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Homework Help: 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

  2. jcsd
  3. Feb 11, 2005 #2
    Is that

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


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

  4. Feb 11, 2005 #3


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    Homework Helper

    Try to arrange it to:
    [tex]\frac{A}{B} <= 0[/tex]
    [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
    do you have to do the "cases" for it?
  6. Feb 11, 2005 #5


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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.

  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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