Simple Inequality with Modulus Question

[kuskus94]

Homework Statement

Determine m :

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

if its final form is :

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

Homework Equations

To remove the modulus, square them...

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)?

 

[haruspex]

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

You can't do that because m might be negative for some x. Try the squaring hint.

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

There's an error in that step.

 

[kuskus94]

You can't do that because m might be negative for some x. Try the squaring hint.

There's an error in that step.

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?

 

[haruspex]

it should be

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

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.

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

So what form should m take to make that guaranteed, and ensure m non-negative?

 

[kuskus94]

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?

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

 

[haruspex]

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

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

 

[kuskus94]

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

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

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

 

[haruspex]

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

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

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

 

[kuskus94]

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

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

I am still in doubt.

 

[haruspex]

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

I am still in doubt.

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