Absolute value |x|: just the « distance to zero »

The definition

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

In practice: $|x|$ strips the sign. $|5| = 5$, $|-5| = 5$, $|0| = 0$.

The interpretation that unlocks everything

$|x|$ = distance from $x$ to $0$ on the number line.

Generalization: $|x - a|$ = distance from $x$ to $a$.

Solve $|x - 3| = 5$

« The distance between $x$ and $3$ is 5. » Two points on the line: $x = 3 + 5 = 8$ or $x = 3 - 5 = -2$.

Solve $|x| \leq 4$

« The distance from $x$ to $0$ is at most 4 » → $x \in [-4; 4]$.

$|x| \geq 4$ → $x \in ]-\infty; -4] \cup [4; +\infty[$.

The 3 properties to remember

• $|x| \geq 0$ always.
• $|xy| = |x| \cdot |y|$.
• Triangle inequality: $|x + y| \leq |x| + |y|$.

The classic mistake

$|x^2 - 9| \ne x^2 - 9$ for all $x$. Only when $x^2 - 9 \geq 0$ (i.e. $|x| \geq 3$).

?Your turn

How many solutions does |x + 2| = 7 have?

A.Just oneB.Two: x = 5 and x = −9C.Two: x = 5 and x = 9D.None

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