To solve an equation containing the absolute value |x|, separate cases according to the sign of the expression inside the absolute value.
Basic Patterns
The solutions of |x| = a（a > 0） are x = ±a。
The solution of |x| = 0 is x = 0。
If the right side is negative, as in |x| = −3, there is no solution.
Example ①
Solve |2x − 1| = 5.
Solution:2x − 1 = 5 or 2x − 1 = −5。
Former: 2x = 6 → x = 3
Latter: 2x = −4 → x = −2
Therefore, x = 3, −2
Solve |2x − 1| = 5.
Solution:2x − 1 = 5 or 2x − 1 = −5。
Former: 2x = 6 → x = 3
Latter: 2x = −4 → x = −2
Therefore, x = 3, −2
Solving by Separating Cases
For |A| = B, separate cases according to the sign of A。
When A ≧ 0, A = B
When A < 0, −A = B
Example ②
Solve |x − 3| = 2x.
Solution:The right side must satisfy 2x ≧ 0 for equality to hold, so x ≧ 0。
(i) When x ≧ 3: x − 3 = 2x → x = −3. This does not satisfy x ≧ 3, so it is invalid。
(ii) When x < 3: −(x − 3) = 2x → 3 − x = 2x → x = 1. This satisfies x < 3 ✓。
Therefore, x = 1
Solve |x − 3| = 2x.
Solution:The right side must satisfy 2x ≧ 0 for equality to hold, so x ≧ 0。
(i) When x ≧ 3: x − 3 = 2x → x = −3. This does not satisfy x ≧ 3, so it is invalid。
(ii) When x < 3: −(x − 3) = 2x → 3 − x = 2x → x = 1. This satisfies x < 3 ✓。
Therefore, x = 1
Key Points
- The solutions of |x| = a（a > 0） are x = ±a
- Solve |A| = B（B is a variable） by separating cases
- Always check whether the solution satisfies the condition for its case
Practice Problems
- Solve |3x + 2| = 7.
- Solve |x − 4| = 6.
- Solve |2x + 1| = x + 4.
Solutions and Explanations
- Solution:x = 5/3, −3
Explanation: 3x+2 = ±7. Thus, 3x = 5 or 3x = −9. - Solution:x = 10, −2
Explanation: x − 4 = ±6. - Solution:x = 3, −5/3
Explanation: The right side must be ≧ 0, so x ≧ −4. (i) When 2x+1 ≧ 0, that is, x ≧ −1/2: 2x+1 = x+4 → x = 3 ✓。 (ii) When x < −1/2: −(2x+1) = x+4 → −2x−1 = x+4 → −3x = 5 → x = −5/3 ✓。