaystyle 0 (12 - 0) = 0$

Yes, $\displaystyle x = 0$ is a root too.

$\displaystyle x = \{0,12\}$

3. Another way (which you may find easier) to solve the equation:

If the product of two values is $\displaystyle 0$, then at least one of them must be $\displaystyle 0$.

$\displaystyle \underbrace{(x^2)}_{x^2=0}\underbrace{(12-x)}_{12-x=0} = 0$

Either $\displaystyle x^2$ or $\displaystyle 12 - x$ or both of them must be $\displaystyle 0$. So,

$\displaystyle x^2 = 0$
$\displaystyle x = 0$

And,

$\displaystyle 12-x = 0$
$\displaystyle x = 12$

Thus,

$\displaystyle x = \{0, 12\}$