Disequazione irrazionale $\sqrt{3x-2}\ge 4$

Text

Solve the irrational inequality $\sqrt{3x-2}\ge 4.$

Solution

Existence condition: $3x-2\ge 0\implies x\ge \tfrac23$. The right-hand side is positive; squaring: $3x-2\ge 16\implies 3x\ge 18\implies x\ge 6.$ Since $x\ge 6$ satisfies the existence condition: $\boxed{x\ge 6.}$