Absolute Value: Distance from Zero
Solving Absolute Value Equations
$|2x - 3| = 7$ creates two cases:
Always check both solutions in the original equation.
$|x + 3| = -5$ has no solution. Absolute value is always ≥ 0, so it can never equal a negative number.
Absolute Value Inequalities
| Form | Solution | Interval Type |
|---|---|---|
| $|x| < a$ | $-a < x < a$ | AND (inner region) |
| $|x| > a$ | $x < -a$ or $x > a$ | OR (outer region) |
| $|x - c| < d$ | $c - d < x < c + d$ | Distance from $c$ < $d$ |
$|x| < a$: values WITHIN $a$ units of origin → single interval. $|x| > a$: values FARTHER than $a$ units → two separate intervals.
Distance Interpretation on Number Line
The minimum of $|x-a| + |x-b|$ is $|a-b|$ (distance between $a$ and $b$), achieved when $x$ is between $a$ and $b$.
10 Absolute Value Traps
1. $|x| = -c$ has NO solution
Absolute value is always non-negative. If the RHS is negative, there's no solution.
2. "Less than" → single interval; "Greater" → two separate
$|x|<3$: one interval $(-3,3)$. $|x|>3$: two intervals $x<-3$ or $x>3$.
3. $|x|^2 = x^2$ (always)
You can drop the absolute value when squaring: $|x|^2 = x^2$.
4. $\sqrt{x^2} = |x|$ not $x$
$\sqrt{x^2} = |x|$, which equals $x$ only when $x \geq 0$.
5. Two-case method: don't forget to check
Both solutions from the two cases must be verified in the original equation.
6. Expressions inside AV could be negative
$|2x-6|$ with $x=1$: expression inside = $-4$, so $|{-4}|=4$.
7. DS: $|x| = x$ only means $x \geq 0$
This doesn't mean $x$ is positive — it could be zero.
8. $|a+b| \leq |a| + |b|$ (triangle inequality)
The AV of a sum is ≤ the sum of the AVs. Equality holds when $a$ and $b$ have the same sign.
9. Midpoint interpretation
$|x-a| = |x-b|$ means $x$ is equidistant from $a$ and $b$: solve for midpoint.
10. Multiply by absolute value unknowns
Can't multiply an inequality by $|x|$ without knowing its sign — but $|x| \geq 0$ always.
10 GMAT Practice Questions
What is the value of $|{-8}| - |{-3}|$?
(C) 5. $|{-8}| = 8$ and $|{-3}| = 3$. $8 - 3 = 5$.
Solve for all values of $x$: $|x - 4| = 9$.
(C) $x = -5$ or $x = 13$. Case 1: $x-4=9$ → $x=13$. Case 2: $x-4=-9$ → $x=-5$.
Solve: $|2x + 3| \leq 7$.
(A) $-5 \leq x \leq 2$. $-7 \leq 2x+3 \leq 7$ → $-10 \leq 2x \leq 4$ → $-5 \leq x \leq 2$.
For what values of $x$ is $|x - 1| > 3$?
(B) $x < -2$ or $x > 4$. $|x-1|>3$ → $x-1 < -3$ or $x-1 > 3$ → $x < -2$ or $x > 4$.
What is the solution to $|x - 3| = |x + 5|$?
(B) $x = -1$. $x$ is equidistant from 3 and −5. Midpoint = $\frac{3+(-5)}{2} = \frac{-2}{2} = -1$. Verify: $|{-1}-3|=4$ and $|{-1}+5|=4$. ✓
Is $|x| > x$?
(1) $x^2 > x$
(2) $x < 0$
(B) Statement (2) alone sufficient. $|x| > x$ is TRUE iff $x < 0$. (2): $x<0$ directly answers YES. Sufficient. (1): $x^2>x$ → $x(x-1)>0$ → $x<0$ or $x>1$. So $x$ could be $-2$ (then YES) or $2$ (then NO: $|2|=2=x$, not greater). Not sufficient. Answer: (B).
If $|x - 5| < 3$, what is the range of $x$?
(A) $2 < x < 8$. $|x-5| < 3$ means $x$ is within 3 units of 5. So $5-3 < x < 5+3$ → $2 < x < 8$.
For how many integer values of $x$ does $|x| + |x - 3| = 5$ hold?
(C) 2 integer solutions. Consider cases. If $x \geq 3$: $x + x-3 = 5$ → $2x=8$ → $x=4$. ✓ If $0 \leq x < 3$: $x + 3-x = 3 \neq 5$. No solutions. If $x < 0$: $-x + 3-x = 5$ → $-2x=2$ → $x=-1$. ✓ Integer solutions: $x=4$ and $x=-1$. Two solutions.
What is the minimum value of $|x - 2| + |x - 8|$?
(B) 6. Minimum of $|x-a|+|x-b|$ is $|a-b|$ (when $x$ is between $a$ and $b$). $|2-8|=6$. Minimum = 6, achieved for any $2 \leq x \leq 8$.
If $\sqrt{x^2} = 7$, what are the possible values of $x$?
(B) $x = \pm 7$. $\sqrt{x^2} = |x| = 7$, so $x = 7$ or $x = -7$.
Lesson Summary — Key Takeaways
|x| = distance from 0; |x−a| = distance from a
The geometric view unlocks most AV problems without algebra.
<: inner interval; >: outer two intervals
$|x|<3$: $(-3,3)$. $|x|>3$: $x<-3$ or $x>3$.
Two-case method for equations
Set inside = positive value AND = negative value. Check both solutions.
√(x²) = |x|, not x
Square root of a square gives absolute value. Don't assume x ≥ 0.