Absolute Value Inequalities Part 1

[nycmathguy]

Set 1.2
Questions
50 & 56
David Cohen

50. Solve | x | < 2

Let me see.

Rule:

If | a | < b, where b > 0, then
-b < a < b.

-2 < x < 2

Answer: -2 < x < 2

56. | x - 1 | ≤ 1/2

Apply this rule

If | a | ≤ b, where b > 0, then
-b ≤ a ≤ b.

-(1/2) ≤ x - 1 ≤ (1/2)

-(1/2) + 1 ≤ x ≤ (1/2) + 1

(1/2) ≤ x ≤ (3/2)

You say?

 

[MathLover1]

way to go!
correct

 

[nycmathguy]

way to go!
correct

Very cool. I did this half awake, half asleep.

 

[MathLover1]

56. | x - 1 | ≤ 1/2 absolute value is same as square root: | x - 1 |=sqrt(x-1), so you have positive and negative root

=> x - 1≤ 1/2
and
=> -(x - 1)≤ 1/2

solutions will be:

x - 1≤ 1/2
x ≤ 1/2+1
x ≤ 3/2

and
-(x - 1)≤ 1/2
-x + 1≤ 1/2
-1/2+ 1≤ x
1/2≤ x

combine solutions: 1/2≤ x ≤ 3/2

 

[nycmathguy]

56. | x - 1 | ≤ 1/2 absolute value is same as square root: | x - 1 |=sqrt(x-1), so you have positive and negative root

=> x - 1≤ 1/2
and
=> -(x - 1)≤ 1/2

solutions will be:

x - 1≤ 1/2
x ≤ 1/2+1
x ≤ 3/2

and
-(x - 1)≤ 1/2
-x + 1≤ 1/2
-1/2+ 1≤ x
1/2≤ x

combine solutions: 1/2≤ x ≤ 3/2

You already said I was right in an earlier reply.

 

[nycmathguy]

56. | x - 1 | ≤ 1/2 absolute value is same as square root: | x - 1 |=sqrt(x-1), so you have positive and negative root

=> x - 1≤ 1/2
and
=> -(x - 1)≤ 1/2

solutions will be:

x - 1≤ 1/2
x ≤ 1/2+1
x ≤ 3/2

and
-(x - 1)≤ 1/2
-x + 1≤ 1/2
-1/2+ 1≤ x
1/2≤ x

combine solutions: 1/2≤ x ≤ 3/2

What about the thread

Rewrite Without Absolute Value Notation Part 1?