# Simple Inequality with Modulus Question

1. Oct 27, 2012

### kuskus94

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

Determine m :

$|x-10|<{1}/{m}$

if its final form is :

$|x^{2}+{4}x-140|<1$

2. Relevant equations

To remove the modulus, square them...

3. The attempt at a solution

I have tried to assume that if

$|x-10|{m}<{1}$

then, I can find

$|x-10|{m}=|x^{2}+{4}x-140|$

$|x-10|{m}=|(x+14)(x-10)|$

but, can I omit (x-10)?

Last edited: Oct 27, 2012
2. Oct 27, 2012

### haruspex

You can't do that because m might be negative for some x. Try the squaring hint.
There's an error in that step.

3. Oct 27, 2012

### kuskus94

Okay, sorry it should be

$|x-10|{m}=|(x+14)(x-10)|$

So, if I do this :

$(x-10)^2{m}^2=((x+14)(x-10))^2$

is it possible?

4. Oct 27, 2012

### haruspex

Not quite. Should be $|x-10||m|=|(x+14)(x-10)|$
But if you go back to |x−10|<1/m you can observe that m is necessarily non-negative, which simplifies the logic.
So what form should m take to make that guaranteed, and ensure m non-negative?

5. Oct 28, 2012

### kuskus94

because $|m|=\frac{|(x+14)(x-10)|}{|(x-10)|}$, is it $|m|=|(x+14)|$ ?

Last edited: Oct 28, 2012
6. Oct 28, 2012

### haruspex

Almost. If m were negative, would it satisfy the requirements?

7. Oct 28, 2012

### kuskus94

Hmmm... so, I need to add $$m>0$$ to the equation or something like that.

$$m=|x+14|; m>0$$?

8. Oct 28, 2012

### haruspex

Yes, except that you don't now need to specify m >= 0. It follows from m=|x+14|.

9. Oct 29, 2012

### kuskus94

So, the final answer is $$m=|x+14|$$?

I am still in doubt.

10. Oct 30, 2012

### haruspex

Looks right to me. Did you try plugging it into the original inequality?