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How to Solve: Absolute Value (Basics)
Hi All,
I have posted a video on YouTube to discuss Absolute Value (Basics)
After this post please go through
Absolute Value Problems and
Absolute Value + Inequality post
Attached pdf of this Article as SPOILER at the top! Happy learning! Absolute Values Theory and Solved Problems Playlist Link hereFollowing is Covered in the Video
Theory
What is Absolute Value / Modulus of a number
Absolute Value on Number Line
Properties of Absolute Values
Absolute Value on Number Line Examples
What is Absolute Value / Modulus of a number• Absolute Value or modulus (|x|) of a real number x is the non-negative value of the number (x), without any consideration to its signEx
|12| = 12
|-12| = 12 (we just the value after ignoring the sign)
• |x| = x for x >0
= -x for x <0
= 0 for x = 0Q1. Find the value of |-3| + | 2*3 – 4*2| + |25|Q2. Find the value of | x+y| where x + z = 20 and y – z = -25Sol1: 3 + | 6-8 | + 25 = 3 + 2 + 25 = 30
Sol2: x + z = 20 and y – z = -25
Adding both of them we get x + y = -5
=> | x+y | = |-5| = 5
Absolute Value on Number Line• Absolute value of a number x can also be imagined as the distance of that number x from 0 on a number lineLet's say we have two numbers x and y and x is positive and y is negative. What we are saying is
|x| = x = distance of x from origin
|y| = -y = distance of y from origin
As, shown in the image below:
Properties of Absolute Values• PROP 1: Absolute value of a number is always Non-negative|a| ≥ 0 for all values of a
Ex: |3| = 3 ≥ 0
|-7| = 7 ≥ 0
• PROP 2: Minimum value of |a| = 0, when a=0Ex: If |x| =0 => x=0
• PROP 3: Square root of a number is always positive\(\sqrt{a^2}\) = |?|
Ex:
If x = \(\sqrt{25}\) => x = +5
But if \(x^2\) = 25 => x = ± \(\sqrt{25}\) => x = ±5
• PROP 4: Absolute value of negative of a number is same as absolute value of the number|-a| = |a|
A derivative of this is
| a-b | = | b-a | because | b-a | = | -(a-b) |
• PROP 5: Product of absolute value of two numbers is same as product of their absolute values|ab| = |a|*|b|
Ex:
|7*3| = |7| * |3| = 21
• PROP 6: Division of absolute value of two numbers is same as division of their absolute values\(|\frac{?}{?}| = \frac{{|?|}}{{|?|} }\)
Ex:
\(|\frac{4}{2}| = \frac{{|4|}}{{|2|} }\) = 2
• PROP 7: Sum of absolute value of two numbers is always ≥ absolute value of their sum|a| + |b| ≥ |a+b|
Ex:
|7| + |3| ≥ |7+3| => 10 ≥ 10
|5| + |-8| ≥ |5 + (-8) | => 13 ≥ 3
• PROP 8: Difference of absolute value of two numbers is always ≤ absolute value of their difference|a| - |b| ≤ |a-b|
Ex:
|7| - |3| ≤ |7-3| => 4 ≤ 4
|5| - |-8| ≤ |5 - (-8) | => -3 ≤ 13
• PROP 9: Taking absolute value multiple times or taking it once gives the same result||a|| = |a|
Ex:
||-4|| = |-4| => |4| = |-4| = 4
• PROP 10: If absolute value of difference of two numbers is zero => both numbers are equal|a-b|=0 => a=b
Ex:
| x-4 | =0 > x=4
Next two will be used a lot to solve absolute values problem!• PROP 11: If |a| ≤ b => -b ≤ a ≤ b• PROP 12: If |a| ≥ b => a ≤ -b or a ≥ b• PROP 13: |\(a^n\)| = \(|a|^n\)Ex:
|\({-2}^4\)| = \(|-2|^4\) = 16
• PROP 14: |a-b| ≥ ||a|-|b||Ex:
|7-4| ≥ ||7|-|4|| => 3 ≥ 3
|8-(-2)| ≥ ||8|-|-2|| => 10 ≥ 6
Q1. If |a-3| ≤ 9 then find the range of values of a. Q2. If |b+5| ≥ 10 then find the range of values of b. Sol1: -6 ≤ a ≤ 12
Check
Video For solution
Sol2: b ≤ -15 or b ≥ 5
Check
Video For solution
Absolute Value on Number Line ExampleIf a and b are two variables given then:
|a-b| always means the distance between points a and b
|a+b| = |a| + |b| when a and b have the same sign and
|a+b| = |b| - |a| when a and b have different sign and |b| > |a|Case 1: a and b are positive and a > b|a-b | = |a| - |b|
|a+b| = |a| + |b|
Case 2: a is positive and b is negativeGiven: |a| > |b|
|a-b| = |a| + |b|
|a+b| = |a| - |b|
Case 3: Both a and b are negativeGiven: |a| > |b|
|a-b| = |a| - |b|
|a+b| = |a| + |b|
Q1. Given the information (below), Simplify |b-a| + |c-b| Sol:
Method 1
|b-a| = Distance between a and b = AB
|c-b| = Distance between c and b = BC
=> |b-a| + |c-b| = AB = BC = AC = |c-a|
Method 2
|b-a| = |b| - |a|
|c-b| = |c| - |b|
=> |b-a| + |c-b| = |b| - |a| + |c| - |b| = |c| - |a| = |c-a|