Can I not assume the modulus of something is the square root of the squared variable

[annaanna]

Ought this not be allowed, I realise now that I didn't have to times both sides by the modulus of that squared since you know it's always going to be positive anyways but does the definition of a modulus not work in practice

 

[lev888]

3/|2| > 1
Does it follow that 3 > 2^2?

 

[annaanna]

3/|2| > 1
Does it follow that 3 > 2^2?

I see..

 

[Steven G]

Squaring just one side of an inequality usually doesn't work.

 

[annaanna]

Squaring just one side of an inequality usually doesn't work.

great input once again steven g!

 

[khansaheb]

Ought this not be allowed, I realise now that I didn't have to times both sides by the modulus of that squared since you know it's always going to be positive anyways but does the definition of a modulus not work in practice

View attachment 37570

You are given:

(5x-1)/|(2x-3)| >= 1

(5x-1)²/(2x-3)² >= 1² >= 1 ......................... "modulus" sign removed

(5x-1)² >= (2x-3)²

25*x² - 10*x + 1 >= 4*x² - 12*x + 1

21*x² + 2*x >= 0

x * (21*x + 2) >= 0

Now investigate the equation above for the set values of x that would satisfy the given condition. Do not forget to check .................

 

[Dr.Peterson]

You are given:

(5x-1)/|(2x-3)| >= 1

(5x-1)²/(2x-3)² >= 1² >= 1 ......................... "modulus" sign removed

(5x-1)² >= (2x-3)²

25*x² - 10*x + 1 >= 4*x² - 12*x + 1

21*x² + 2*x >= 0

x * (21*x + 2) >= 0

Now investigate the equation above for the set values of x that would satisfy the given condition. Do not forget to check .................

Squaring an inequality troubles me; this will result in extraneous solutions.

There's also an arithmetic error.

My preferred approach is to multiply both sides by the (positive) denominator, and then take it in two cases, according to whether 2x-3>0 or 2x-3<0.

 

[annaanna]

Squaring an inequality troubles me; this will result in extraneous solutions.

There's also an arithmetic error.

My preferred approach is to multiply both sides by the (positive) denominator, and then take it in two cases, according to whether 2x-3>0 or 2x-3<0.

There’s an error in mine as well but wait so then can I actually treat the modulus as just the square root of a squared term? even if it is more complicated?

 

[Dr.Peterson]

so then can I actually treat the modulus as just the square root of a squared term? even if it is more complicated?

Neither of us is recommending using square roots; we're doing very different things from that. The closest we come is in saying that if you square both sides, the square of |2x-3| is (2x-3)^2, because squaring loses information about the sign.

I'm strongly recommending against squaring at all.

But, yes, the square root of the square of anything is its absolute value (modulus), so you could do that if you wanted.

 

[khansaheb]

arithmetic error.

Sorry abut that .....

(5x-1)² >= (2x-3)²

25*x² - 10*x + 1 >= 4*x² - 12*x + 9

21*x² + 2*x - 8 >= 0

Up above is a quadratic relation

Now investigate the equation above for the set of values of x that would satisfy the given "original condition". Do not forget to check......