Pre-Cal; rational inequalities

1. Sep 18, 2010

priscilla98

1. The problem statement, all variables and given/known data

1. x-2|x| < 3

2. Relevant equations

3. The attempt at a solution

Okay, I attempted this equation below but I don't know if I'm right.

x-2|x|- 3 < 0

x^2 - 2x - 3 < 0

(x - 3) (x + 1) < 0

x = 3 x = -1

2. Sep 18, 2010

eumyang

Why do you have an "x" in the first line and then a "x2" in the 2nd? What is the original problem? Is it this:
$$x - 2|x| - 3 < 0$$
or this?
$$x^2 - 2|x| - 3 < 0$$

BTW: this is not a rational inequality.

69

3. Sep 18, 2010

epenguin

There is just no justification for the step from the first inequality to the second.

Your last line, an equality is unlikely to be a correct answer to this question about an inequality.

But presuming what you mean is at that at these points it changes from < to > or vice versa, even if finding an answer is difficult, checking whether it is the right answer - whether your starting inequalities change at those points you gave - is not and you should not need to ask us.

This is really a very simple question as long as you get straight what it means, particularly what |x| means.

I suggest it is just slightly more obvious if you rewrite the problem in the form

x - 3 < 2|x|

and draw a graph of the two sides.

4. Sep 18, 2010

priscilla98

The original problem is

x-2|x|< 3

5. Sep 18, 2010

priscilla98

x-2|x| < 3

My last line which was x^2 - 2x < 3. I treated this problem x-2|x| < 3 as x(x - 2), I don't know if I did it right or not, I at least attempted to answer this question because I don't like to ask questions and I didn't at least attempted it. Okay, then you might think its an easy question but I don't know, I admit I'm not perfect. I'm trying to figure it out and take it step by step. For example, I'm pretty sure |x| just means x

x - 3 < 2|x|

Therefore, how would you start off to get x separately? Therefore,

x - 3 < 2x is the same as x - 3 < 2|x|, am i right?

6. Sep 18, 2010

epenguin

They are equivalent. You get mine by adding 2|x| to both sides, and subtracting 3 from both sides. You already did something like that yourself and that bit was not wrong.

If something < something else then if you subtract the same amount from (or add the same amount to) both the something and the something else, then the < remains true in the result. Basically all you ever do with equations and inequalities is do the same thing to both sides so that you change one true statement into another. But you were not doing that in the step I criticised.

7. Sep 18, 2010

HallsofIvy

Staff Emeritus
The simplest way to do this problem, and any problems involving absolute value, it to break it into cases:

case I: $x\ge 0$ so |x|= x.
The inequality is just x- 2x= -x< 3. Then x> -3 but since we are requiring that $x\ge 0[/math] that says the inequality is true for all [itex]x\ge 0$.

case II: x< 0 so |x|= -x.
The inequality is x+ 2x= 3x< 3 so that x< 1. Since we are requiring that x< 0, that says the inequality is true for all x< 0.

Putting those together, x- 2|x|< 3 is true for all x!

8. Sep 18, 2010

priscilla98

So, this is how you did it --> x - 3 < 2|x|

As you stated you just subtract 3 and add the 2x over the other side. Then, this is an equality can you state x cannot equal to 2. I guess your right that the problem is similar but I don't really understand how you can get x by it self. For example x > 2.

Wait, would you have to get the LCD? But I'm confused, usually when I see fractions, I use the LCD

9. Sep 18, 2010

epenguin

OK you are trying it out, but you are very handicapped if you don't know what the question means. Which you don't if you don't know what |x| means (where x is any real number).

It just means the quantity x in absolute amount, i.e. always a non-negative number, i.e. |x| = x when x > 0 or x = 0, |x| = -x when x < 0.

Problems involving |x| are usually solved by treating the two cases separately. Edit: as HOI has also said.

(Can I ask has anything I said reminded you of anything said in class? - because with some of the problems I have seen brought here I wonder if students had no notes or no textbooks to look up things.)

Last edited: Sep 18, 2010
10. Sep 18, 2010

epenguin

Hope my previous post helps. There are no fractions, LCD etc. in this problem it is much simpler, only additions/subtractions and knowing what | | means.

11. Sep 18, 2010

epenguin

Sorry there was a typo in #9 making it incomprehensible , now corrected.

12. Sep 18, 2010

priscilla98

Okay, your right, x has two values since it has the brackets, it would be x and -x. Yes, what your talking about now reminds me of something in class. I even brought the Pre-calculus by J.S. Ratti and I have so many notes that I can't even count how much I must have now since school started last week. This is one of the problems from my hw and my teacher counted this as an extra credit since she didn't teach it much but thanks for helping me in this problem. I really do hope with my devotion to studying pre-calculus that I will pass the test.

Thanks a lot

13. Sep 18, 2010

priscilla98

Thanks a lot for showing me step by step on how to solve this problem. Some people are so understanding to others when they need help and are quite unsure. You just made me have extra points for my test which i guess i need. But I'm going to study for this because I have calculus along with pre-calculus but i brought some books and will go through them.