Inequality with absolute values and squares (|x²-4| > |x-2|)

Statement

Solve the inequality $|x^2-4|>|x-2|$.

Solution

Both sides are absolute values: we square. Observing that $x^2-4=(x-2)(x+2)$: $(x^2-4)^2>(x-2)^2 \iff (x-2)^2(x+2)^2-(x-2)^2>0 \iff (x-2)^2\bigl[(x+2)^2-1\bigr]>0.$ The factor $(x-2)^2\ge0$ vanishes at $x=2$ (where the strict inequality does not hold). The factor $(x+2)^2-1=(x+1)(x+3)>0$ gives $x<-3\ \lor\ x>-1$.

Solution: $\boxed{x<-3\ \lor\ (x>-1\ \land\ x\neq2)}$.

Links

Topics: Parabola
Concepts: Absolute value inequalities · Absolute value
Methods: Absolute value mixed cases
Skills: Solving inequalities
Exercise type: Solving an inequality