BPM MATH0 – Week 2
Absolute Value and Inequalities
Week 2 – Absolute Value and Inequalities
Weekly Goals
ˆ Understand the definition and graph of the absolute value function.
ˆ Solve equations and inequalities involving absolute value.
ˆ Practice solving linear and quadratic inequalities.
Solved Examples – With Detailed Steps
Example 1: Solve the equation:
2|x − 3| − 4 = 0
Steps:
ˆ Isolate the absolute value: |x − 3| = 2
ˆ Solve two cases:
x − 3 = 2 ⇒ x = 5, x − 3 = −2 ⇒ x = 1
x = 1 or x = 5
Example 2: Solve the inequality:
|2x + 1| < 5
Steps:
ˆ Rewrite as compound inequality: −5 < 2x + 1 < 5
ˆ Subtract 1: −6 < 2x < 4
ˆ Divide by 2: −3 < x < 2
x ∈ (−3, 2)
Example 3: Solve the quadratic inequality:
x2
− 5x + 6 ≤ 0
Steps:
ˆ Factor: (x − 2)(x − 3) ≤ 0
ˆ Determine sign changes on intervals:
ˆ both terms are negative so the product is positive for x < 2
ˆ x − 2 > 0 and x − 3 < 0 for x ∈ [2, 3]
ˆ both terms and their product are positive for x > 3
BPM MATH0 – Week 2
Absolute Value and Inequalities
x ∈ [2, 3]
Example 4: Graph the function:
f(x) = |x + 2| − |x − 1|
Steps:
ˆ Identify critical points: x = −2, x = 1
ˆ Split into intervals and analyze:
ˆ x < −2: f(x) = −(x + 2) − (1 − x) = −3
ˆ −2 ≤ x < 1: f(x) = x + 2 − (1 − x) = 2x + 1
ˆ x ≥ 1: f(x) = x + 2 − (x − 1) = 3
We obtain piecewise function: f(x) =



−3 x < −2
2x + 1 −2 ≤ x < 1
3 x ≥ 1
Example 5: Solve the equation:
|2x − 3| = |x + 1|
Steps:
ˆ Find critical points where expressions inside absolute values are zero:
2x − 3 = 0 ⇒ x =
3
2
, x + 1 = 0 ⇒ x = −1
BPM MATH0 – Week 2
Absolute Value and Inequalities
ˆ Divide the real line into intervals:
1. Case 1: x < −1
|2x − 3| = −(2x − 3), |x + 1| = −(x + 1)
−2x + 3 = −x − 1 ⇒ −x = −4 ⇒ x = 4 /∈ (−∞, −1) discard
2. Case 2: −1 ≤ x < 3
2
|2x − 3| = −(2x − 3), |x + 1| = x + 1
−2x + 3 = x + 1 ⇒ −3x = −2 ⇒ x = 2
3
∈ [−1, 3
2
) accept
3. Case 3: x ≥ 3
2
|2x − 3| = 2x − 3, |x + 1| = x + 1
2x − 3 = x + 1 ⇒ x = 4 ∈ [3
2
, ∞) accept
x =
2
3
, x = 4
Practice Problems for Seminar
Absolute Value
1. Solve: |3x − 6| = 9
2. Solve: 2|x − 1| + 3 = 9
3. Solve: |x − 2| = |3x + 1|
4. Graph: f(x) = |x| − |x − 3|
5. Graph: f(x) = 2x + |x − 2|
Inequalities
6. Solve: x2
− 4x > 5
7. Solve: (x + 3)(x + 1) ≤ −1
8. Solve: |x + 2| ≥ 4
9. Solve: |x − 4| < x