Hi Forum.

I've asked a lot of times about absolute values.

But here I go again:

Consider two conditions

$\displaystyle x^2-3x-10$ and $\displaystyle |x-2|<a$

Where x is a real number, and a is a positive real number.

The range of a, so that $\displaystyle |x-2|<a$ is a __necessary__ condition for $\displaystyle x^2-3x-10<0$

The range of a, so that $\displaystyle |x-2|<a$ is a __sufficient__ condition for $\displaystyle x^2-3x-10<0$

This problem is mainly about logic, right?

Here is something that I thought, after calculating the range of $\displaystyle x^2-3x-10<0$, that is $\displaystyle -2<x<5$

If we have $\displaystyle -2<x<5$

-2________________________________5

|________________________________|

___________________________a=0 ______________________________ a=3

|_________________________________________________ _| <-|x-2|<a

Edit: This didn't work so well but, the a=0 is right below -2 and the a=3 is right below 5

This is our __sufficient condition__: 0<a<3

I'm not sure what the absolute value does here.

What does the **absolute value** mean?

Absolute values are only the distance to x=0, right?

After calculating -correctly- you get a nicer solution that is:

$\displaystyle 0<a\leqslant 3$

But how do we get such solution?

$\displaystyle |x-2|<a $ is

$\displaystyle x-2$** if ** $\displaystyle x-2\geqslant a$ and

$\displaystyle -x+2$** if** $\displaystyle x-2<a$

This relates to __two range of values__ so this got a little out of control.

Somehow confusing, I wish I knew somewhere to find exercises of this kind!

I'm confused about that necessary/sufficient part

**Sufficient**

If you mow the lawn, you receive 10 dollars.

**____________________Necessary**

The $\displaystyle 0<a\leqslant 3$ range is sufficient for f(x) to work, or something like that?

But where does necessity comes in this? I know that if something is necessary it is not sufficient.

A--->B

A implies B

A sufficient, B necessary

I appreciate any help!

Thanks!