Math: Absolute value (Modulus)

ABSOLUTE VALUE
(Modulus)

This post is a part of [GMAT MATH BOOK]

created by: walker
edited by: bb, Bunuel



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:



Important properties:

\(|x|\geq0\)

\(|0|=0\)

\(|-x|=|x|\)

\(|x|+|y|\geq|x+y|\)


How to approach equations with moduli

It's not easy to manipulate with moduli in equations. There are two basic approaches that will help you out. Both of them are based on two ways of representing modulus as an algebraic expression.

1) \(|x| = \sqrt{x^2}\). This approach might be helpful if an equation has × and /.

2) |x| equals x if x>=0 or -x if x<0. It looks a bit complicated but it's very powerful in dealing with moduli and the most popular approach too (see below).

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, \(|x-1|=4\)
a) Positive: if \((x-1)\geq0\), 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 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 problems

Let’s consider following examples,

Example #1
Q.: \(|x+3| - |4-x| = |8+x|\). 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 (9 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 #2
Q.: \(|x^2-4| = 1\). 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 & 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?

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.


Pitfalls

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




Practice Questions

Easy:

Medium:


Hard:



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