Equations Involving Absolute Value
Lawrence Tsang
01/31/2004
1
Definition
Definition 1 (Absolute Value Function).
(
x
if x ≥ 0,
|x| =
−x if x < 0.
2
(1)
Properties
Theorem 2. For M ≥ 0, |x| ≤ M if and only if
−M ≤ x ≤ M.
(2)
Proof. If x ≥ 0, then 0 ≤ x = |x| ≤ M . Also, if x < 0, then 0 < −x = |x| ≤ M ,
which implies 0 > x ≥ −M . Both cases imply −M ≤ x ≤ M .
Conversely, if x ≥ 0, then |x| = x. Using the right side inequality x ≤ M , we
have |x| ≤ M . If x < 0, then |x| = −x. Using the left side inequality −M ≤ x,
which implies −x ≤ M , we have |x| ≤ M . Both cases imply |x| ≤ M .
Theorem 3 (Positive Definite). For all a ∈ R,
|a| ≥ 0,
|a| = 0 if and only if a = 0.
Proof. We have three possibilities:
1. If a = 0, then |a| = a = 0.
2. If a > 0, then |a| = a > 0.
3. If a < 0, then |a| = −a and −a > 0 imply |a| > 0.
Theorem 4. |ab| = |a| |b|.
Proof. There are four possibilities:
1. If a ≥ 0,b ≥ 0, then |ab| = ab = |a||b|.
2. If a ≥ 0,b < 0, then |ab| = a(−b) = |a||b|.
1
(3)
(4)
3
EQN W/ ABS
3. If a < 0,b ≥ 0, then |ab| = (−a)b = |a||b|.
4. If a < 0,b < 0, then |ab| = (−a)(−b) = |a||b|.
Theorem 5 (Triangle inequality). For every a, b ∈ R,
|a + b| ≤ |a| + |b| .
(5)
Proof. Note that |a| ≤ |a|. Thus, by Theorem 2,
−|a| ≤ a ≤ |a|.
(6)
−|b| ≤ b ≤ |b|.
(7)
−|a| − |b| ≤ a + b ≤ |a| + |b|.
(8)
Similarly,
Adding (7) and (7) gives
Again by Theorem 2, we have
|a + b| ≤ |a| + |b|.
3
(9)
Equations Involving Absolute Value
Example 1. Solve
|x − 5| = 3.
(10)
Solution. Look for the root of the term inside the absolute function.
x−5=0
x = 5.
Then break up the equation into two intervals x ≤ 5 and x ≥ 5.
(
−(x − 5) = 3 for x ≤ 5,
|x − 5| = 3
⇒
(x − 5) = 3
for x ≥ 5.
(11)
(12)
(13)
For x ≤ 5, we have −(x − 5) = 3 and
5−x=3
x = 2.
(14)
(15)
Since x = 2 ≤ 5, we accept this solution.
For x ≥ 5, we have x − 5 = 3 and
x−5=3
x = 8.
Since x = 8 ≥ 5, we accept this solution too.
The solutions are 2 and 8.
2
(16)
(17)
3
EQN W/ ABS
Example 2. Solve
|x2 − 3x + 2| − 2x + 3 = 0.
(18)
Solution. Find the zeros of each of the term in the absolute function. For
x2 − 3x + 2 = (x − 2)(x − 1), the roots are 2 and 1. So, we have 3 intervals:
x≤1
1≤x≤2
x≥2
(19)
(20)
(21)
For x ≤ 1, we have x2 − 3x + 2 ≥ 0. Thus, (18) becomes
x2 − 3x + 2 − 2x + 3 = 0.
(22)
x2 − 5x + 5 = 0.
(23)
Simplifying this, we get
The roots are
p
√
25 − 4(5)
5+ 5
x1 =
=
> 1,
2
p2
√
5 − 25 − 4(5)
5− 5
x2 =
=
> 1.
2
2
5+
(24)
(25)
Since both roots are out of range x ≤ 1, they are rejected as solutions to (18).
For 1 ≤ x ≤ 2, we have x2 − 3x + 2 ≤ 0. Thus, (18) becomes
−(x2 − 3x + 2) − 2x + 3 = 0.
(26)
−x2 + x + 1 = 0.
(27)
Solving this gives
The roots are
√
√
1− 2
−1 + 1 + 1
x3 =
=
< 0,
−2
2
√
√
1+ 2
−1 − 1 + 1
x4 =
=
.
−2
2
(28)
(29)
Clearly x3 is outside of the range 1 ≤ x ≤ 2, but x4 is within range. So we have
found one solution x4 in this range.
Lastly, for x ≥ 2, we have x2 − 3x + 2 ≥ 0. The equation is the same as in
the first case. However, only x1 is within the range x ≥ 2. that only one of the
two solution is out of range.
Thus, the solutions to (18) are
√
5+ 5
,
(30)
x1 =
2√
1− 2
.
(31)
x3 =
2
3
4
EXERCISE
Example 3. Solve
|x + 1| − |x| + 2|x − 1| = 2x − 1.
(32)
Solution. For each absolute value expression, find the zero of the expression
inside.
For |x + 1|, the zero of x + 1 is -1.
For |x|, the zero of x is 0.
For |x − 1|, the zero of x − 1 is 1.
So, we break up the real line to 4 intervals:
x ≤ −1
−1 ≤ x ≤ 0
0≤x≤1
1 ≤ x.
(33)
(34)
(35)
(36)
For each interval rewrite (32) without the absolute values as
−x − 1 + x + 2(1 − x) = 2x − 1
x + 1 + x + 2(1 − x) = 2x − 1
x + 1 − x + 2(1 − x) = 2x − 1
x + 1 − x + 2(x − 1) = 2x − 1
for
for
for
for
x ≤ −1
−1 ≤ x ≤ 0
0≤x≤1
1 ≤ x.
(37)
(38)
(39)
(40)
Now, solve each equation and accepting only solution within the corresponding
domain of definition.
1
(reject)
2
x = 2 (reject)
x=1
−1 = −1 (all x ≥ 1)
x=
for x ≤ −1
(41)
for −1 ≤ x ≤ 0
for 0 ≤ x ≤ 1
for 1 ≤ x.
(42)
(43)
(44)
Since only solutions in (43) and (44) are valid, the solution to (32) is
x ≥ 1.
4
Exercise
Exercise 1. Solve |x − 7| = 2.
Exercise 2. Solve 3|5 − 2x| + 4 = 9x − 1.
Exercise 3. Solve |1 − 2x + x2 + 5x3 | = −3.
Exercise 4. Solve |x + 2| = 3|x − 1|.
Exercise 5. Solve |2x + 3| = |2x − 4|.
Exercise 6. Solve |2x + 4| = |3x + 6|.
4
(45)
4
Exercise 7. Solve |2x + 4| + 1 = |3x + 6|.
Exercise 8. Solve |2x + 4| = |3x + 6| + 1.
Exercise 9. Solve 3|x + 3| + |2x − 1| = |x − 4|.
Exercise 10. Solve 2|x2 − 3x + 2| − |2x − 3| = 0.
Exercise 11. Solve |x − 3||x − 2| = 5 − x.
5
EXERCISE