Simplify Absolute Value Expressions

Definition of the Absolute Value Function

This is a tutorial on how to simplify expressions with absolute value. It is important to understand the definition of the absolute value first.
Expressions with absolute value can be simplified only when the sign of the expression inside the absolute value is known
If x ≥ 0 then | x | = x
if x < 0 then | x | = (-1) x
In order to simplify an expression with absolute value, we examine the sign of the quantity inside the absolute value. If that quantity is positive or equal to zero, its absolute value is the quantity itself.
| 2 | = 2 because the quantity 2 which inside the abolsute value is positive.
It that quantity is negative, we multiply it by -1.
| - 5 | = (- 1)(- 5) = 5 because the quantity - 5 which is inside the absolute value is negative.
NOTE that the absolute value is either positive or equal to zero and is NEVER negative.

Examples with Detailed Solutions

Example 1

Simplify the expressions and rewrite them without absolute value
  1. | -10 |
  2. | 0 |
  3. | -2 + 10 |
  4. | 1/2 -20 |
  5. | √3 - 5 |
  6. | √(14) - 3 pi + 10 |
  7. | (-2) / 5 |
Solution to Example1
  1. -10 is negative and according to the definition above | - 10 | can be simplified as follows
    | -10 | = (-1)(-10) = 10
  2. according to the definition above | 0 | can be simplified as follows
    | 0 | = 0
  3. - 2 + 10 = 8 is positive and according to the definition above | -2 + 10 | can be simplified as follows
    | -2 + 10 | = | 8 | = 8
  4. 1/2 - 20 = -39/2 is negative, according to the definition above | 1/2 -20 | can be simplified as follows
    | 1/2 - 20 | = | -39/2 | = -(-39/2) = 39/2
  5. √3 - 5 is approximately equal to -3.27 which is negative, according to the definition above | √3 - 5 | can be simplified as follows
    | √3 - 5 | = - ( √3 - 5) = 5 - √3
  6. √(14) - 3 pi + 10 is approximately equal to 4.32 which is positive, according to the definition above | √(14) - 3 pi + 10 | can be simplified as follows
    | √(14) - 3 pi + 10 | = √(14) - 3 pi + 10
  7. (-2)/5 = -2/5 is negative, according to the definition above | (-2) / 5 | can be simplified as follows
    | (-2) / 5 | = | -2 / 5 | =(-1)(-2/5) = 2/5

Examples with algebraic expressions are now presented.

Example 2

Simplify the algebraic expressions and rewrite tem without absolute value.
  1. | x2 + 1 |
  2. | x + 3 | , if x < -3
  3. | - x + 2 | , if x > 2

Solution to Example 2

  1. x2 + 1 is always positive and according to the definition above | x2 + 1 | can be simplified as follows
    | x2 + 1 | = x2 + 1 because the quantity x2 + 1 inside the absolute value is positive.
  2. if x < - 3 then x + 3 < 0. According to the definition above | x + 3 | can be simplified as follows
    | x + 3 | = - (x + 3) = - x - 3 because the quantity x + 3 inside the absolute value is negative.
  3. if x > 2 then x - 2 > 0 and - x + 2 < 0. According to the definition above | -x + 2 | can be simplified as follows
    | -x + 2 | = - ( - x + 2 ) = x - 2 because the quantity - x + 2 inside the absolute value is negative.

Important Rules for the Absolute Value Expressions

  1. Product rule: | A B | = | A | | B |
    Example: | (-2) x2 | = | -2 | | x2 | = 2 x2 , x2 is either positive or zero.
  2. Quotient rule: | A / B | = | A | / | B |
    Example: | -10 / (x2+1) | = | -10 | / | (x2+1) | = 10 / (x2 + 1) , x2 + 1 is positive.
  3. Square root of a square rule: √(A2) = | A |
    Example: √(x2 + 5)2 = | x2 + 5 | = x2 + 5 , x2 + 5 is positive.

More Questions on Absolute Value Expressions with Solutions

Rewrite the following expressions without absolute value.
  1. | -2 (-19 + 7) | =
  2. If x < 9, then | x - 9 | =
  3. If -3 < x < 3, then | x2 - 9 | =
  4. If | x | > 2 , then | x2 - 4 | =
  5. If x > 1, then | | - x | / ( x2 - 1) | =
  6. | (- x2 - 4) ( - x4 - 9) | =
  7. Rewrite without square root or absolute value.
  8. If x < 2, then √(x2 - 4 x + 4) =

Solutions to the Above Questions

  1. | -2 (-19 + 7) | = | - 2 (-12) | = |24| = 24
  2. x < 9 gives x - 9 < 0
    hence
    | x - 9 | = - (x - 9) = - x + 9
  3. If -3 < x < 3, then x2 - 9 < 0
    hence
    | x2 - 9 | = - (x2 - 9) = - x2 + 9
  4. If | x | > 2 , then x2 - 4 > 0
    hence
    | x2 - 4 | = - ( x2 - 4) = - x2 + 4
  5. Use the quotient rule to write
    | | - x | / ( x2 - 1) | = | | - x | | / |( x2 - 1) |
    If x > 1, then x > 0 and therefore | | - x | | = | x | = x
    If x > 1, then x2 - 1 > 0 and therefore |( x2 - 1) | = x2 - 1
    Finally
    | | - x | / ( x2 - 1) | = | | - x | | / |( x2 - 1) | = x / ( x2 - 1 )
  6. Use the product rule to write
    | (- x2 - 4) ( - x4 - 9) | = | (- x2 - 4) | | ( - x4 - 9) |
    - x2 - 4 and - x4 - 9 are both negative hence
    | (- x2 - 4) ( - x4 - 9) | = (-1) (- x2 - 4) (-1) ( - x4 - 9)
    = ( x2 + 4 ) ( x4 + 9)
  7. Write x2 - 4 x + 4 as a perfect square
    x2 - 4 x + 4 = (x - 2)2
    hence
    √(x2 - 4 x + 4) = √(x - 2)2
    use the square root of a square rule to write
    √(x2 - 4 x + 4) = √(x - 2)2 = | x - 2 |
    If x < 2, then x - 2 < 0
    hence
    √(x2 - 4 x + 4) = √(x - 2)2 = | x - 2 | = - (x - 2) = - x + 2

More links and references to absolute value functions

Definition of the absolute value.
Absolute value Functions.
Solve Equations with Absolute Value.
Absolute Value Equations And Inequalities Problems.
Tutorial on Absolute Value Inequalities.