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Math Help - urgent . tmorrow sch

  1. #1
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    urgent . tmorrow sch

    find the range of values of x for each of the following inequalities

    x^2 + 2x +1 >0
    (x+1)^2 > 0
    (x+1) >0
    x>1


    how come the answer given is ( x is a subset of IR, x is not equal to -1)
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  2. #2
    Moo
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    Hello,

    This is because
    (x+1)^2 > 0
    is always TRUE.

    Unless (x+1)=0, because it's >, not \geq

    And (x+1)=0 if and only if x=... ?
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  3. #3
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    Quote Originally Posted by Moo View Post
    Hello,

    This is because

    is always TRUE.

    Unless (x+1)=0, because it's >, not \geq

    And (x+1)=0 if and only if x=... ?

    ok thx for clarification.

    here is another question.

    (x-2)^2 (x+1) less than or equal 0

    how to find the range of values of x

    x less than or equal to -1, while x=2 ( why)??
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  4. #4
    Super Member Matt Westwood's Avatar
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    (x-2)^2 (x+1) \le 0

    When x=2 then (x-2)^2 = 0 so (x-2)^2 (x+1) = 0 so that gives you the x=2 solution.

    Otherwise (x-2)^2 > 0 and so (x-2)^2 (x+1) \le 0 whenever (x+1) \le 0 or when x \le -1.

    Put them together:

    (x-2)^2 (x+1) \le 0 \Longrightarrow x=2 \textrm{ or } x \le -1.

    So it's:
    "x less than or equal to -1, or x=2" not "x less than or equal to -1, while x=2".
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