ABSOLUTE VALUE
(Modulus)


Definition

The absolute value (or modulus) |x| of a real number x is x's numerical value without regard to its sign.

For example, |3| = 3; |-12| = 12; |-1.3|=1.3



Graph:
Image


Important properties:

|x|\geq0

|x| = \sqrt{x^2}

|0|=0

|-x|=|x|

|x|+|y|\geq|x+y|


3-steps approach:

General approach to solving equalities and inequalities with absolute value:

1. Open modulus and set conditions.
To solve/open a modulus, you need to consider 2 situations to find all roots:

For example, |x-1|=4
a) Positive: if (x-1)>=0, we can rewrite the equation as: x-1=4
b) Negative: if (x-1)<0, we can rewrite the equation as: -(x-1)=4
We can also think about conditions like graphics. x=1 is a key point in which the expression under modulus equals zero. All points right are the first conditions (x>1) and all points left are second conditions (x<1).

2. Solve new equations:
a) x-1=4 ==> x=5
b) -x+1=4 ==> x=-3

3. Check conditions for each solution:
a) x=5 has to satisfy initial condition x-1>=0. 5-1=4>0. It satisfies. Otherwise, we would have to reject x=5.
b) x=-3 has to satisfy initial condition x-1<0. -3-1=-4<0. It satisfies. Otherwise, we would have to reject x=-3.


3-steps approach for more than one modulus

Let's consider following example,

Problem: |x+3| - |4-x| = |8+x|. How many solutions does the equation have?

Solution: There are 3 key points here: -8, -3, 4. So we have 4 conditions:

a) x < -8. -(x+3) - (4-x) = -(8+x) ==> x = -1. We reject the solution because our condition is not satisfied (-1 is not less than -8)

b) -8 <= x < -3. -(x+3) - (4-x) = (8+x) ==> x = -15. We reject the solution because our condition is not satisfied (-15 is not within (-8,-3) interval.)

c) -3 <= x < 4. (x+3) - (4-x) = (8+x) ==> x = 9. We reject the solution because our condition is not satisfied (-15 is not within (-3,4) interval.)

d) x >= 4. (x+3) + (4-x) = (8+x) ==> x = -1. We reject the solution because our condition is not satisfied (-1 is not more than 4)

Answer: 0


Tip & Tricks

The 3-steps method works in almost all cases. At the same time, often there are shortcuts and tricks that allow you to solve absolute value problems in 10-20 sec.

I. Thinking of inequality with modulus as a segment at the number line.

For example,
Problem: 1<x<9. What inequality represents this condition?
Image
A. |x|<3
B. |x+5|<4
C. |x-1|<9
D. |-5+x|<4
E. |3+x|<5
Solution: 10sec. Traditional 3-steps method is too time-consume technique. First of all we find length (9-1)=8 and center (1+8/2=5) of the segment represented by 1<x<9. Now, let's look at our options. Only B and D has 8/2=4 on the right side and D had left site 0 at x=5. Therefore, answer is D.

II. Converting inequalities with modulus into range expression.
In many cases, especially in DS problems, it helps avoid silly mistakes.

For example,
|x|<5 is equal to x e (-5,5).
|x+3|>3 is equal to x e (-inf,-6)&(0,+inf)

III. Thinking about absolute values as distance between points at the number line.

For example,
Problem: A<X<Y<B. Is |A-X| <|X-B|?
1) |Y-A|<|B-Y|
Solution:
Image
We can think about absolute values here as distance between points. Statement 1 means than distance between Y and A is less than Y and B. Because X is between A and Y, distance between |X-A| < |Y-A| and at the same time distance between X and B will be larger than that between Y and B (|B-Y|<|B-X|). Therefore, statement 1 is sufficient.


Pitfalls

The most typical pitfall is ignoring third step in opening modulus - always check whether your solution satisfies conditions.


Official GMAC Books:

The Official Guide, 12th Edition: PS #22; PS #50; PS #130; DS #1; DS #153;

The Official Guide, Quantitative 2th Edition: PS #152; PS #156; DS #96; DS #120;

The Official Guide, 11th Edition: DT #9; PS #20; PS #130; DS #3; DS #105; DS #128;