inequation with complex solutions

Question

Could somebody please help me solve this inequality: $|x-2| < x|x|$ I tried to solve it by using three different values of $x$: 1. $x < 0$ Solution : $1/2 - \sqrt{7}i/2 < x < 1/2 + \sqrt{7}i/2$

2. $0 \leq x \leq 2$ Solution: $x \in (-\infty, -2] \cup [1,\infty)$

3. $x > 2$ Solution: $x \in (-\infty,1/2 + \sqrt{7}i/2) \cup (1/2 + \sqrt{7}i/2, \infty)$

I don't know how to choose the right solution so I would really apreciate your help. Thanks

Answer 1

Clearly, $x$ is real and $x\ne0$

$$x>\frac{|x-2|}{|x|}\ge0$$

So, we have $x^2>|x-2|$

If $x-2<0\iff x<2, x^2>-(x-2)$ $\iff x^2+x-2>0\iff(x+2)(x-1)>0$

$\implies$ either $x>1$ or $x<-2$ but $x<2\implies x<-2$

Check if $x-2\ge0\iff x\ge2$ and $|x-2|=+(x-2)$

Answer 2

squaring the given inequality we get $x^2-4x+4<x^2$ from here we get $1<x$