For equations involving |x| and $x^2$: since $x^2$=|x|^2, substitute t=|x| (t>=0). Convert to a standard quadratic in t. Solve for t, reject negative values. Each positive t gives x=+/-t (2 solutions). t=0 gives x=0 (1 solution). Example: |x|^2-5|x|+6=0 → $t^{2-5t+6}$=0 → t=2,3. Solutions: x=+/-2, +/-3 (4 solutions). For inequalities like |x|^2-3|x|-4<0: $t^{2-3t-4}$<0 → (t-4)(t+1)<0 → -1<t<4. Since t>=0: 0<=t<4. So -4<x<4 (but x!=0 is allowed since |0|=0<4).