Inequalities - What is happening here?

**Kataangel** · December 8th 2009, 10:21 AM

My teacher wrote:

1. $\frac{4}{(x+2)^3}>0$
2. $x+2>0$
3. $x>-2$

How can 1. become 2. ???

Is it even a valid move?

**alibond07** · December 8th 2009, 10:27 AM

Maybe you have read what your teacher has written wrong, the last two lines make sense. Maybe the top line was a seperate question?

**masters** · December 8th 2009, 10:30 AM

Originally Posted by Kataangel

My teacher wrote:

1. $\frac{4}{(x+2)^3}>0$
2. $x+2>0$
3. $x>-2$

How can 1. become 2. ???

Is it even a valid move?

Hi Kataangel,

The denominator cannot = 0, therefore x cannot be -2.

The numerator is already positive, so $(x-2)^3$ must be positive as well.

$(x+2)^3>0$

$x+2>0$

$x>-2$

**Kataangel** · December 8th 2009, 11:54 AM

I'm confused about where $(x-2)^4$ comes from?

What's happening is my teacher is going from step 1. to step 2. to step 3.

I wanted to know if those moves are algebraically correct because I'm not quite sure on inequalities and I'm worried that going from step 1. to step 2. might be wrong. Is it wrong?

I think this might be what's happening:

$\frac{4}{(x+2)^3}>0$

$\frac{4}{(x+2)^3}*(x+2)^4>0*(x+2)^4$

$4(x+2)>0$

$(x+2)>0$

$x>-2$

Is this correct?

**Plato** · December 8th 2009, 12:34 PM

Originally Posted by Kataangel

I'm confused about where $(x-2)^4$ comes from?

Actully $(x-2)^4$ does not come into it.
It is a logical solution not operational.

If we know that $\frac{4}{a^3}>0$ then we know at once $a>0$ .
It can be no other way.
If $a<0$ then $\frac{4}{a^3}>0$ is false..

So your teacher just knew that $(x-2)>0$ from the logic of inequalities.

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