Worked exercise — Type II with a semicircle

Statement

Classify and solve: $\sqrt{9-x^2}\ge x-1.$

Solution

It is a Type II ($\sqrt{A}\ge B$): a union of two cases. The CE is $9-x^2\ge 0 \iff -3\le x\le 3$.

Case 1 ($x-1<0$, that is $x<1$): the inequality is automatically true within the CE. We obtain $-3\le x<1$.

Case 2 ($x-1\ge 0$, that is $x\ge 1$): we square. $9-x^2 \ge (x-1)^2 = x^2-2x+1 \implies 2x^2-2x-8\le 0 \implies x^2-x-4\le 0.$ The zeros are $x=\dfrac{1\pm\sqrt{17}}{2}\approx -1{,}56 \ \text{e}\ 2{,}56$, so $x^2-x-4\le 0$ for $\dfrac{1-\sqrt{17}}{2}\le x\le \dfrac{1+\sqrt{17}}{2}$. Intersecting with $x\ge 1$ and with the CE $x\le 3$: $1\le x\le \dfrac{1+\sqrt{17}}{2}$.

Union: $\bigl[-3,\,1\bigr) \cup \Bigl[1,\ \tfrac{1+\sqrt{17}}{2}\Bigr]$.

Solution: $\boxed{\,-3\le x\le \dfrac{1+\sqrt{17}}{2}\,}$ (with $\tfrac{1+\sqrt{17}}{2}\approx 2{,}56$).

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