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Absolute Value Inequalities - Math Question

Does anyone know how to solve this?
|x+2k| > |x-k|
I need to find what k is
and k is a positive constant
Yes, but I don't believe we can just provide solutions.
What have you attempted?
Original post by Maths Fan
Yes, but I don't believe we can just provide solutions.
What have you attempted?

hold on, I'll upload my working out
Reply 3
It may help you to also have a read through Chapter 0, Section 0.5 of Stitz and Zeager, which goes through inequalities of absolute terms:

https://www.stitz-zeager.com/
https://www.stitz-zeager.com/ch_0_links.pdf

Page 55 has some worked examples.
Original post by Maths Fan
Yes, but I don't believe we can just provide solutions.
What have you attempted?

Screen Shot 2021-05-16 at 8.10.31 PM.png
Original post by 0le
It may help you to also have a read through Chapter 0, Section 0.5 of Stitz and Zeager, which goes through inequalities of absolute terms:

https://www.stitz-zeager.com/
https://www.stitz-zeager.com/ch_0_links.pdf

Page 55 has some worked examples.

Thank you! I'll check them out
Original post by 0le
It may help you to also have a read through Chapter 0, Section 0.5 of Stitz and Zeager, which goes through inequalities of absolute terms:

https://www.stitz-zeager.com/
https://www.stitz-zeager.com/ch_0_links.pdf

Page 55 has some worked examples.

There's unfortunately nothing for absolute value inequalities WITH constants...
In red: The first line is correct, the second isn't. You have to multiply BOTH sides by -1 when flipping the inequality.

Note: you could divide through by k at this point as it's a positive quantity.

Screen Shot 2021-05-16 at 8.10.31 PM.png
(edited 2 years ago)
Original post by ghostwalker
In red: The first line is correct, the second isn't. You have to multiply BOTH sides by -1 when flipping the inequality.

Note: you could divide through by k at this point as it's non-zero.

Screen Shot 2021-05-16 at 8.10.31 PM.png

Thank you so much for pointing that out!!
I was finally left with k>-2x
What do I do next?
Reply 9
Original post by Kundana Amudala
Thank you so much for pointing that out!!
I was finally left with k>-2x
What do I do next?

What does the original question actually say? Can you upload an image?
Original post by Kundana Amudala
Thank you so much for pointing that out!!
I was finally left with k>-2x
What do I do next?


Note: I should have said you can cancel the k as it's a positive quantity. Just being non-zero isn't good enough with inequalities.

Well you can't go any further, other than rearrange to x> -k/2 perhaps.
Original post by mqb2766
What does the original question actually say? Can you upload an image?

sure
Screen Shot 2021-05-16 at 8.40.36 PM.png
So they want you to find a range of x (in terms of k) which you have done?

Without squaring, you could have noted one inequality is
x+2k > x-k
which is trivially satisfied, as well as the other combiations.
x+2k > -(x-k)
etc ...

If you thought about it on the number line, they want the xs which are closer to the point "k" than to the point "-2k". The midpoint -k/2 represents the boundary (equal distance from these two points). Just do a quick sketch.
(edited 2 years ago)
Reply 13
Original post by Kundana Amudala
There's unfortunately nothing for absolute value inequalities WITH constants...

It does not matter. The constant is not really the issue here but rather the logic. The solution using the methods described in that book is as follows:

If x>c| x | > c , then the solution is either:
x<cx < - c or x>cx > c

By doing it this way, you avoid the need to work with squared terms. Sometimes, as in the case here, when you work through one of those inequalities it becomes apparent that it leads to a "dead-end" so to speak. So in your question, doing it this way leads to the two inequalities:
3k>03k > 0 and k>2xk > -2x

Obviously the first one does not provide any useful information since we know that k was positive anyway.

EDIT: Beaten to it above! :tongue:
(edited 2 years ago)
Original post by 0le
It does not matter. The constant is not really the issue here but rather the logic. The solution using the methods described in that book is as follows:

If x>c| x | > c , then the solution is either:
x>cx > - c or x>cx > c

By doing it this way, you avoid the need to work with squared terms. Sometimes, as in the case here, when you work through one of those inequalities it becomes apparent that it leads to a "dead-end" so to speak. So in your question, doing it this way leads to the two inequalities:
3k>03k > 0 and k>2xk > -2x

Obviously the first one does not provide any useful information since we know that k was positive anyway.

EDIT: Beaten to it above! :tongue:

Yup and they got a free sketch as well :-)
Thanks a lot!!

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