What is the necessary and sufficient condition?

[nuuskur]

If you pick $a=4$ then
<br /> \lvert x-2 \rvert &lt; 4 \iff -4 &lt; x-2 &lt; 4 \iff -2 &lt;x&lt;6<br />

 

[nuuskur]

f you pick, say, $a=1$, then the implication
<br /> (x+2)(x-5)&lt;0 \implies -1&lt;x&lt;3<br />
is false. We can pick $x = -1.5$, it still stands that $(x+2)(x-5)<0$, but at the same time $x\leq -1$.

Ok, the implication is still false and the same counter-example works, but I should go back to primary school, because elementary arithmetic is, at times, an impossible task. It should be for $a=1$
<br /> (x+2)(x-5)&lt;0 \implies 1&lt;x&lt;3<br />
Now, any $x\in (-2,1]\cup [3,5)$ works as a counter-example.

but I am certain now that this is correct

 

[Helly123]

Ok, the implication is still false and the same counter-example works, but I should go back to primary school, because elementary arithmetic is, at times, an impossible task. It should be for $a=1$
<br /> (x+2)(x-5)&lt;0 \implies 1&lt;x&lt;3<br />
Now, any $x\in (-2,1]\cup [3,5)$ works as a counter-example.

but I am certain now that this is correct

Ok thanks

 

[tnich]

Yes.
For | x − 2 | < a to be a sufficient condition so that x² − 3x − 10 < 0, the range for a is 0 < a ≤ 3.
You have not been very clear in showing that this is the case.

This does not give all of the values of x which satisfy the given quadratic inequality. Rather any x fulfilling | x − 2 | < a will satisfy the quadratic inequality.for the specified range of a values. So being in that interval is a sufficient condition for x to satisfy the quadratic inequality..

I'm still hung up on the original problem statement which says x² − 3x − 10 < 10. Did we decide somewhere that this was in error?

 

[Helly123]

I'm still hung up on the original problem statement which says x² − 3x − 10 < 10. Did we decide somewhere that this was in error?

Yes. I made mistake. It supposed to be x^2 - 3x - 10 < 0