Disequazione irrazionale $\sqrt{x^2-3x}\le 2$

Text

Solve the irrational inequality $\sqrt{x^2-3x}\le 2.$

Solution

Existence condition: $x^2-3x\ge 0\implies x\le 0\ \vee\ x\ge 3$. The right-hand side is positive; squaring: $x^2-3x\le 4\implies x^2-3x-4\le 0\implies (x-4)(x+1)\le 0\implies -1\le x\le 4.$ Intersecting with the existence condition: $\boxed{-1\le x\le 0\ \vee\ 3\le x\le 4.}$