<s>[INST] solve $|x-6|>|x^2-5x+9|$ 
solve $|x-6|>|x^2-5x+9|,\ \ x\in \mathbb{R}$

I have done $4$ cases.
$1.)\ x-6>x^2-5x+9\ \ ,\implies  x\in \emptyset \\
2.)\ x-6<x^2-5x+9\ \ ,\implies x\in \mathbb{R} \\
3.)\  -(x-6)>x^2-5x+9\ ,\implies 1<x<3\\
4.)\  (x-6)>-(x^2-5x+9),\ \implies x>3\cup x<1 $
I am confused on how I proceed.
Or if their is any other short way than making $4$ cases than I would like to know. 
I have studied maths up to $12$th grade. Thanks.
 [/INST] Another way is to note the inequality is equivalent to
$$(x-6)^2>(x^2-5x+9)^2\iff -(x-1)(x-3)(x^2-6x+15)>0$$
The quadratic is always positive, so this is the same as $(x-1)(x-3)<0$ which means $x\in (1,3)$. 
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