Worksheet

Worksheet: absolute value of a real number

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Summary

The absolute value $|a|$ of a real number $a$ is its distance to the origin: $|a| = a$ if $a \ge 0$ and $|a| = -a$ if $a < 0$. The computation rules follow from the definition: $|ab| = |a|\cdot|b|$ and $|a|^2 = a^2$. The worksheet with $15$ exercises trains direct evaluation, the properties and simple absolute-value equations.

Press “Check” next to an exercise. Your answers stay on this device.

I. Evaluating absolute values. Removing the absolute value bars

  1. Calculate:

    a) $| - 7| + |3 - 8| - | - 2| \cdot | - 4|$;

    b) $\left| 2 - \sqrt{5} \right| + \left| \sqrt{5} - 3 \right|$;

    c) $\left| 1 - \sqrt{2} \right| + \left| \sqrt{2} - \sqrt{3} \right| + \left| \sqrt{3} - 2 \right|$.

  2. Show that the number $a = \left| 2\sqrt{3} - 4 \right| + \left| 2\sqrt{3} - 3 \right|$ is a natural number.

  3. Show that $|x - 2| + |x + 3| = 5$, for any $x \in \lbrack - 3,\ 2\rbrack$.

  4. Rewrite without the absolute value the expression $E(x) = |x - 3| + |2x + 4|$, where $x$ is a real number.

II. Properties of the absolute value

  1. Show that the number $N = \sqrt{\left( 1 - \sqrt{2} \right)^{2}} + \sqrt{\left( 3 - \sqrt{2} \right)^{2}}$ is a natural number.

  2. Find the real numbers $x$ for which:

    a) $|x - 5| = x - 5$;

    b) $|2x + 6| = - 2x - 6$.

  3. Find the real numbers $x$ and $y$ for which $|x - 2y| + |x + y - 6| = 0$.

  4. Show that $\sqrt{x^{2} - 4x + 4} + \sqrt{x^{2} - 10x + 25} = 3$, for any $x \in \lbrack 2,\ 5\rbrack$.

  5. Find the sets:

    a) $A = \left\{ x \in {\mathbb{Z}} \mid |2x - 1| \leq 5 \right\}$;

    b) $B = \left\{ x \in {\mathbb{R}} \mid |x + 2| > 3 \right\}$.

  6. Let $x$ be a real number such that $|x - 1| \leq 2$. Show that $|3x + 1| \leq 10$.

III. Equations with absolute values

  1. Solve in $\mathbb{R}$ the equations:

    a) $|x - 3| = 5$;

    b) $|2x + 1| = 7$;

    c) $|3x - 2| = - 1$;

    d) $\sqrt{x^{2} - 6x + 9} = 2$.

  2. Solve in $\mathbb{R}$ the equations:

    a) $|x - 1| = |2x + 3|$;

    b) $|x - 2| = 2x - 7$.

  3. Solve in $\mathbb{R}$ the equations:

    a) $|x - 1| + |x - 3| = 4$;

    b) $\left| |x - 2| - 3 \right| = 1$.

  4. Solve in $\mathbb{R}$ the equations:

    a) $x^{2} - 5|x| + 6 = 0$;

    b) $\left| x^{2} - 4 \right| = 3x$.

  5. Consider the expression $E(x) = |x - 1| + |x - 5|$, where $x$ is a real number.

    a) Show that $E(x) \geq 4$, for any real number $x$.

    b) Solve in $\mathbb{R}$ the equation $E(x) = 4$.

Frequently asked questions

What is $|-5|$ and why?

$5$, because the absolute value measures the distance to zero, and a distance is always positive. Since $-5 < 0$, the definition gives $|-5| = -(-5) = 5$. In other words, the absolute value erases the minus sign of a negative number.

Is $|a + b| = |a| + |b|$ true?

No, not in general: equality holds only when $a$ and $b$ share the same sign or one of them is zero. The general rule is the triangle inequality, $|a + b| \le |a| + |b|$. For example, $|{-3} + 5| = 2$, while $|-3| + |5| = 8$.

How do you solve the equation $|x - 1| = 3$?

Look for the numbers at distance $3$ from $1$ on the real line, so $x - 1 = 3$ or $x - 1 = -3$, giving $x = 4$ or $x = -2$. Checking is quick: $|4 - 1| = 3$ and $|-2 - 1| = 3$. The method works for any equation of the form $|x - a| = r$ with $r \ge 0$.