Inequations with two absolute values (1 Viewer)

[WEMG]

For equations with two absolute values, eg |3x+6|>|5x+2|, do I just square the whole thing?

And would the answer for the example be -1< x<2?

Are there any other alternative methods?

 

[hscishard]

Sketching by pretending they are = signs is much easier.

 

[WEMG]

Sketching by pretending they are = signs is much easier.

thanks

 

[ohexploitable]

Answer is x<2, x=/-1.

 

[WEMG]

Answer is x<2, x=/-1.

how do u know if x cannot equal -1? do u have to test?

 

[ohexploitable]

how do u know if x cannot equal -1? do u have to test?

Logic?

 

[ohexploitable]

Oh wait my answers are wrong, disregard everything I've said. I'm thinking weird today

 

[WEMG]

Logic?

yeh i thought u could usually tell from the beginning through making condition

 

[ohexploitable]

As hscishard said, just graph it.

-1 < x < 2

 

[WEMG]

As hscishard said, just graph it.

-1 < x < 2

Yeh graphing prob easier but my graphs tend to get messy, I'll stick with the square meth . Thanks for your help.

 

[hscishard]

Just get a fair idea, then just find the intersecting points, then use logic to tell if its greater than or lesser than.

 

[twistedrebel]

take 4 cases

 

[Trebla]

For equations with two absolute values, eg |3x+6|>|5x+2|, do I just square the whole thing?

And would the answer for the example be -1< x<2?

Are there any other alternative methods?

Since both sides of the inequality are non-negative, you are allowed to simply square both sides whilst still preserving the unequality sign.
(3x + 6)² > (5x + 2)²
9x² + 36x + 36 > 25x² + 20x + 4
16x² - 16x - 32 < 0
x² - x - 2 < 0
(x - 2)(x + 1) < 0
=> - 1 < x < 2 which confirms your answer
Sketching is also fine, but you still you need to solve for the intersection points algebraically by considering the two cases. In my opinion, squaring both sides for this problem is the quickest method.

 

[bouncing]

since both sides of the inequality are non-negative, you are allowed to simply square both sides whilst still preserving the unequality sign.
(3x + 6)² > (5x + 2)²
9x² + 36x + 36 > 25x² + 20x + 4
16x² - 16x - 32 < 0
x² - x - 2 < 0
(x - 2)(x + 1) < 0
=> - 1 < x < 2 which confirms your answer
sketching is also fine, but you still you need to solve for the intersection points algebraically by considering the two cases. In my opinion, squaring both sides for this problem is the quickest method.

this is the best way (y)

and its almost impossible to get it wrong this way unless there are sign errors !

 

[spakfilla]

As hscishard said, just graph it.

-1 < x < 2

Where did you get the graphing program?