Absolute Value Equations And Inequalities
Questions on solving equations and inequalities with absolute value are presented along with detailed solutions. You may want to work through the tutorials on solving equations with absolute value before solving the questions below.
Review of Equations and Inequalities with Absolute Value
In order to solve equations or inequalities with absolute value, we need to rewrite them without absolute value, solve them and if necessary check the solutions obtained.
The most important about the absolute value of an algebraic expression, is that we can simplify them if we know the sign of the expression inside the absolute value symbol according to the following definition.
\[ | x | = x \; \text{if} \; x \ge 0 \; , \; \text{and} \; | x | = - x \; \text{if} x \lt 0 \]
Equivalences: Some specific equations and inequalities with absolute value may be written without the absolute value symbol as follows:
\[ 1) \quad |x| = k , \quad \text{where k is a positive real number} \Leftrightarrow \quad x = k \; \text{or} \; x = - k \]
\[ 2) \quad |x| \lt k , \quad \text{where k is a positive real number} \Leftrightarrow \quad - k \lt x \lt k \]
\[ 3) \quad |x| \gt k ,\quad \text{where k is a positive real number} \Leftrightarrow \quad x \gt k \; \text{or} \; x \lt - k \]
Absolute Value Equations
Question 1
Find all real solutions to the equation with absolute value.
\[ 2|-2x - 2| - 3 = 13 \]
Solution to Question 1
Rewrite the equation with the absolute value expression on one side
, add 3 to both sides
\[ 2| - 2 x - 2| = 16 \]
divide both sides by 2
\[ | - 2 x - 2| = 8 \]
Use the equivalence 1) above to write the equation without absolute value: a) \( -2 x - 2 = 8 \) and b) \( -2 x - 2 = -8 \)
\[ a) \quad -2 x - 2 = 8 \]
Solve to obtain
\[ x = - 5 \]
Second equation
\[ b) \quad - 2 x - 2 = - 8 \]
Solve to obtain
\[ x = 3 \]
As an exercise, check by substitution that \( x = - 5 \) and \( x = 3 \) are solutions to the given equation.
Conclusion Solution set of the given equation is: \[ \{-5,3\} \]
Question 2
Find all real solutions to the equation with absolute value.
\[ |x - 1| = 2x + 1 \]
Solution to Question 2
The signs of \( x - 1 \) and \( 2x + 1 \) change with \( x \), and one way to solve this equation is to consider two cases.
a) Assume \( x - 1 \geq 0 \) , hence \( |x-1| = x - 1 \) and rewrite the equation as
\[
x - 1 = 2x + 1
\]
Solve for \( x \):
\[
x = -2
\]
b) Assume \( x - 1 \leq 0 \) , , hence \( |x-1| = -( x - 1) \) and rewrite the equation as
\[
-(x - 1) = 2x + 1
\]
\[
- x + 1 = 2x + 1
\]
Solve for \( x \):
\[
x = 0
\]
Important: Because we *assumed* the sign of the expression inside the absolute value symbol, we need to check the two solutions found above.
Check \( x = -2 \):
Substitute \( x = -2 \) in both sides of the original equation.
Left side:
\[
|(-2) - 1| = |-3| = 3
\]
Right side:
\[
2(-2) + 1 = -4 + 1 = -3
\]
The two sides are not equal, so \( x = -2 \) is NOT a solution.
Check \( x = 0 \):
Substitute \( x = 0 \) in both sides of the original equation.
Left side:
\[
|0 - 1| = |-1| = 1
\]
Right side:
\[
2(0) + 1 = 0 + 1 = 1
\]
The two sides are equal, so \( x = 0 \) is a solution.
Solution set of the given equation is:
\[
\{ 0 \}
\]
Question 3
Find all real solutions to the equation
\[ |2x - 1| = x^2 \]
Solution to Question 3
\( x^2 \) is either 0 or positive, but the sign of \( 2x - 1 \) changes with \( x \). Let us consider two cases:
a) Assume \( 2x - 1 \geq 0 \) and rewrite the equation as
\[
2x - 1 = x^2
\]
which may be written as
\[
x^2 - 2x + 1 = 0
\]
Solve for \( x \):
\[
x = 1
\]
b) \( 2x - 1 \leq 0 \) and rewrite the equation as
\[
- (2x - 1) = x^2
\]
which may be written as
\[
x^2 + 2x - 1 = 0
\]
Solve for \( x \):
\[
x = -1 \pm \sqrt{2}
\]
Check all solutions found above (detailed calculations will be left out for you as an exercise).
All solutions found : \( x = 1 \), \( x = -1 - \sqrt{2} \), and \( x = -1 + \sqrt{2} \) — are solutions to the original equation given above.
Question 4
Find all real solutions to the equation with absolute value.
\[ -2|x + 1| -2 = 4 \]
Solution to Question 4
Rewrite the given equation with the term containing the absolute value on one side.
\[ -2|x + 1| = 6 \]
Divide both sides by \( -2 \):
\[ |x + 1| = - 3 \]
The absolute value of an algebraic expression cannot be negative and therefore the given equation has no solutions.
Question 5
Find all real solutions to the equation with absolute value.
\[ |x - 1| = |x^2 -2x + 1| \]
Solution to Question 5
Note that \( x^2 - 2x + 1 \) can be written as
\[
x^2 - 2x + 1 = (x - 1)^2
\]
an expression that is positive or equal to zero. Our equation may be written as:
\[
|x - 1| = |(x-1)^2| = |x - 1|^2
\]
Rewrite the above as
\[
|x - 1|^2 - |x - 1| = 0
\]
Factor:
\[
|x - 1| (|x - 1| - 1) = 0
\]
Which gives:
\[
|x - 1| = 0 \quad \text{and solution } x = 1
\]
or
\[
|x - 1| - 1 = 0 \quad \Rightarrow \quad |x - 1| = 1
\]
Solve the above to obtain two solutions:
\[
x = 0 \quad \text{and} \quad x = 2
\]
We did not make any assumptions as in the above questions, and we therefore do not need to check our solutions (except perhaps for mistakes made).
\[
\text{Solution set: } \{0, 1, 2\}
\]
Absolute Value Inequalities
Question 6
Solve the inequality.
\[
|x + 2| \lt 3
\]
Solution to Question 6
The above inequality is solved by writing a double inequality equivalent (see equivalency 2) above) to the given inequality but without absolute value
\[
|x + 2| \lt 3 \Leftrightarrow - 3 \lt x + 2 \lt 3
\]
Solve the double inequality to obtain
\[
- 5 \lt x \lt 1
\]
The above solution set is written in interval form as follows:
\[
(-5 , 1)
\]
Question 7
Solve the inequality with absolute value..
\[ - 3|-2x + 4| \le 4 \]
Solution to Question 7
Divide both sides of the given inequality by \( - 3 \) and change the symbol of the inequality to obtain
\[
|-2x + 4| \ge - 4 / 3
\]
The absolute value of an expression is always positive or equal to zero. Hence any value of \( x \) in the above inequality will satisfy it and therefore the solution set of the given inequality is the set of all the real numbers given by
\[
(-\infty , \infty) \]
Question 8
Solve the inequality.
\[
| - 2 x - 4 | \gt 9
\]
Solution to Example 8
Solving the above inequality is equivalent to solving the inequalities (see equivalence 3 above)
\[
-2 x - 4 \gt 9 \quad \text{or} \quad -2 x - 4 \lt - 9
\]
Which gives
\[
x \lt -13 / 2 \quad \text{or} \quad x \gt 5 / 2
\]
The above solution set is written in interval form as follows
\[ (-\infty , -13 / 2) \cup (5 / 2 , + \infty) \]
Question 9
Solve the inequality with absolute values.
\[ |x + 1| \le |x^2 + 2x + 1| \]
Solution to Example 9
Note that the algebraic expression \( |x^2 + 2x + 1| \) may be written as
\[
|x^2 + 2x + 1| = |(x + 1)^2| = |x + 1|^2
\]
Hence the inequality may be written as
\[
|x + 1| - |x + 1|^2 \leq 0
\]
\( |x + 1| \) is positive or equal to zero, hence the left side of the above inequality may be factored as
\[
|x + 1|(1 - |x + 1|) \leq 0
\]
We now use a table to study the sign of the two factors on the left side of the above inequality to determine the solution set.
\( |x + 1| \) is positive for all values of \( x \) except at \( x = -1 \), where it is equal to zero.
The zeros of \( 1 - |x + 1| \) are \( -2 \) and \( 0 \).
The table of signs is shown below.
The solution set of the given inequality is given by
\[ (-\infty , - 2] \cup {\{-1\}} \cup [0 , + \infty) \]
Check the above solution set to the inequality graphically using the graphs of the two sides ( \( |x + 1| \) in blue and \( |x^2 + 2x + 1| \) in red) of the given inequality shown below.
Question 10
Solve the inequality.
\[ x + 2 \lt |x^2 - 4| \]
Solution to Question 10
One way to take out the absolute value is to study the sign of the expression inside the absolute value symbol. There are two cases
Case 1
For \( x^2 - 4 \ge 0 \), or \( x\) in the interval \( (-\infty , -2] \cup [2 , +\infty)\), we can write
\[ |x^2 - 4| = x^2 - 4 \]
Substitute the expression with the absolute value in the given inequality and solve
\[ x + 2 \lt x^2 - 4 \]
which may written as
\[ x^2 - x - 6 \gt 0 \]
The solution set to above inequality is given by the interval
\[ (-\infty , -2) \cup (3 , +\infty) \]
We need to select only the solutions within the interval \( (-\infty , -2] \cup [2 , +\infty)\). The intersection of intervals \((-\infty , -2] \cup [2 , +\infty)\) and \((-\infty , -2) \cup (3 , +\infty)\) gives the solution set
\[ (-\infty , -2) \cup (3 , +\infty) \]
Case 2
For \(x^2 - 4 \lt 0\), or \(x\) in the interval \((-2 , 2)\), we can write
\[ |x^2 - 4| = -(x^2 - 4) \]
Substitute the expression with the absolute value in the given inequality and solve
\[ x + 2 \lt -(x^2 - 4) \]
which may be rewritten as
\[ x^2 + x - 2 \lt 0 \]
Solution set to the above inequality is given by the interval: \( (-2 , 1) \)
We need to select only the solutions within the interval \((-2 , 2)\). The solution set to above inequality is given by the intersection of the intervals \((-2 , 1)\) and \((-2 , 2)\) which gives
\[ (-2 , 1) \]
Conclusion: The solution set to the given inequality is :
\[ (-\infty, -2) \cup (-2 , 1) \cup (3 , +\infty) \]
Check the above answer to the inequality graphically using the graphs of the two sides of the given inequality shown below.
References and links to absolute value equations and Inequalities.