Understanding and Solving Inequalities Involving Square Roots and Fractions

  • Thread starter Thread starter gillgill
  • Start date Start date
  • Tags Tags
    Inequality
Click For Summary

Homework Help Overview

The discussion revolves around solving an inequality involving square roots and fractions, specifically the expression √(x+2) + 1/x+2 > 0. Participants are clarifying the notation and exploring the implications of different interpretations of the inequality.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to clarify the correct interpretation of the inequality, questioning whether it should be written as √(x+2) + 1/x + 2 > 0 or √(x+2) + 1/(x+2) > 0. There are suggestions to arrange the inequality into a fraction format and to consider the domain of x.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations and approaches to the problem. Some guidance has been offered regarding the arrangement of the inequality and the importance of defining the domain, but no consensus has been reached on the correct form of the inequality.

Contextual Notes

Participants are noting the importance of careful notation to avoid ambiguity and are considering the implications of the inequality being "equal or bigger," which adds complexity to the problem.

gillgill
Messages
128
Reaction score
0
Can you guys help me solve this inequality

√(x+2) + 1/x+2 >0

thanks
 
Physics news on Phys.org
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
 
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?
 
You may want to 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.
 
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 ©)
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
Replies
2
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
6
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
25
Views
2K
  • · Replies 32 ·
2
Replies
32
Views
3K