Inequality with the root on the greater side

Statement

Solve $\sqrt{2x+1}>x-1$.

Solution

The root is on the greater side: this is Type I, a union of two cases.

Case (a): $\begin{cases}2x+1\ge 0 \\ x-1 < 0\end{cases} \iff -\tfrac{1}{2}\le x<1$.

Case (b): $\begin{cases}x-1\ge 0 \\ 2x+1 > (x-1)^2\end{cases}$. The second: $2x+1>x^2-2x+3 \iff x^2-4x+2<0 \iff 2-\sqrt{2}<x<2+\sqrt{2}$. Intersecting with $x\ge 1$: $1\le x<2+\sqrt{2}$.

Union: $\boxed{-\dfrac{1}{2}\le x < 2+\sqrt{2}}$.

Links

Topics: Radicals
Concepts: Irrational inequality
Methods: Irrational inequality case analysis
Skills: Reasoning by cases · Solving inequalities
Exercise type: Solving an inequality