Absolute value inequality (1 Viewer)

[YBK]

Please tell me if what i'm doing is correct or not

|x - 2| + |x + 2| > 6

(x - 2)^2 + (x + 2)^2 > 36

2x^2 + 8 > 36
x^2 > 14
x > + - root 14

Therefore
x > root 14
x < - root 14


Does that make sense...

Thanks a lot!

 

[word.]

you must remember that (a + b)² = a² + 2ab + b²
also, when a + b = c,
squaring both sides gives (a + b)² = c², not a² + b² = c²

((x - 2) + (x + 2))² = (x² - 4x + 4) + 2(x² - 4) + (x² + 4x + 4)

2x² + 8 + 2x² - 8 > 36
x² > 9
x < -3, x > 3

 

[YBK]

you must remember that (a + b)² = a² + 2ab + b²
also, when a + b = c,
squaring both sides gives (a + b)² = c², not a² + b² = c²

((x - 2) + (x + 2))² = (x² - 4x + 4) + 2(x² - 4) + (x² + 4x + 4)

2x² + 8 + 2x² - 8 > 36
x² > 9
x < -3, x > 3

wooooooooo!!
thanks man !!!!!

*super relief*

 

[Antwan23q]

err
just be easier if you wrote

|x - 2| + |x + 2| > 6
|x - 2 + x + 2| > 6
|2x| > 6
|x| > 3
x > 3 , -x > 3
x > 3 , x < -3

 

[YBK]

err
just be easier if you wrote

|x - 2| + |x + 2| > 6
|x - 2 + x + 2| > 6
|2x| > 6
|x| > 3
x > 3 , -x > 3
x > 3 , x < -3

woah...

u sure that works for everything?

 

[pLuvia]

It should, but in this case

Example
|x-2| - |x+2|>6
|x-2 - x-2|>6

Remember the -ve sign

But just write it the way you are comfortable with, I would prefer the conventional way

 

[YBK]

It should, but in this case

Example
|x-2| - |x+2|>6
|x-2 - x-2|>6

Remember the -ve sign

But just write it the way you are comfortable with, I would prefer the conventional way

what's the conventional way ?

|x-2| - |x+2|>6
|x-2 - x-2|>6
|0-4| > 6

that doesn't work ! :S :S

 

[Antwan23q]

thats stupid... that equation was like made not to work.
u cant solve it any other way

u cant just change it and expect it to work.

 

[Antwan23q]

and plus, they are both absolute powers, why cant u add them?

 

[YBK]

lol, it just seems too good to be true

why wouldn't our teacher teach us that way..... makes things much easier!!!

 

[YBK]

what's the conventional way ?

|x-2| - |x+2|>6
|x-2 - x-2|>6
|0-4| > 6

that doesn't work ! :S :S

ummm... ignore what i said....

i'm probably just freaked out cause we have maths 3unit test tomorrow!!!!!

 

[Antwan23q]

hahah, well in that case, ask them, just to be sure

 

[Trebla]

what's the conventional way ?

I dunno what the hell you did, but the most common way to solve this inequality is by using the 3 regions method. (i.e. for |x - 2| + |x + 2| > 6, solve for regions x < -2, -2 ≤ x ≤ 2 and x > 2)
Question:
|x - 2| + |x + 2| > 6

At x < -2
-(x - 2) + -(x + 2) > 6
-2x > 6
.: x < -3
Check if it satisfies domain and where the solution intersects with the domain...
.: one solution is x < -3

At -2 ≤ x ≤ 2
-(x - 2) + (x + 2) > 6
4 > 6
This is false, .: no solutions for that domain

At x > 2
(x - 2) + (x + 2) > 6
2x > 6
.: x > 3
Check if it satisfies domain and where the solution intersects with the domain...
.: one solution is x > 3

.: solutions are x > 3 and x < -3

It's a long method, but if you're careful it should work for any double absolute value equation/inequality.

Also, if you square both sides of an inequality, you may introduce new and incorrect solutions so you may have to check your solutions in the end.
e.g. For x > 4
If you SBS, then x² > 16
Now there are two solutions for that, which are x > 4 and x < -4.
But the only solutions for x are x > 4.....! So x < -4 is incorrect.
You've introduced incorrect solutions into the inequality if you square both sides in this case. In other cases you may not introduce incorrect solutions and you would be safe, but you have to check your solutions if you use SBS for an inequality.

Another common method, if you're dealing with nice whole numbers is by graph. Just make sure you draw it to an appropriate scale.
So, change: |x - 2| + |x + 2| > 6
into: |x - 2| > 6 - |x + 2|
Graph:
y = |x - 2|
and
y = 6 - |x + 2|
and see where the graph y = |x - 2| is ABOVE the graph y = 6 - |x + 2|
Your domains or x-values which satisify the inequality from the graph are your solutions.

 

[YBK]

I dunno what the hell you did, but the most common way to solve this inequality is by using the 3 regions method. (i.e. for |x - 2| + |x + 2| > 6, solve for regions x < -2, -2 ≤ x ≤ 2 and x > 2)
Question:
|x - 2| + |x + 2| > 6

At x < -2
-(x - 2) + -(x + 2) > 6
-2x > 6
.: x < -3
Check if it satisfies domain and where the solution intersects with the domain...
.: one solution is x < -3

At -2 ≤ x ≤ 2
-(x - 2) + (x + 2) > 6
4 > 6
This is false, .: no solutions for that domain

At x > 2
(x - 2) + (x + 2) > 6
2x > 6
.: x > 3
Check if it satisfies domain and where the solution intersects with the domain...
.: one solution is x > 3

.: solutions are x > 3 and x < -3

It's a long method, but if you're careful it should work for any double absolute value equation/inequality.

Also, if you square both sides of an inequality, you may introduce new and incorrect solutions so you may have to check your solutions in the end.
e.g. For x > 4
If you SBS, then x² > 16
Now there are two solutions for that, which are x > 4 and x < -4.
But the only solutions for x are x > 4.....! So x < -4 is incorrect.
You've introduced incorrect solutions into the inequality if you square both sides in this case. In other cases you may not introduce incorrect solutions and you would be safe, but you have to check your solutions if you use SBS for an inequality.

Another common method, if you're dealing with nice whole numbers is by graph. Just make sure you draw it to an appropriate scale.
So, change: |x - 2| + |x + 2| > 6
into: |x - 2| > 6 - |x + 2|
Graph:
y = |x - 2|
and
y = 6 - |x + 2|
and see where the graph y = |x - 2| is ABOVE the graph y = 6 - |x + 2|
Your domains or x-values which satisify the inequality from the graph are your solutions.

Thanks for that reply!

Ummm.. as for what I did, it was basically a method I made up... lol

yeah, i know about the graphing method... but it seems too long....

But with the three region method, how do u know which regions it should be restricted to?

Does the easy method that was mentioned work?

How do you check if your results are right when u do square both sides method... just put the values back in the equation?

 

[Trebla]

But with the three region method, how do u know which regions it should be restricted to?

The critical points of the absolute value (i.e. when an absolute value = 0) determine the points you're working with. The three regions are the numbers less than the lowest critical point, the numbers between the highest and lowest critical points and the number greater than the highest ciritical point.

How do you check if your results are right when u do square both sides method... just put the values back in the equation?

Yep and check if they equal or satisfy the equation/inequality.

Does the easy method that was mentioned work?

I'm not too sure about that. You might want to be careful. If you use the 3 regions method, you won't have to taken into account any exceptions, but you have to be careful with your algebra....because it is easy to make unnecessary errors in that method.

 

[word.]

it's perfectly legal to square both sides with absolute values

x > 4
why the hell would you square both sides here?

the reason you can square both sides with absolute values is because absolute values are defined as the square root of the square and it allows the inequality to be unchanged. this does not create any extra solutions...

i.e. |x| = Sqrt{x²}

hence consider |x + 4| > 7
Sqrt{(x + 4)²} > 7
squaring both sides gives (x + 4)² > 49 etc.

although with one absolute value grouping it's easier just to consider when it is negative or positive

|x + 4| > 7
so x + 4 > 7 or -x - 4 > 7
x > 3, x < -11

oh and |x + 2| + |x - 2| and |x + 2 + x - 2| are two very different things.

 

[YBK]

The critical points of the absolute value (i.e. when an absolute value = 0) determine the points you're working with. The three regions are the numbers less than the lowest critical point, the numbers between the highest and lowest critical points and the number greater than the highest ciritical point.

Yep and check if they equal or satisfy the equation/inequality.

I'm not too sure about that. You might want to be careful. If you use the 3 regions method, you won't have to taken into account any exceptions, but you have to be careful with your algebra....because it is easy to make unnecessary errors in that method.

sweet... i think i'll ask our teacher about the square both sides method... because I can see myself doing a lot of stupid mistakes with the three region one...

for our test we had an equation that looked something like this: |x-5| > |x-10|
I just graphed it on a number line, seemed easier than the algebra. :uhhuh:

 

[YBK]

oh and |x + 2| + |x - 2| and |x + 2 + x - 2| are two very different things.

yeah, I agree... it just doesn't look right at all.