Worked exercise — Fractional inequality with a radical

Statement

Classify and solve: $\dfrac{\sqrt{x^2-4}}{x-3}\le 0.$

Solution

It is a fractional inequality with a radical: one studies the sign of the numerator and of the denominator.

Overall CE: $x^2-4\ge 0 \iff x\le -2 \vee x\ge 2$, and $x-3\ne 0 \iff x\ne 3$.

Numerator $\sqrt{x^2-4}$: it is $\ge 0$ wherever it exists, and equals $0$ only at $x=\pm 2$.

Denominator $x-3$: negative for $x<3$, positive for $x>3$.

The fraction is $\le 0$ when the numerator is zero (points $x=\pm 2$, which give the value $0$), or when the numerator is positive and the denominator negative. Combining with the CE:

• $x\le -2$: numerator $\ge 0$, denominator $<0$ $\Rightarrow$ fraction $\le 0$ ✓
• $2\le x<3$: numerator $\ge 0$, denominator $<0$ $\Rightarrow$ fraction $\le 0$ ✓
• $x>3$: numerator $>0$, denominator $>0$ $\Rightarrow$ fraction $>0$ ✗ (and $x=3$ is excluded by the CE)

Solution: $\boxed{\,x\le -2 \ \vee\ 2\le x<3\,}$.

Links

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