A complete inequality from start to finish

The concluding example brings together all the steps: factorisation of numerator and denominator, four cornerstones, a full sign table and reading off the solution while excluding the points of non-existence.

Example — A complete inequality, from start to finish

Solve $\dfrac{x^2+x-2}{x^2-9} \ge 0$.

Factorisation: $\frac{x^2+x-2}{x^2-9} = \frac{(x+2)(x-1)}{(x+3)(x-3)}.$ Cornerstones: $-3$ (denominator), $-2$ (numerator), $1$ (numerator), $3$ (denominator).

	$x<-3$	$x=-3$	$-3<x<-2$	$x=-2$	$-2<x<1$	$x=1$	$1<x<3$	$x=3$	$x>3$
$x+3$	$-$	$0$	$+$	$+$	$+$	$+$	$+$	$+$	$+$
$x+2$	$-$	$-$	$-$	$0$	$+$	$+$	$+$	$+$	$+$
$x-1$	$-$	$-$	$-$	$-$	$-$	$0$	$+$	$+$	$+$
$x-3$	$-$	$-$	$-$	$-$	$-$	$-$	$-$	$0$	$+$
product	$+$	$\nexists$	$-$	$0$	$+$	$0$	$-$	$\nexists$	$+$

Solution: $\ge 0$ (sign $+$ or zero), excluding $x=\pm 3$: $\boxed{x < -3 \;\;\text{or}\;\; -2\le x\le 1 \;\;\text{or}\;\; x > 3}$

Links

Topics: Inequalities
Concepts: Fractional inequality · Point of non-existence · Factorisation · Sign analysis · Sign table
Skills: Solving inequalities · Factorising