Yes jam2011,

good method, a variation of Plato's.

However, you've mistyped it on the 2nd line...

$\displaystyle |x+2|<|x(x-1)|$

$\displaystyle |x+2|-|x(x-1)|<0$

$\displaystyle (x+2)^2-\left[x(x-1)\right]^2<0$

$\displaystyle \left[x+2+x(x-1)\right]\left[x+2-x(x-1)\right]<0$

$\displaystyle \left(2+x^2\right)\left(2x+2-x^2\right)<0$

$\displaystyle 2+x^2>0$

therefore

$\displaystyle 2x+2-x^2<0\Rightarrow\ x^2-2x-2>0$

Find the roots and the inequality is true for x to the left of the smaller root

and to the right of the larger root.