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Inequality √(x+2) + 1/x+2 >0

  • Thread starter gillgill
  • Start date
Can you guys help me solve this inequality

√(x+2) + 1/x+2 >0

thanks
 
Is that

[tex]\sqrt{x+2} + \frac{1}{x} + 2 > 0[/tex]
or
[tex]\sqrt{x+2} + \frac{1}{x+2} > 0[/tex]

?

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

--J
 

VietDao29

Homework Helper
1,417
1
Hi,
Try to arrange it to:
[tex]\frac{A}{B} <= 0[/tex]
Or
[tex]\frac{A}{B} >= 0[/tex]
Where [itex]A = a_{1} \times a_{2} \times a_{3} \times ... \times a_{n}[/itex]
and [itex]B = b_{1} \times b_{2} \times b_{3} \times ... \times b_{k}[/itex]
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 [tex]\frac{A}{B}[/tex] is positive or negative or zero in each case.
Hope this help,
Viet Dao,
 
o..icic..thx...
do you have to do the "cases" for it?
 

dextercioby

Science Advisor
Homework Helper
Insights Author
12,920
500
You may wanna begin by stating clearly the domain of the "x"...And then look for those "x" which would satisfy your inequation.

Daniel.
 
If the question is designed as "equal or bigger",it is more tricky.
 
1,100
0
I still want to know what √(x+2) + 1/x+2 >0 is.

How we decided yet???

Is it:

[tex]\sqrt{\frac{(x + 2) + 1}{x + 2}} > 0[/tex]

Is it:

[tex]\sqrt{(x + 2) + \frac{1}{x + 2}} > 0[/tex]

or something else???

The Bob (2004 ©)
 

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