# Inequality √(x+2) + 1/x+2 >0

1. Feb 11, 2005

### gillgill

Can you guys help me solve this inequality

√(x+2) + 1/x+2 >0

thanks

2. Feb 11, 2005

### Justin Lazear

Is that

$$\sqrt{x+2} + \frac{1}{x} + 2 > 0$$
or
$$\sqrt{x+2} + \frac{1}{x+2} > 0$$

?

You must be careful with your notation to ensure that there isn't ambiguity!

--J

3. Feb 11, 2005

### VietDao29

Hi,
Try to arrange it to:
$$\frac{A}{B} <= 0$$
Or
$$\frac{A}{B} >= 0$$
Where $A = a_{1} \times a_{2} \times a_{3} \times ... \times a_{n}$
and $B = b_{1} \times b_{2} \times b_{3} \times ... \times b_{k}$
Then just simply solve the inequation by drawing a chart to see if each element is positive or negative or zero. And finally, see if $$\frac{A}{B}$$ is positive or negative or zero in each case.
Hope this help,
Viet Dao,

4. Feb 11, 2005

### gillgill

o..icic..thx...
do you have to do the "cases" for it?

5. Feb 11, 2005

### dextercioby

You may wanna begin by stating clearly the domain of the "x"...And then look for those "x" which would satisfy your inequation.

Daniel.

6. Feb 11, 2005

### primarygun

If the question is designed as "equal or bigger",it is more tricky.

7. Feb 12, 2005

### The Bob

I still want to know what √(x+2) + 1/x+2 >0 is.

How we decided yet???

Is it:

$$\sqrt{\frac{(x + 2) + 1}{x + 2}} > 0$$

Is it:

$$\sqrt{(x + 2) + \frac{1}{x + 2}} > 0$$

or something else???