# Homework Help: Absolute equation problems.

1. Mar 4, 2005

### maxpayne_lhp

Hello,
Well, I have the equation:
$$x^2-2x-m|x-1|+m^2=0 (1)$$ They ask me what m is so that the equation has solution(s)
Well, I let $$t=|1+x|$$ and solve the problem with t (no more x) and I have the last equation: $$t^2-mt+m^2-1=0$$ (2)
in order that (1) has solution(s), (2) must have solution(s) t grater or equal 0
Is that right?, is the action of let t is something make my solution wrong?
Thanks

2. Mar 4, 2005

### dextercioby

The substitution $$t=:|x-1|$$ is much better.

Daniel.

3. Mar 4, 2005

### maxpayne_lhp

Why? So, is my solution wrong?
Thanks

4. Mar 4, 2005

### dextercioby

Because it substitutes exactly the problematic term,namely the absolute value...

Try to compute the new equation using my solution.I think it's much easier.

Daniel.

5. Mar 4, 2005

### VietDao29

If you call t = |x + 1|, you obviously WON'T get $t^2-mt+m^2-1=0$. You can just get that if you substitude t = |x - 1|. And if I am correct, that problem does have solution.
Viet Dao,

6. Mar 4, 2005

### maxpayne_lhp

Okay, thanks Dextercioby. Viet dao, you meant that I compute incorrectly? Hmmm let me do it again.
Or, we can 'break out' the absolute by devide into two situation: x-1 >= 0 or x-1 <0 it's long, but not so confusing.

7. Mar 5, 2005

### saltydog

Well, "what up" then? Excuse my American urban getto slang as I use poor grammar to express emotions. Personally, I'd like to see closure on questions but that's just me. Anyway, seems to me one way to solve this is to break it up into two cases: x-1$\geq 0$ and x-1<0. Doing that gives the same result for the bounds on m considering real solutions:

$$-\frac{2}{\sqrt{3}}\leq m \leq \frac{2}{\sqrt{3}}$$

What did you get?

8. Mar 5, 2005

### Hurkyl

Staff Emeritus
I wonder if it would be easier to look at it more abstractly...

Instead of asking "for which m does this equation have a solution", you could instead say "solve this equation for m and x" or "for which x does this equation have a solution".

It's somewhat easier to solve this equation for m than for x... then once you have a bound on x, you could turn it into a bound on m.

But maybe that's the hard way. *shrug*