Learn how to solve equations involving absolute value with step-by-step examples, detailed solutions, and explanations.
Review of Absolute Value
Here are the key rules to solve absolute value equations:
- \(|x| = 0\) if \(x = 0\)
- \(|x| = x\) if \(x > 0\)
- \(|x| = -x\) if \(x < 0\)
- The equation \(|x| = k\) with \(k < 0\) has no real solutions.
- The equation \(|x| = k\) with \(k \ge 0\) is equivalent to \(x = k\) or \(x = -k\).
Examples with Solutions
Example 1: Solve \(|x + 6| = 7\)
Solution:
- By rule 5: \[ x + 6 = 7 \quad \text{or} \quad x + 6 = -7 \]
- Solve the first equation: \[ x + 6 = 7 \implies x = 1 \]
- Solve the second equation: \[ x + 6 = -7 \implies x = -13 \]
Check solutions:
- \(x = 1\): \(|1 + 6| = |7| = 7\), matches right-hand side.
- \(x = -13\): \(|-13 + 6| = |-7| = 7\), matches right-hand side.
The solutions are \(x = 1\) and \(x = -13\).
Example 2: Solve \(-2 |x/2 + 3| - 4 = -10\)
Solution:
- Rewrite in the form \(|A| = B\): \[ -2 |x/2 + 3| = -6 \implies |x/2 + 3| = 3 \]
- Split into two equations: \[ x/2 + 3 = 3 \quad \text{or} \quad x/2 + 3 = -3 \]
- Solve first: \(x/2 + 3 = 3 \implies x = 0\)
- Solve second: \(x/2 + 3 = -3 \implies x = -12\)
Check solutions:
- \(x = 0\): LHS = \(-2|0/2 + 3| - 4 = -10\), matches RHS
- \(x = -12\): LHS = \(-2|-6 + 3| - 4 = -10\), matches RHS
The solutions are \(x = 0\) and \(x = -12\).
Example 3: Solve \(|2x - 2| = x + 1\)
Solution:
- Case 1: \(2x - 2 \ge 0 \implies x \ge 1\), then \(|2x-2| = 2x-2\): \[ 2x - 2 = x + 1 \implies x = 3 \]
- Case 2: \(2x - 2 < 0 \implies x < 1\), then \(|2x-2| = -(2x-2)\): \[ -(2x-2) = x + 1 \implies x = \frac{1}{3} \]
Check solutions:
- \(x = 3\): \(|2(3)-2| = 4 = 3+1\)
- \(x = 1/3\): \(|2(1/3)-2| = 4/3 = 1/3 + 1\)
The solutions are \(x = 3\) and \(x = 1/3\).
Example 4: Solve \(|x^2 - 4| = x + 2\)
Solution:
- Case 1: \(x^2 - 4 \ge 0 \implies x^2 \ge 4\), then \(|x^2-4| = x^2-4\): \[ x^2 - 4 = x + 2 \implies (x+2)(x-3) = 0 \implies x = -2, 3 \]
- Case 2: \(x^2 - 4 < 0 \implies x^2 < 4\), then \(|x^2-4| = -(x^2-4)\): \[ -(x^2-4) = x + 2 \implies (x+2)(x-1) = 0 \implies x = 1 \]
Check solutions:
- \(x = -2\): \(|(-2)^2 - 4| = 0 = -2 + 2\)
- \(x = 3\): \(|3^2 - 4| = 5 = 3 + 2\)
- \(x = 1\): \(|1^2 - 4| = 3 = 1 + 2\)
The solutions are \(x = -2, 1, 3\).
Solutions to Matched Exercises
Matched Exercise 1: Solve \(|4x - 2| = 10\)
Solution:
- Split into two equations: \[ 4x - 2 = 10 \quad \text{or} \quad 4x - 2 = -10 \]
- Solve first: \(4x = 12 \implies x = 3\)
- Solve second: \(4x = -8 \implies x = -2\)
Solutions: \(x = 3, -2\)
Matched Exercise 2: Solve \(-3|2x + 1| + 5 = -13\)
Solution:
- Isolate the absolute value: \[ -3|2x + 1| = -18 \implies |2x + 1| = 6 \]
- Split into two equations: \[ 2x + 1 = 6 \quad \text{or} \quad 2x + 1 = -6 \]
- Solve first: \(2x = 5 \implies x = \frac{5}{2}\)
- Solve second: \(2x = -7 \implies x = -\frac{7}{2}\)
Solutions: \(x = \frac{5}{2}, -\frac{7}{2}\)
Matched Exercise 3: Solve \(|x + 2| = 2x - 1\)
Solution:
- Case 1 (\(x + 2 \ge 0 \implies x \ge -2\)): \[ x + 2 = 2x - 1 \implies x = 3 \] (Valid since \(3 \ge -2\))
- Case 2 (\(x + 2 < 0 \implies x < -2\)): \[ -(x + 2) = 2x - 1 \implies -x - 2 = 2x - 1 \implies 3x = -1 \implies x = -\frac{1}{3} \] (Invalid since \(-\frac{1}{3}\) is not \(< -2\))
Solution: \(x = 3\)
Matched Exercise 4: Solve \(|x^2 - 1| = x + 1\)
Solution:
- Case 1 (\(x^2 - 1 \ge 0 \implies x \le -1\) or \(x \ge 1\)): \[ x^2 - 1 = x + 1 \implies x^2 - x - 2 = 0 \implies (x - 2)(x + 1) = 0 \implies x = 2, -1 \] (Both satisfy \(x^2 - 1 \ge 0\))
- Case 2 (\(x^2 - 1 < 0 \implies -1 < x < 1\)): \[ -(x^2 - 1) = x + 1 \implies -x^2 + 1 = x + 1 \implies x^2 + x = 0 \implies x(x + 1) = 0 \implies x = 0, -1 \] Within \(-1 < x < 1\), only \(x = 0\) is valid.
Solutions: \(x = -1, 0, 2\)
More Exercises with Answers
- \(|x-4|=9 \implies x = -5, 13\)
- \(|x^2 + 4| = 5 \implies x = -1, 1\)
- \(|x^2 - 9| = x + 3 \implies x = -3, 2, 4\)
- \(|x + 1| = x - 3 \implies \text{no real solutions}\)
- \(|-x| = 2 \implies x = -2, 2\)