How do I solve this inequality? I have |x+2k|>|x-k|; squared both sides to get 6kx+3k^2 > 0

Kulla_9289 said:
Yeah, it was stated. Sorry for the confusion. How would one solve |x+3k|<4|x-k|? I squared both sides and ended up with x>7k/3, x>k/5 and x<7k/3, x<k/5. As usual, k is a positive constant.
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Please show the details of your work. You have the right pieces in your answer (x>7k/3, x>k/5, x<7k/3, x<k/5), but have not put them together in the right way. We need to see the work in order to show you what to change.
 
I used ab<0, a<0,b>0 and a>0,b<0. Initially, I squared both sides and then brought them over to LHS and used difference of two squares and got to (-3x+7k)(5x-k)<0.
 
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Kulla_9289 said:
I used ab<0, a<0,b>0 and a>0,b<0. Initially, I squared both sides and then brought them over to LHS and used difference of two squares and got to (-3x+7k)(5x-k)<0.
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I asked for more detail than that; the issue is how the four inequalities you listed fit together, and you didn't go that far. But I'll work with what you said.

The main problem in what you said is the "and": "I used ab<0, a<0,b>0 and a>0,b<0." You have to be more precise than that.

From (-3x + 7k)(5x - k) < 0, the next step is to carefully break down the cases:

Either
  • -3x + 7k < 0 and 5x - k > 0, OR
  • -3x + 7k > 0 and 5x - k < 0
so either
  • x > 7k/3 and x > k/5, OR
  • x < 7k/3 and x < k/5
Now, what must be true if x > 7k/3 and x > k/5? You can turn this into a single inequality. Do the same with the second case.

(Actually, I would have first changed (-3x + 7k)(5x - k) < 0 to (3x - 7k)(5x - k) > 0, because positive leading terms are easier to think about.)
 
Kulla_9289 said:
I have |x+2k|>|x-k|. I squared both sides and got 6kx + 3k² > 0. On factorising, 3k(2x + k) > 0. So, k = 0 and x = -k/2. But I am not sure how one would sketch this graph to solve the inequality because there is only one value of x.
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|x+2k| = x+2k if x+2k>0 and -(x+2k) if x+2k<0
|x-k| = x-k if x-k>0 and -(x-k) if x-k<0

OR

|x+2k| = x+2k if x>-2k and -(x+2k) if x<-2k
|x-k| = x-k if x>k and -(x-k) if x<k

Assume k>0:
Solve for |x+2k|>|x-k| when x is in each of these intervals: (-oo, -2k), (-2k,0), (0,k), (k, oo)
 
Kulla_9289 said:
k<0, when you substitute the k/5 into the x in 7k/3
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No; k is a parameter, with a fixed value (though we don't know it). And it's known to be positive, isn't it (post #20)? (Did you ever tell us the exact wording of the actual problem, to confirm this?)

If x is greater than 7/3 and also greater than 1/5, how do we say that with a single inequality? (Hint: if its greater than 7/3, isn't it necessarily greater than 1/5?)

To put it another way, graph x > 7k/3 and x > k/5, then find the intersection (because both have to be true). What is the result? I've been assuming you've seen this idea before, since you're studying inequalities.
 
Yes. But why are we finding the intersection for this problem but not for the first problem (the one this thread is on)?
 
Dr.Peterson said:
No; k is a parameter, with a fixed value (though we don't know it). And it's known to be positive, isn't it (post #20)? (Did you ever tell us the exact wording of the actual problem, to confirm this?)
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I have to assume then k=1
x>7/3 and x>1/5: the intersection is x>7/3 so x>7k/3
x < 7/3 and x < 1/5: the intersection is x<1/5 so x<k/5
 
Kulla_9289 said:
Yes. But why are we finding the intersection for this problem but not for the first problem (the one this thread is on)?
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Because it's a different problem!

For this one, you got (-3x + 7k)(5x - k) < 0. For the first one, you got 3k(2x + k) > 0. These have different behaviors. Do you see the key difference?

And looking at the problems, the difference in |x+3k|<4|x-k| is that the coefficients are different, whereas they were the same in the first problem. If you graph both sides this time, you'll see why there are two intervals in the solution.

Kulla_9289 said:
I have to assume then k=1
x>7/3 and x>1/5: the intersection is x>7/3 so x>7k/3
x < 7/3 and x < 1/5: the intersection is x<1/5 so x<k/5
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No, I just suggested temporarily replacing k with 1 because you seemed confused having an unknown constant. Your conclusions here are correct.
 
Kulla_9289 said:
How would you do it without assuming 1?
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Like I said:
Dr.Peterson said:
To put it another way, graph x > 7k/3 and x > k/5, then find the intersection (because both have to be true). What is the result?
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Since x > 7k/3 > k/5 implies x > k/5, x > 7k/3 and x > k/5 is equivalent to x > 7k/3. So the answer for that case is x > 7k/3.

Or, again, draw the graph:

Code: 
                   k/5      7k/3
              <-----+---------+------->
x>7k/3:                       o=======>
x>k/5:              o=================>
intersection:                 o=======>
which is x>7k/3

The other case gives you x < k/5, and you take the union of the two results: The final answer is x > 7k/3 or x < k/5.

Where do you need to assume a value for k? As long as it is a positive constant, everything works the same.
 
It is very dangerous to assume that an unknown number is one (or zero or minus 1). Those numbers have special properties, which means that using them as examples may be misleading.

In any case, the problem involves an unspecified, and therefore unknown, parameter. You are missing the power of generalization when you think in terms of a specific example.

This SECOND problem (which should be in its own thread) involves nothing more than simple algebra.

[math] \text {Given: } k > 0 \text { and } |x + 3k| < 4|x - k| \implies\\ x^2 + 6kx + 9k^2 < 16x^2 - 32kx + 16k^2 \implies 0 < 15x^2 - 38kx + 7k^2 = (3x - 7k)(5x - k) \implies\\ \text {CASE I: } 3x - 7k < 0 \text { and } 5x - k < 0 \implies x < \dfrac{7k}{3} \text { AND } x < \dfrac{k}{5} \implies\\ x < \dfrac{k}{5} \ \because \dfrac{1}{5} < 1 < \dfrac{7}{3} \text { and } k > 0 \implies \dfrac{k}{5} < \dfrac{7k}{3}.\\ \text {CASE II: } 3x - 7k > 0 \text { and } 5x - k > 0 \implies x > \dfrac{7k}{3} \text { AND } x > \dfrac{k}{5} \implies\\ x > \dfrac{7k}{3} \ \because \dfrac{7k}{3} > \dfrac{k}{5} \text { as previously shown.}\\ \text {Combining cases, we get } x < \dfrac{k}{5} \text { OR } x >\dfrac{7k}{3}. [/math]
You are making this way too hard for yourself.
 
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