Find all values for $x$ such that $|x^2|>|3x-2|$

Question

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!

Answer 1

Hint: Consider two cases: $3x - 2 \ge 0$, i.e. $x \ge \frac{2}{3}$ and $x < \frac{2}{3}$. Note we have $|x^2| = x^2$.

Answer 2

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.

Answer 3

Squaring gives: $$x^4>(3x-2)^2$$ or $$x^4-(3x-2)^2>0$$ or $$(x^2-3x+2)(x^2+3x-2)>0$$ or $$(x-1)(x-2)\left(x-\frac{-3+\sqrt{17}}{2}\right)\left(x-\frac{-3-\sqrt{17}}{2}\right)>0,$$ which gives the answer: $$\left(-\infty,\frac{-3-\sqrt{17}}{2}\right)\cup\left(\frac{-3+\sqrt{17}}{2},1\right)\cup(2,+\infty)$$