Worked exercise — Type I with a strict root

Statement

Classify and solve: $\sqrt{2x+1}<x-1.$

Solution

It is a Type I ($\sqrt{A}<B$), with a strict inequality. Equivalent system (the right-hand side must be strictly positive, because a root cannot be less than $0$): $\begin{cases} 2x+1\ge 0 \\ x-1> 0 \\ 2x+1 < (x-1)^2 \end{cases}$ First (CE): $x\ge -\tfrac12$. Second: $x>1$. Third: $2x+1 < x^2-2x+1 \implies 0 < x^2-4x \implies x(x-4)>0 \implies x<0 \vee x>4$. Intersection: $x\ge -\tfrac12 \ \wedge\ x>1 \ \wedge\ (x<0 \vee x>4)$. The first two give $x>1$; with the third, only $x>4$ survives.

Solution: $\boxed{\,x>4\,}$.

Links

Topics: Irrational inequalities
Concepts: Existence conditions · Irrational inequality
Methods: Irrational inequality by cases
Skills: Reasoning by cases · Solving inequalities
Exercise type: Solving an inequality