Inequality with two roots

Statement

Solve $\sqrt{x+3}<\sqrt{x-1}+1$ (reducible to Type II after manipulation).

Solution

Existence conditions: both roots are needed: $x+3\ge 0$ and $x-1\ge 0$, that is $x\ge 1$.

Both sides are $\ge 0$, so we can square: $x+3 < (x-1) + 2\sqrt{x-1} + 1 \iff x+3 < x + 2\sqrt{x-1} \iff 3 < 2\sqrt{x-1}.$ Hence $\sqrt{x-1} > \dfrac{3}{2}$; squaring again (both sides $\ge 0$): $x-1 > \dfrac{9}{4} \iff x > \dfrac{13}{4}$.

Intersecting with the existence condition $x\ge 1$: $\boxed{x > \dfrac{13}{4}}$.

Links

Topics: Radicals
Concepts: Existence conditions · Irrational inequality
Methods: Irrational inequality case analysis
Skills: Solving inequalities
Exercise type: Solving an inequality