Disequazione irrazionale $\sqrt{x^2-5}<2$

Text

Solve the irrational inequality $\sqrt{x^2-5}<2.$

Solution

Existence condition: $x^2-5\ge 0\implies x\le -\sqrt5\ \vee\ x\ge \sqrt5$. The right-hand side is positive; squaring: $x^2-5<4\implies x^2<9\implies -3<x<3.$ Intersecting with the existence condition: $\boxed{-3<x\le -\sqrt5\ \vee\ \sqrt5\le x<3.}$