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Inequality √(x+2) + 1/x+2 >0

  1. Feb 11, 2005 #1
    Can you guys help me solve this inequality

    √(x+2) + 1/x+2 >0

    thanks
     
  2. jcsd
  3. Feb 11, 2005 #2
    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
     
  4. Feb 11, 2005 #3

    VietDao29

    User Avatar
    Homework Helper

    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,
     
  5. Feb 11, 2005 #4
    o..icic..thx...
    do you have to do the "cases" for it?
     
  6. Feb 11, 2005 #5

    dextercioby

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    Science Advisor
    Homework Helper

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

    Daniel.
     
  7. Feb 11, 2005 #6
    If the question is designed as "equal or bigger",it is more tricky.
     
  8. Feb 12, 2005 #7
    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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