<s>[INST] Find all values for $x$ such that $|x^2|>|3x-2|$ I have absolutely no idea how to even start this inequality - I have seen some methods involving squaring both sides and rearranging to get $x^4 - 9x^2 + 12x - 4$, but have no idea where to go from there.
Any help would be greatly appreciated!
 [/INST] Clearly $x^2\geq0$ for all real $x$ (and $x=0$ doesn't satisfy the inequality).
So you only need to consider the inequalities $x^2>3x-2$ or $x^2>2-3x$.
From the first you have $x^2-3x+2=(x-2)(x-1)>0$, so $x<1$ or $x>2.$
The second gives $x^2+3x-2>0,$ so  $x<\frac{1}{2}(-3-\sqrt{17})$ or $x>\frac{1}{2}(-3+\sqrt{17}).$ This is a quadratic, so find the roots and determine which points $x$ satisfy the inequality (you can draw a sketch).
Thus combining the solutions we have $x<\frac{1}{2}(-3-\sqrt{17}),$ $x>2$ or $\frac{1}{2}(\sqrt{17}-3)<x<1.$
See the graph here.
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