ABSOLUTE VALUE(Modulus)

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DefinitionThe 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|=1.3Graph: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:

- Positive (or rather non-negative)
- Negative

For example,

|x-1|=4a) Positive: if

(x-1)\geq0, we can rewrite the equation as:

x-1=4b) Negative: if

(x-1)<0, we can rewrite the equation as:

-(x-1)=4We 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 condition

(x>1) and all points left are second condition

(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 complex problemsLet’s consider following examples,

Example #1Q.: |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 \leq 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 \leq 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 \geq 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)

(Optional) The following illustration may help you understand how to open modulus at different conditions.

Answer: 0

Example #2Q.: |x^2-4| = 1. What is x?

Solution: There are 2 conditions:

a)

(x^2-4)\geq0 -->

x \leq -2 or

x\geq2.

x^2-4=1 -->

x^2 = 5. x e {

-\sqrt{5},

\sqrt{5}} and both solutions satisfy the condition.

b)

(x^2-4)<0 -->

-2 < x < 2.

-(x^2-4) = 1 -->

x^2 = 3. x e {

-\sqrt{3},

\sqrt{3}} and both solutions satisfy the condition.

(Optional) The following illustration may help you understand how to open modulus at different conditions.

Answer:

-\sqrt{5},

-\sqrt{3},

\sqrt{3},

\sqrt{5}Tip & TricksThe 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?

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

PitfallsThe most typical pitfall is ignoring the 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;

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Fig's post with absolute value problems: [

Absolute Value Problems]

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