Thread: Mathmatics Question?? Powers and inequalities.

  1. #1
    essence of digital xddxogm3's Avatar
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    Mathmatics Question?? Powers and inequalities.

    I'm going to attempt to ask an educated algebra question. Please forgive a bad translation.

    Inequalities and Powers/Roots.
    Legend:
    () <-Power
    [] <-Square Root

    Solve:
    the square of x is less than 4
    x(2)<4

    would this translate to
    x is less then the positive/negative of the square root of 4.
    x<+/>-[4]
    or
    x<2 and x>-2
    [edit1]
    -2<x<2
    [/edit1]

    ??????????????????????????
    My book doesn't address this type of inequality, and would really like to know this.
    Last edited by xviddivxoggmp3; 09-26-2004 at 04:56 PM.
    "Hence to fight and conquer in all your battles is not supreme excellence;
    supreme excellence consists in breaking the enemy's resistance without fighting."
    Art of War Sun Tzu

  2. #2
    Cheesy Poofs! PJYelton's Avatar
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    Yes, that is correct.

  3. #3
    essence of digital xddxogm3's Avatar
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    note the change of the greater than and less than symbol on the
    x<2 and x>-2
    is that change of symbol accurate as well?
    "Hence to fight and conquer in all your battles is not supreme excellence;
    supreme excellence consists in breaking the enemy's resistance without fighting."
    Art of War Sun Tzu

  4. #4
    Crazy Fool Perspective's Avatar
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    its usually written this way
    Quote Originally Posted by me
    -2 < x < 2

  5. #5
    essence of digital xddxogm3's Avatar
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    yes, I understand that, but when you have an inequality like the one I was working, does it follow the muliplication/division rule when working with negatives? Do we always switch the > or < upon noticing that the square root will result in a negative value? again I appologize if i'm not to clear. I'm new to saying math. I think math, but never have to explain it much. I'm sorry if I'm repeating myself. Did you put the -2<x<2 to state that we do follow the above mentioned multiplication/devision rule of negatives and inequalities?
    Last edited by xviddivxoggmp3; 09-26-2004 at 04:59 PM.
    "Hence to fight and conquer in all your battles is not supreme excellence;
    supreme excellence consists in breaking the enemy's resistance without fighting."
    Art of War Sun Tzu

  6. #6
    S Sang-drax's Avatar
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    We usually write it using absolute value:

    x^2 < 4 <=>

    |x| < 2

    The inverse of x^2 is sqrt(x) when x >= 0 and
    sqrt( -x ) when x < 0


    As for switching signs, remember that:
    -x < y <=> x > -y
    Last edited by Sang-drax; 09-26-2004 at 05:04 PM.
    Last edited by Sang-drax : Tomorrow at 02:21 AM. Reason: Time travelling

  7. #7
    essence of digital xddxogm3's Avatar
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    Thank you all for the help.
    One last attempt to clarify.
    Please do not think I'm an idiot.
    I have an exam and want to make sure all concepts are know front and back.

    x^2 < 4

    {x|-2<x<2}

    Correct? Yes or No
    "Hence to fight and conquer in all your battles is not supreme excellence;
    supreme excellence consists in breaking the enemy's resistance without fighting."
    Art of War Sun Tzu

  8. #8
    S Sang-drax's Avatar
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    Yes, that is correct.

    Here's how to derive it:

    x^2 < 4
    |x| < 2
    +- x < 2
    (i) x < 2
    (ii) -x < 2 <=> x > -2

    (i) & (ii) <=> -2 < x < 2 <=> |x| < 2
    Last edited by Sang-drax; 09-26-2004 at 05:24 PM.
    Last edited by Sang-drax : Tomorrow at 02:21 AM. Reason: Time travelling

  9. #9
    essence of digital xddxogm3's Avatar
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    Have I said how much you people on this board rock lately?
    You all kick some major @$$.
    "Hence to fight and conquer in all your battles is not supreme excellence;
    supreme excellence consists in breaking the enemy's resistance without fighting."
    Art of War Sun Tzu

  10. #10
    Crazy Fool Perspective's Avatar
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    On an exam be sure to follow Sang-drax's solution. If you leave out the absolute value profs will mash your marks down like gravy soaked potatoes.....*drool* im sooo hungry.

  11. #11
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    Using the squareroot on both sides of the inequality looks correct, but unless if you have theorem in your book that tells you it is correct, you should not rely on it without first proving it. Ie., you must first show that if x^2 < a, then sqrt(x^2) < sqrt(a).

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