Absolute Value Calculator

The absolute value of a real number $x$, written $|x|$, is its distance from $0$ on the number line:

$|x| = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases}$

Key properties:

• $|x| \geq 0$ for all $x$, with equality iff $x = 0$.
• $|xy| = |x||y|$ (multiplicative).
• $|x + y| \leq |x| + |y|$ (triangle inequality).
• $|x|^2 = x^2$, so $|x| = \sqrt{x^2}$.

Geometric interpretation: $|a - b|$ is the distance between the numbers $a$ and $b$ on the number line. This is why absolute value inequalities translate cleanly into distance statements.

Absolute value extends to complex numbers ($|a + bi| = \sqrt{a^2 + b^2}$) and to vectors (Euclidean norm), but here we focus on the real-valued case used in most homework.

Type 1: Absolute Value Equation

$|f(x)| = c$ where $c$ is a constant.

• If $c < 0$: no solution (absolute value can never be negative).
• If $c = 0$: solve $f(x) = 0$.
• If $c > 0$: split into two cases: $f(x) = c$ or $f(x) = -c$. Solve each, keep all valid solutions.

Example: $|2x - 3| = 7$ splits into $2x - 3 = 7$ or $2x - 3 = -7$, giving $x = 5$ or $x = -2$.

Type 2: Less-Than Inequality

$|f(x)| < c$ (or $\leq$) where $c > 0$.

Equivalent to: $-c < f(x) < c$ (a compound inequality, AND).

Geometric meaning: $f(x)$ is within distance $c$ of $0$.

Example: $|2x + 1| < 7$ becomes $-7 < 2x + 1 < 7$, giving $-4 < x < 3$.

If $c \leq 0$, there's no solution (or only $f(x) = 0$ if $c = 0$).

Type 3: Greater-Than Inequality

$|f(x)| > c$ (or $\geq$) where $c \geq 0$.

Equivalent to: $f(x) < -c$ or $f(x) > c$ (a disjunction, OR).

Example: $|3x - 6| \geq 9$ becomes $3x - 6 \leq -9$ or $3x - 6 \geq 9$, giving $x \leq -1$ or $x \geq 5$.

If $c < 0$, every real number satisfies the inequality.

Tricky: Absolute Value on Both Sides

$|f(x)| = |g(x)|$ splits into $f(x) = g(x)$ or $f(x) = -g(x)$.

Verifying Solutions

Always plug back into the original equation. Squaring or splitting can introduce extraneous solutions in some contexts.

• Dropping the negative case: $|x| = 5$ has two solutions, $x = 5$ and $x = -5$. Beginners often only write the positive one.
• Using AND vs OR backwards: $|x| < c$ uses AND (between $-c$ and $c$); $|x| > c$ uses OR (less than $-c$ or greater than $c$). Swapping them gives wrong answers.
• Forgetting that $c$ must be non-negative: $|f(x)| = -3$ has no solution because $|f(x)| \geq 0$ always.
• Sign confusion in the negative case: $|2x - 3| = 7$ gives $2x - 3 = -7$, not $-(2x) - 3 = 7$. Negate the whole expression equal to $-c$.
• Missing extraneous solutions: After solving, always plug back into the original equation. If the absolute value structure relied on $f(x)$ being non-negative, check that.