Inequality with the root on the smaller side

Statement

Solve $\sqrt{9-x^2}<x+1$.

Solution

The root is on the smaller side: this is Type II, a single system of three conditions. $\begin{cases}9-x^2\ge 0 \\ x+1 > 0 \\ 9-x^2 < (x+1)^2\end{cases}$

• First: $-3\le x\le 3$.
• Second: $x>-1$.
• Third: $9-x^2 < x^2+2x+1 \iff 2x^2+2x-8>0 \iff x^2+x-4>0 \iff x<\dfrac{-1-\sqrt{17}}{2}\ \vee\ x>\dfrac{-1+\sqrt{17}}{2}$.

Intersecting the three conditions: $\dfrac{-1+\sqrt{17}}{2}<x\le 3$.

Solution: $\boxed{\dfrac{-1+\sqrt{17}}{2}<x\le 3}$.

Links

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