Another question about absolute value inequalities:

|x - 3| |2x + 1|
How can you go from the above step, to:

(|x - 3|)^2 (|2x + 1|)^2 ?

Question

Another question about absolute value inequalities:

|x - 3| > |2x + 1| 
How can you go from the above step, to:

(|x - 3|)^2 > (|2x + 1|)^2 ?

hero · Answer

You don't. You realize that |x-3| and |2x+1| are both absolute value and absolute value, also known as distance, is positive, so the result is always positive. So you re-write it like this: 

(x-3)^2 > (2x+1)^2

hero · Answer

myininaya, are you following me?

myininaya · Answer

would after multiplying and bringing everything to one side we get
(x+4)(3x-2)<0

yes hero lol

anonymous · Answer

Wait, you mean you "do", not "dont" right? Because you are saying that you don't go from the first step to the next. So you can square both sides because the argument within the absolute value will always be positive?

myininaya · Answer

what?
are you talking about my sentence?
somehow my sentence got cut off
this would be what you get after multiplying and bringing everything to one side:*

hero · Answer

$$(\pm (x-3))^2 > (\pm (2x+1))^2$$

hero · Answer

If that makes any sense

myininaya · Answer

it makes sense to me :)

hero · Answer

Well that's good

hero · Answer

When are you going to tutor me?

anonymous · Answer

Hero I don't understand how you got that plus minus statement.