A progressive collection of 20 practice problems on absolute value, designed to help develop a correct understanding of the definition of absolute value, its geometric meaning, and its main operational properties.
In each exercise, absolute value is not treated as a simple mechanical rule for "dropping the minus sign," but as a quantity that measures a distance. For this reason, every step is carefully justified, distinguishing the cases where the expression inside the absolute value is positive, zero, or negative.
The exercises begin with the direct evaluation of numerical absolute values and gradually progress to algebraic expressions, distances on the real number line, simplifications involving absolute values, and introductory equations with absolute value.
Exercise 1 — level ★☆☆☆☆
Evaluate:
\[ |7| \]
Answer
\[ |7|=7 \]
Solution
The absolute value of a real number measures its distance from \(0\) on the real number line. Since distance cannot be negative, the absolute value is always a number greater than or equal to zero.
In this case the number inside the absolute value is \(7\). Since:
\[ 7>0, \]
we apply the first case of the definition:
\[ |x|=x \qquad \text{if } x\geq 0. \]
Therefore:
\[ |7|=7. \]
Geometrically, this means that the number \(7\) lies at a distance of \(7\) from \(0\) on the real number line.
Exercise 2 — level ★☆☆☆☆
Evaluate:
\[ |-9| \]
Answer
\[ |-9|=9 \]
Solution
The number inside the absolute value is \(-9\), which is a negative number.
In this case we must apply the second branch of the definition of absolute value:
\[ |x|=-x \qquad \text{if } x<0. \]
Here \(x=-9\). Therefore:
\[ |-9|=-(-9). \]
Since the opposite of \(-9\) is \(9\), we obtain:
\[ |-9|=9. \]
This does not mean that the absolute value "always changes the sign." It means, rather, that it returns the distance of the number from \(0\). The number \(-9\) lies \(9\) units away from \(0\), so its absolute value is \(9\).
Exercise 3 — level ★☆☆☆☆
Evaluate:
\[ |0| \]
Answer
\[ |0|=0 \]
Solution
The absolute value of \(0\) is \(0\), because \(0\) has zero distance from itself.
We can also verify this directly from the definition. Since:
\[ 0\geq 0, \]
the first case applies:
\[ |x|=x \qquad \text{if } x\geq 0. \]
Substituting \(x=0\), we obtain:
\[ |0|=0. \]
This example is important because it shows that absolute value does not always return a positive number, but a non-negative number. Indeed, \(0\) is not positive: it is zero.
Exercise 4 — level ★☆☆☆☆
Evaluate:
\[ |3-8| \]
Answer
\[ |3-8|=5 \]
Solution
Before applying the absolute value, we must evaluate the expression inside.
We have:
\[ 3-8=-5. \]
So the expression becomes:
\[ |3-8|=|-5|. \]
The number \(-5\) is negative. By definition, if \(x<0\), then:
\[ |x|=-x. \]
Applying this rule to \(x=-5\), we obtain:
\[ |-5|=-(-5)=5. \]
Therefore:
\[ |3-8|=5. \]
A common mistake to avoid is immediately writing \(|3-8|=3-8\). This would be incorrect, because one must first determine whether the inner expression is positive, zero, or negative.
Exercise 5 — level ★★☆☆☆
Evaluate:
\[ |-4|+|6|-|{-2}| \]
Answer
\[ |-4|+|6|-|{-2}|=8 \]
Solution
The expression contains multiple absolute values. It is best to evaluate them one at a time, by examining the sign of the number inside each one.
Consider the first absolute value:
\[ |-4|. \]
Since \(-4\) is negative, its absolute value is its opposite:
\[ |-4|=4. \]
Now consider:
\[ |6|. \]
Since \(6\) is positive, the absolute value equals the number itself:
\[ |6|=6. \]
Finally:
\[ |{-2}|. \]
Since \(-2\) is negative, we have:
\[ |{-2}|=2. \]
Substituting these values into the original expression:
\[ |-4|+|6|-|{-2}|=4+6-2. \]
Carrying out the calculation:
\[ 4+6-2=10-2=8. \]
Therefore:
\[ |-4|+|6|-|{-2}|=8. \]
This exercise shows that absolute values must be evaluated before performing external operations. Only after correctly resolving the absolute values can we carry out additions and subtractions.
Exercise 6 — level ★★☆☆☆
Evaluate:
\[ |2-7|+|-3| \]
Answer
\[ |2-7|+|-3|=8 \]
Solution
To correctly evaluate the expression, we must first simplify what lies inside each absolute value.
Consider the first absolute value:
\[ |2-7|. \]
Evaluating the subtraction:
\[ 2-7=-5. \]
Therefore:
\[ |2-7|=|-5|. \]
Since \(-5\) is negative, its absolute value is:
\[ |-5|=5. \]
Now consider the second absolute value:
\[ |-3|. \]
Since \(-3\) is also negative:
\[ |-3|=3. \]
Substituting these results into the original expression:
\[ |2-7|+|-3|=5+3. \]
Performing the addition:
\[ 5+3=8. \]
Therefore:
\[ |2-7|+|-3|=8. \]
From a geometric standpoint, the absolute values here represent distances on the real number line. Since distances are always non-negative quantities, the absolute values are replaced by positive or zero numbers.
Exercise 7 — level ★★☆☆☆
Simplify:
\[ |x| \qquad \text{given that } x<0 \]
Answer
\[ |x|=-x \]
Solution
The expression contains a variable, so we cannot evaluate the absolute value directly as we did in the numerical exercises. We must instead apply the definition.
We are told that:
\[ x<0. \]
This means that \(x\) is a negative number.
By definition:
\[ |x|= \begin{cases} x & \text{if } x\geq 0,\\ -x & \text{if } x<0. \end{cases} \]
Since we are in the case \(x<0\), we apply the second branch of the definition:
\[ |x|=-x. \]
It is important to understand the meaning of this expression. If \(x\) is negative, then \(-x\) is positive. For example, if:
\[ x=-4, \]
then:
\[ |x|=-(-4)=4. \]
Therefore:
\[ |x|=-x \qquad \text{when } x<0. \]
Exercise 8 — level ★★☆☆☆
Simplify:
\[ |x-3| \qquad \text{given that } x>3 \]
Answer
\[ |x-3|=x-3 \]
Solution
To remove the absolute value, we must determine the sign of the expression inside.
The expression inside the absolute value is:
\[ x-3. \]
We are told that:
\[ x>3. \]
Subtracting \(3\) from both sides of the inequality gives:
\[ x-3>0. \]
Therefore the expression inside the absolute value is positive.
When a quantity is positive or zero, the absolute value equals the quantity itself:
\[ |a|=a \qquad \text{if } a\geq 0. \]
Applying this property with \(a=x-3\), we obtain:
\[ |x-3|=x-3. \]
Geometrically, this means that for values of \(x\) greater than \(3\), the distance between \(x\) and \(3\) is simply \(x-3\).
Exercise 9 — level ★★☆☆☆
Simplify:
\[ |x-3| \qquad \text{given that } x<3 \]
Answer
\[ |x-3|=3-x \]
Solution
In this exercise too, we must examine the sign of the expression inside the absolute value.
The expression is:
\[ x-3. \]
We are told that:
\[ x<3. \]
Subtracting \(3\) from both sides, we obtain:
\[ x-3<0. \]
Therefore the expression inside the absolute value is negative.
When a quantity is negative, the absolute value equals its opposite:
\[ |a|=-a \qquad \text{if } a<0. \]
Applying this rule with \(a=x-3\), we obtain:
\[ |x-3|=-(x-3). \]
Expanding by distributing the negative sign:
\[ -(x-3)=-x+3. \]
This can also be written as:
\[ -x+3=3-x. \]
Therefore:
\[ |x-3|=3-x. \]
This result is consistent with the geometric interpretation of absolute value. If \(x\) lies to the left of \(3\) on the real number line, the distance between \(x\) and \(3\) is given by \(3-x\).
Exercise 10 — level ★★★☆☆
Evaluate:
\[ |{-3}\cdot 4| \]
Answer
\[ |{-3}\cdot 4|=12 \]
Solution
We can solve this exercise in two different ways.
The first method consists of evaluating the product inside the absolute value first.
We have:
\[ -3\cdot 4=-12. \]
Therefore:
\[ |{-3}\cdot 4|=|-12|. \]
Since \(-12\) is negative:
\[ |-12|=12. \]
Thus:
\[ |{-3}\cdot 4|=12. \]
Alternatively, we can use the product property of absolute value:
\[ |ab|=|a|\cdot |b|. \]
Applying it, we obtain:
\[ |{-3}\cdot 4|=|{-3}|\cdot |4|. \]
Now:
\[ |{-3}|=3 \qquad \text{and} \qquad |4|=4. \]
Therefore:
\[ |{-3}\cdot 4|=3\cdot 4=12. \]
Both methods lead to the same result:
\[ |{-3}\cdot 4|=12. \]
Exercise 11 — level ★★★☆☆
Evaluate:
\[ \left|\frac{-12}{3}\right| \]
Answer
\[ \left|\frac{-12}{3}\right|=4 \]
Solution
Here too, we can proceed in two different ways.
The first method consists of evaluating the division inside the absolute value first.
We have:
\[ \frac{-12}{3}=-4. \]
The expression therefore becomes:
\[ \left|\frac{-12}{3}\right|=|-4|. \]
Since \(-4\) is negative, its absolute value is its opposite:
\[ |-4|=4. \]
Therefore:
\[ \left|\frac{-12}{3}\right|=4. \]
Alternatively, we can use the quotient property of absolute value:
\[ \left|\frac{a}{b}\right|=\frac{|a|}{|b|} \qquad \text{with } b\neq 0. \]
Applying this property, we obtain:
\[ \left|\frac{-12}{3}\right| = \frac{|-12|}{|3|}. \]
Now:
\[ |-12|=12 \qquad \text{and} \qquad |3|=3. \]
Therefore:
\[ \frac{|-12|}{|3|} = \frac{12}{3}=4. \]
This method also gives:
\[ \left|\frac{-12}{3}\right|=4. \]
Exercise 12 — level ★★★☆☆
Simplify:
\[ \sqrt{(-5)^2} \]
Answer
\[ \sqrt{(-5)^2}=5 \]
Solution
This exercise is very important because it clarifies one of the fundamental properties of absolute value:
\[ \sqrt{x^2}=|x|. \]
Many students incorrectly write:
\[ \sqrt{x^2}=x, \]
but this equality does not hold for all real numbers. The principal square root always returns a non-negative number.
In our case:
\[ \sqrt{(-5)^2}=|-5|. \]
Now we evaluate the absolute value:
\[ |-5|=5. \]
Therefore:
\[ \sqrt{(-5)^2}=5. \]
We can also verify the result directly:
\[ (-5)^2=25. \]
Therefore:
\[ \sqrt{25}=5. \]
The result is \(5\), not \(-5\), because the principal square root is always non-negative.
Exercise 13 — level ★★★☆☆
Find the distance between the numbers:
\[ -2 \qquad \text{and} \qquad 5 \]
Answer
\[ 7 \]
Solution
The distance between two real numbers \(a\) and \(b\) is given by the absolute value of their difference:
\[ |a-b|. \]
In this exercise the two numbers are:
\[ a=-2 \qquad \text{and} \qquad b=5. \]
We can therefore write:
\[ |-2-5|. \]
Evaluating the subtraction:
\[ -2-5=-7. \]
We obtain:
\[ |-7|. \]
Since \(-7\) is negative:
\[ |-7|=7. \]
Therefore the distance between \(-2\) and \(5\) is:
\[ 7. \]
Geometrically, this means that on the real number line it takes \(7\) units to go from \(-2\) to \(5\).
Exercise 14 — level ★★★☆☆
Simplify:
\[ |x|^2 \]
Answer
\[ |x|^2=x^2 \]
Solution
A fundamental property of absolute value states that:
\[ |x|^2=x^2. \]
This equality holds for every real number \(x\).
To understand why, we distinguish two cases.
First case: \(x\geq 0\).
In this case:
\[ |x|=x. \]
Squaring:
\[ |x|^2=x^2. \]
Second case: \(x<0\).
In this case:
\[ |x|=-x. \]
Squaring:
\[ |x|^2=(-x)^2. \]
But the square of a number and the square of its opposite are equal:
\[ (-x)^2=x^2. \]
Therefore in this case too:
\[ |x|^2=x^2. \]
We can therefore conclude that:
\[ |x|^2=x^2 \]
for every real number \(x\).
Exercise 15 — level ★★★★☆
Solve:
\[ |x|=4 \]
Answer
\[ S=\{-4,4\} \]
Solution
The equation:
\[ |x|=4 \]
asks us to find all numbers that are at a distance of \(4\) from \(0\) on the real number line.
There are exactly two numbers with this property:
\[ 4 \qquad \text{and} \qquad -4. \]
Indeed:
\[ |4|=4 \]
and:
\[ |-4|=4. \]
We can therefore write:
\[ x=4 \qquad \text{or} \qquad x=-4. \]
The solution set is therefore:
\[ S=\{-4,4\}. \]
In general, when:
\[ |x|=a \qquad \text{with } a>0, \]
the solutions are:
\[ x=\pm a. \]
Exercise 16 — level ★★★★☆
Solve:
\[ |x|=0 \]
Answer
\[ S=\{0\} \]
Solution
The absolute value of a number represents its distance from \(0\) on the real number line.
The equation:
\[ |x|=0 \]
therefore asks us to find all numbers at zero distance from \(0\).
There is only one number with this property:
\[ x=0. \]
Indeed:
\[ |0|=0. \]
No other number satisfies the equation, because the absolute value of any nonzero number is always strictly positive.
We can therefore conclude that:
\[ S=\{0\}. \]
Exercise 17 — level ★★★★☆
Solve:
\[ |x|=-3 \]
Answer
\[ S=\varnothing \]
Solution
The absolute value of a real number is always greater than or equal to zero. Indeed:
\[ |x|\geq 0 \]
for every real number \(x\).
In the equation:
\[ |x|=-3 \]
the right-hand side is negative.
This is impossible, since a distance cannot be negative.
Therefore, there is no real number whose absolute value equals \(-3\).
Hence the equation has no solution:
\[ S=\varnothing. \]
Exercise 18 — level ★★★★☆
Solve:
\[ |x-2|=5 \]
Answer
\[ S=\{-3,7\} \]
Solution
The equation:
\[ |x-2|=5 \]
expresses a distance. In particular, \(|x-2|\) represents the distance between \(x\) and \(2\) on the real number line.
The equation therefore asks us to find all numbers that are \(5\) units away from \(2\).
On the real number line there are two possibilities:
- the number lies \(5\) units to the right of \(2\);
- the number lies \(5\) units to the left of \(2\).
Algebraically, this means solving the two equations:
\[ x-2=5 \]
or:
\[ x-2=-5. \]
Solving the first:
\[ x-2=5. \]
Adding \(2\) to both sides:
\[ x=7. \]
Now solving the second:
\[ x-2=-5. \]
Adding \(2\) to both sides:
\[ x=-3. \]
The solutions of the equation are therefore:
\[ x=-3 \qquad \text{or} \qquad x=7. \]
Therefore:
\[ S=\{-3,7\}. \]
Exercise 19 — level ★★★★★
Solve:
\[ |2x-1|=7 \]
Answer
\[ S=\left\{-3,4\right\} \]
Solution
When an equation has the form:
\[ |A|=k \qquad \text{with } k>0, \]
we must consider two cases:
\[ A=k \]
or:
\[ A=-k. \]
In our case:
\[ A=2x-1 \qquad \text{and} \qquad k=7. \]
We must therefore solve the two equations:
\[ 2x-1=7 \]
or:
\[ 2x-1=-7. \]
Solving the first:
\[ 2x-1=7. \]
Adding \(1\) to both sides:
\[ 2x=8. \]
Dividing by \(2\):
\[ x=4. \]
Now turning to the second equation:
\[ 2x-1=-7. \]
Adding \(1\):
\[ 2x=-6. \]
Dividing by \(2\):
\[ x=-3. \]
The solutions of the equation are therefore:
\[ x=-3 \qquad \text{or} \qquad x=4. \]
Therefore:
\[ S=\left\{-3,4\right\}. \]
Exercise 20 — level ★★★★★
Solve:
\[ |x+4|=|x-2| \]
Answer
\[ S=\{-1\} \]
Solution
The equation:
\[ |x+4|=|x-2| \]
equates two distances.
The expression:
\[ |x+4|=|x-(-4)| \]
represents the distance between \(x\) and \(-4\).
The expression:
\[ |x-2| \]
represents the distance between \(x\) and \(2\).
The equation therefore asks us to find the point on the real number line that is equidistant from \(-4\) and \(2\).
Intuitively, this point is the midpoint between \(-4\) and \(2\).
We now find the solution algebraically.
Squaring both sides:
\[ |x+4|^2=|x-2|^2. \]
Since:
\[ |a|^2=a^2, \]
we obtain:
\[ (x+4)^2=(x-2)^2. \]
Expanding both squares:
\[ x^2+8x+16=x^2-4x+4. \]
Cancelling \(x^2\) from both sides:
\[ 8x+16=-4x+4. \]
Collecting the \(x\)-terms on the left and the constant terms on the right:
\[ 8x+4x=4-16. \]
We obtain:
\[ 12x=-12. \]
Dividing by \(12\):
\[ x=-1. \]
Verification:
\[ |-1+4|=|3|=3 \]
and:
\[ |-1-2|=|-3|=3. \]
Both sides are equal, confirming the solution.
Therefore:
\[ S=\{-1\}. \]