Worked exercise — Sum of two radicals

Statement

Classify and solve: $\sqrt{x-1}+\sqrt{x+1}\le 0.$

Solution

It is an always non-negative sum: two radicals, each $\ge 0$. CE: $x-1\ge 0$ and $x+1\ge 0$, that is $x\ge 1$.

Within the CE the sum is $\ge 0$, and it equals $0$ only if both terms are zero simultaneously: $x-1=0$ and $x+1=0$, incompatible conditions. Indeed, for every $x\ge 1$ we have $\sqrt{x+1}\ge\sqrt2>0$, so the sum is strictly positive. There is no $x$ for which it is $\le 0$.

Solution: $\boxed{\,\text{nessuna soluzione}\,}$ ($\varnothing$).

Links

Topics: Irrational inequalities
Concepts: Existence conditions · Irrational inequality
Methods: Non-negative sides shortcut
Skills: Reasoning by cases · Solving inequalities
Exercise type: Solving an inequality