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

[Kulla_9289]

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.

 

[Zermelo]

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.

What do you mean by k=0 and x=-k/2?
Those would be the solutions to the equation LS = 0, you have LS>0

This inequality depends on two variables, x and k. If you want to graph all the possible solutions, you would need to graph the 3D surface z=3k(2x + k), in the coordinate system with axes (x,k,z), and see when is z>0.

If you want to solve for x, and fix k as a constant, you can solve for x in the upper inequality (but be careful, k can be negative too), and this could then be graphed if you take some concrete k, like k=1

But this inequality isn’t something you would graph. This is a product of 2 numbers, which is very nice: when can x*y be positive? What happens when x < 0 and y < 0? What about x<0 and y> 0? Etc, consider all combinations of the left factorand and right factorand, and keep the ones that make x*y>0 as the solution

 

[Kulla_9289]

I squared both sides. Not sure how you would do this since there are two unknowns.

 

[Dr.Peterson]

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.

I would presume that k is intended to be a constant (that is, a fixed parameter).

If you knew the value of k, how would you graph y = |x+2k|? How would you graph y = |x-k|?

Try graphing it for a specific case, say k=1. And what happens if k=0?

 

[Kulla_9289]

Let's get this done through algebra. I wanted to graph after squaring both sides of the inequality.

 

[BigBeachBanana]

Let's get this done through algebra. I wanted to graph after squaring both sides of the inequality.

$$|f(x)| > |g(x)| \iff |f(x)|^2 > |g(x)|^2 \iff f(x)^2>g(x)^2\\$$
$$f(x)^2-g(x)^2 >0$$
$$\tag{Difference of squares} [f(x)-g(x)] [f(x)+g(x)]>0$$
What can you do with this?

 

[Kulla_9289]

@BigBeachBanana Gets me to 3k(2x + k) > 0

 

[BigBeachBanana]

@BigBeachBanana Gets me to 3k(2x + k) > 0

Good. Think about when this inequality is true.
You have two factors i.e. 3k and (2x+k) that are being multiplied. When do you multiply two factors to give you a positive number?

 

[Kulla_9289]

Hmm. This is rather an interesting one. Say, I have such: xy>0
If I have x<0, say -2, and y<0, say -7, I get 14>0, which is true.
If I have x>0, say 2, and y>0, say 7, I get 14>0 which is true.
We can conclude that for xy>0, x<0 and y<0 or x>0 and y>0

 

[Kulla_9289]

Let 3k = x
Let 2x+k = y
3k>0, 2x+k>0 or 3k<0, 2x+k<0
k>0, k>-2x or k<0, k<-2x

 

[BigBeachBanana]

k>0, k>-2x or k<0, k<-2x

Can you get the intervals for $x$ now?

 

[Kulla_9289]

k>0, x>-k/2 or k<0, x<-k/2

 

[JeffM]

k>0, x>-k/2 or k<0, x<-k/2

Yes

 

[Kulla_9289]

Then?

 

[Dr.Peterson]

k>0, x>-k/2 or k<0, x<-k/2

Then?

Then you've solved the problem. I'd put it this way:

• If k > 0, the solution is x > -k/2;
• if k < 0, the solution is x < -k/2.
• If k=0, there is no solution.

And this agrees with what you find by graphing. (Though if you graph, you might forget to consider different cases for k.)

Keep in mind, again, that k is a parameter that is presumed to be given.

 

[JeffM]

Then what?

$$|x + 2k| > |x - k| \ge 0 \implies \sqrt{(x + 2k)^2} > \sqrt{(x - k)^2} \ge 0 \implies\\ x^2 + 4kx + 4k^2 > x^2 - 2kx + k^2 \implies 6kx + 3k^2 > 0 \implies\\ 3k(2x + k) > 0 \implies k(2x + k) > 0.\\ \therefore k \ne 0 \text { and } 2x + k \ne 0 \text { by zero product property.}\\ k < 0 \implies 2x + k < 0 \implies - \infty < x < - \dfrac{k}{2}. \\ k > 0 \implies 2x + k > 0 \implies \infty > x > - \dfrac{k}{2}.$$
What more is to be said?

 

[Kulla_9289]

The mark scheme says that x > -k/2 is the only solution for some unknown reason

 

[Dr.Peterson]

The mark scheme says that x > -k/2 is the only solution for some unknown reason

It would have been very helpful if you had said this at the start! Showing the entire problem, and the fact that makes you doubt your answer, saves a lot of time.

I would guess that they either said, or meant to imply, that k is positive.

 

[JeffM]

Hmm.

As Dr. Peterson says, either they meant to say initially that k was defined to be positive, or they meant to give as the correct answer
$|x| > \left | - \dfrac{k}{2} \right|$, which is an odd way to answer the problem correctly. In the latter case, I'd prefer to show the answer as $|x| > \left | \dfrac{k}{2} \right|$.

 

[Kulla_9289]

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.