EQN-03-11

Absolute-Value Equations

Split an absolute-value equation into two cases and discard impossible ones.

~5 min read•Difficulty 3/5

Because |expression $∣ = k$ means the inside is $+ k$ or $- k :$

∣ x - 3∣ = 5

x - 3 = 5 \to x = 8

x - 3 = - 5 \to x = - 2

So $x = 8$ or $x = - 2$ — two solutions.

Get the absolute value alone before splitting.

2∣ x ∣ + 1 = 9 \to 2∣ x ∣ = 8 \to ∣ x ∣ = 4 \to x = 4 or x = - 4

An absolute value can never equal a negative number.

∣ x + 1∣ = - 3 \to no solution (a distance can’ t be negative)

Substitute both candidate answers back into the original equation; a careless sign in one case is the usual slip.

Worked examples

Example 1 (difficulty 3/5)

Solve

∣2 x - 1∣ = 7.

◎ Predict the answer first:

Example 2 (difficulty 2/5)

How many solutions does

∣ x + 4∣ = - 2

have?

◎ Predict the answer first:

Only solving the positive case

Why: The first case is the obvious one.

Fix: Always set the inside equal to

+ k

AND

- k .

Splitting before isolating the bars

Why: Eagerness to remove the absolute value.

Fix: Get |...| alone first; otherwise the cases are set up wrong.