Mixed cases with two absolute values

When two absolute values appear in an inequality, of the form $|A(x)|$ and $|B(x)|$, we cannot apply the schemes for a single absolute value: we need either a case analysis on the signs of the two arguments, or an algebraic trick.

The “squaring” algebraic trick. When both sides are absolute values, $|A|\lessgtr|B|$ (or $|A|=|B|$), squaring preserves the direction, because both are $\ge0$: the inequality is equivalent to $A^2\lessgtr B^2$, that is $(A-B)(A+B)\lessgtr0$, which is studied with the sign table. The idea of the “square as a shortcut for getting rid of the absolute value” is one of the basic heuristics discussed by Esty.

Property — Casework for $|A|<|B|$ and $|A|=|B|$

• $|A|=|B| \iff A=B\ \lor\ A=-B$ (the two absolute values coincide if the arguments are equal or opposite);
• $|A|<|B| \iff A^2<B^2 \iff (A-B)(A+B)<0$;
• $|A|\le|B| \iff (A-B)(A+B)\le0$.

Example — $|x-1|=|2x+3|$

Equivalent to $x-1=2x+3\ \lor\ x-1=-(2x+3)$:

• $x-1=2x+3 \implies x=-4$;
• $x-1=-2x-3 \implies 3x=-2 \implies x=-\frac23$.

Solution: $\boxed{x=-4\ \lor\ x=-\frac23}$.

Example — $|x-1|<|2x+1|$ by squaring

$(x-1)^2<(2x+1)^2 \iff (x-1)^2-(2x+1)^2<0$. Using the difference of squares: $\bigl[(x-1)-(2x+1)\bigr]\bigl[(x-1)+(2x+1)\bigr]=(-x-2)(3x)=-3x(x+2)<0,$ that is $x(x+2)>0$. Solution: $\boxed{x<-2\ \lor\ x>0}$.

Case discussion on the signs. When the squaring trick does not apply (for example $|A|+|B|=C$ with $C$ constant), one fills in the sign table of $A$ and $B$: each region singles out a “without absolute values” version of the inequality.

Example — $|x-2|+|x+1|=5$

Study of the signs: $A=x-2$ changes sign at $x=2$; $B=x+1$ changes sign at $x=-1$. Three regions:

• $x<-1$ ($A<0,\ B<0$): $-(x-2)-(x+1)=5 \implies -2x+1=5 \implies x=-2$ (compatible);
• $-1\le x<2$ ($A<0,\ B\ge0$): $-(x-2)+(x+1)=5 \implies 3=5$, false, no solution;
• $x\ge2$ (both $\ge0$): $(x-2)+(x+1)=5 \implies 2x-1=5 \implies x=3$ (compatible).

Solution: $\boxed{x=-2\ \lor\ x=3}$.

Geometric interpretation: $|x-2|+|x+1|$ is the sum of the distances of $x$ from the points $2$ and $-1$ on the real line. Since between $2$ and $-1$ the distance is $3$, the sum equals $5$ when $x$ lies outside the interval $[-1,2]$, at a distance $\frac{5-3}{2}=1$: we obtain $x=-2$ or $x=3$.

Example — $|x|+|x-3|<5$

Sign table of $x$ and $x-3$:

• $x<0$: $-x-(x-3)<5 \iff -2x+3<5 \iff x>-1$, that is $-1<x<0$;
• $0\le x<3$: $x-(x-3)<5 \iff 3<5$, always true, the whole of $[0,3)$;
• $x\ge3$: $x+(x-3)<5 \iff 2x<8 \iff x<4$, that is $3\le x<4$.

Union: $\boxed{-1<x<4}$.

In summary — "Mixed cases" decision table

Form of the inequality	Recommended method
$\lvert A\rvert \lessgtr \lvert B\rvert$	squaring $\to (A-B)(A+B)\lessgtr0$
$\lvert A\rvert = \lvert B\rvert$	$A=B\ \lor\ A=-B$
$\lvert A\rvert+\lvert B\rvert \lessgtr C$	sign table of $A$ and $B$
$\lvert A\rvert\cdot\lvert B\rvert \lessgtr C$	replace with $\lvert AB\rvert \lessgtr C$ (scheme $\lvert F\rvert \lessgtr k$)
$\lvert A\rvert \lessgtr B$ with $B$ arbitrary	casework on the sign of $B$ (special cases 1 and 2)

Links

Topics: Parabola
Concepts: Absolute value inequalities · Distance on the line · Line · Sign table · Absolute value
Methods: Absolute value mixed cases
Skills: Reasoning by cases · Solving inequalities