Absolute Value
Read absolute value as distance first; the algebra and graph will follow.
Absolute Value
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Absolute value measures distance from a reference point
|x − a| is the distance between x and a. Because distance cannot be negative, an equation such as |x − 3| = 5 usually has two solutions: one 5 units left of 3 and one 5 units right.
Two boundary points when d > 0.
Inside or within the boundaries.
Outside or at least that far away.
Vertex at (h, k).
Distance creates two directions
A positive distance d from a lies at a − d and a + d.
Isolate first
Before splitting into cases, get the absolute − value expression alone.
Center ± distance
Example: Solve |x − 4| = 7.
The center is 4 and the distance is 7:
Check: both − 3 and 11 are exactly 7 units from 4.
Why it works
Thinking in terms of distance explains the two solutions and prevents the common mistake of simply removing the absolute − value bars.
Five forms you should recognize
x is 5 units from 3, so move in both directions:
x = 3 − 5 = −2x = 3 + 5 = 8A 12 cm rod may differ by at most 0.04 cm.
|L − 12| ≤ 0.0411.96 ≤ L ≤ 12.04“Within” means the measurements between the two limits are accepted.
The boundary points are 6 and 14. “At least 4 units away” means the outside regions:
x ≤ 6 or x ≥ 14The left side is a distance and can never be negative.
No real solution.The vertex is (4, 7). The negative coefficient turns the V downward, so 7 is the maximum value.
Check before you commit
- Treating |x| as “make x positive” without considering distance
- Forgetting the second case in an equation
- Splitting before isolating the absolute value
- Using “and” for an outside inequality
- Using “or” for a within inequality
- Trying to set an absolute value equal to a negative number
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Foundations
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Core Practice
Use mixed forms with less scaffolding.
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Absolute Value FAQ
Why are there usually two solutions?
Two points can be the same positive distance from a center—one on each side.
When does an absolute-value equation have one solution?
When the isolated absolute value equals zero. The only possible value inside the bars is then zero.
How do “within” and “at least” differ?
“Within d” uses the interval between the boundaries. “At least d away” uses the two outer regions.
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