When solving an inequality involving an absolute value, what is the standard approach?

When solving an inequality involving an absolute value, what is the standard approach?

• A. Isolate inside the absolute value and solve the two cases
• B. Square both sides
• C. Ignore the absolute value and solve as a normal inequality
• D. Replace with its square root

Answer: (A)

The standard way to solve inequalities with absolute value is to isolate what’s inside the absolute value, then consider the two possibilities that the absolute value represents. Since |u| is the distance of u from zero, the inequality is really about where u lies relative to zero. If you have |u| ≤ a with a ≥ 0, you must have -a ≤ u ≤ a. If you have |u| ≥ a, you must have u ≤ -a or u ≥ a. This split comes from the fact that the absolute value is nonnegative and captures both positive and negative versions of the inner expression. For example, solve |2x - 3| ≤ 5. Let u = 2x - 3. Then -5 ≤ u ≤ 5, so -5 ≤ 2x - 3 ≤ 5. Adding 3 gives -2 ≤ 2x ≤ 8, and dividing by 2 yields -1 ≤ x ≤ 4. If you instead try squaring both sides or ignoring the absolute value, you can easily miss solutions or include invalid ones, because those methods don’t faithfully reflect the two-direction nature of the inside expression that the absolute value encodes.

The standard way to solve inequalities with absolute value is to isolate what’s inside the absolute value, then consider the two possibilities that the absolute value represents. Since |u| is the distance of u from zero, the inequality is really about where u lies relative to zero.

If you have |u| ≤ a with a ≥ 0, you must have -a ≤ u ≤ a. If you have |u| ≥ a, you must have u ≤ -a or u ≥ a. This split comes from the fact that the absolute value is nonnegative and captures both positive and negative versions of the inner expression.

For example, solve |2x - 3| ≤ 5. Let u = 2x - 3. Then -5 ≤ u ≤ 5, so -5 ≤ 2x - 3 ≤ 5. Adding 3 gives -2 ≤ 2x ≤ 8, and dividing by 2 yields -1 ≤ x ≤ 4.

If you instead try squaring both sides or ignoring the absolute value, you can easily miss solutions or include invalid ones, because those methods don’t faithfully reflect the two-direction nature of the inside expression that the absolute value encodes.