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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! Following 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|
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