Special case - modulus greater than or equal to B(x)

When the modulus is greater than something, $|A(x)|\ge B(x)$, the shortcut is different: you distinguish the two cases according to the sign of the right-hand side $B(x)$.

Property — $|A(x)|\ge B(x)$

The inequality $|A(x)|\ge B(x)$ is solved as the union of two cases:

• Case (a): $B(x)<0$. An absolute value is always $\ge0$, so it is automatically greater than a negative number: the inequality is always true. The solution of the case is $\{x:\ B(x)<0\}$.
• Case (b): $B(x)\ge0$. Then $|A(x)|\ge B(x)$ is equivalent to $A(x)\ge B(x)\ \lor\ A(x)\le-B(x)$.

Final solution: the union of $S_a$ and $S_b$.

Diagram for $|A(x)|\ge B(x)$: if the right-hand side is negative the inequality is already true.

Example

$|x-1|\ge2$. Here $A(x)=x-1$ and $B(x)=2$ is a positive constant: case (a) is empty and case (b) remains: $x-1\ge2\ \lor\ x-1\le-2 \iff x\ge3\ \lor\ x\le-1.$ Solution: $\boxed{x\le-1\ \lor\ x\ge3}$.

Example

$|x^2-3x|\ge x-4$. Here $A(x)=x^2-3x$ and $B(x)=x-4$.

Case (a): $x-4<0 \iff x<4$. In this zone the inequality is always true, so $S_a=(-\infty;4)$.

Case (b): $x\ge4$. We must solve $x^2-3x\ge x-4\ \lor\ x^2-3x\le-(x-4)$, that is:

• $x^2-4x+4\ge0 \iff (x-2)^2\ge0$: always true;
• $x^2-2x-4\le0 \iff 1-\sqrt5\le x\le1+\sqrt5$, which intersected with $x\ge4$ gives the empty set.

The first is always true, so $S_b=[4;+\infty)$.

Union: $S=S_a\cup S_b=\mathbb{R}$: the inequality is always satisfied.

Warning — A frequent mistake

When in $|A(x)|\ge B(x)$ the term $B(x)$ is always non-negative (for example a positive constant or a sum of squares), case (a) is empty and you go straight to case (b): $A(x)\ge B(x)\ \lor\ A(x)\le-B(x)$. But if $B(x)$ can take negative values (for example it is a polynomial), do not forget case (a), otherwise you lose all the solutions in which the right-hand side is negative.

Links

Topics: Parabola
Concepts: Inequalities with absolute value · Absolute value
Skills: Reasoning by cases · Solving inequalities