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Math Help - math problems

  1. #1
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    Hello! I have a problem.

    Hello I have a problem.
    We have X+Y+Z+W=X+Y+Z+W=0
    What the value of x,y,z,w?
    Last edited by CaptainBlack; May 11th 2006 at 11:41 PM.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by sophea
    Hello I have a problem.
    We have X+Y+Z+W=X+Y+Z+W=0
    What the value of x,y,z,w?

    Since X^2,\ Y^2,\ Z^2, W^2 \ge 0 if their sum is 0 they must individualy all be [math0[/tex], so:

    <br />
X=Y=Z=W=0<br />

    RonL

    PS You should have made your post a new thread not tacked it on the the
    end of an old unrelated thread
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  3. #3
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    Arrow

    Quote Originally Posted by CaptainBlack
    Since X^2,\ Y^2,\ Z^2, W^2 \ge 0 if their sum is 0 they must individualy all be [math0[/tex], so:

    <br />
X=Y=Z=W=0<br />

    RonL

    PS You should have made your post a new thread not tacked it on the the
    end of an old unrelated thread
    Ok! thank you very for you solution.
    now i have one more problem:
    X.X=2X +[X]
    Note: X-1<[X] <X
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by sophea
    Ok! thank you very for you solution.
    now i have one more problem:
    X.X=2X +[X]
    Note: X-1<[X] <X
    There is a limited range of  X over which this can have a solution. For
    instance the inequality cannot hold if  |X| \ge 2.5 . So plug in  -2,-1,0,1,2 in
    turn for  [X] and then solve for  X . Keep any solutions for which  [X] is
    equal to the assumed value used.

    As an example we will work the case where we assume  [X]=-2 .

    Then we want solutions of:

     <br />
X^4=2 X^2-2<br />

    which is a quadratic in  X^2 , so using the quadratic formula:

     <br />
X^2=1 \pm \sqrt{-1}<br />
,

    which is complex and so not an admissible solution (I will assume we want
    real solutions).

    Now lets try assuming  [X]=-1 , then:

     <br />
X^2=1 \pm \sqrt{0}=1,<br />

    and so  X=\pm 1 are solutions of this equation and  X=-1 is consistent with
    our assumption and so is a solution of the original problem.

    Carrying on like this you will find all of the remaining solutions.

    RonL
    Last edited by CaptainBlack; May 22nd 2006 at 09:10 PM.
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  5. #5
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    Thumbs down

    Quote Originally Posted by CaptainBlack
    There is a limited range of  X over which this can have a solution. For
    instance the inequality cannot hold if  |X| \ge 2.5 . So plug in  -2,-1,0,1,2 in
    turn for  [X] and then solve for  X . Keep any solutions for which  [X] is
    equal to the assumed value used.

    As an example we will work the case where we assume  [X]=-2 .

    Then we want solutions of:

     <br />
X^4=2 X^2-2<br />

    which is a quadratic in  X^2 , so using the quadratic formula:

     <br />
X^2=1 \pm \sqrt{-1}<br />
,

    which is complex and so not an admissible solution (I will assume we want
    real solutions).

    Now lets try assuming  [X]=-1 , then:

     <br />
X^2=1 \pm \sqrt{0}=1,<br />

    and so  X=\pm 1 are solutions of this equation and  X=-1 is consistent with
    our assumption and so is a solution of the original problem.

    Carrying on like this you will find all of the remaining solutions.

    RonL
    Dear sir!
    How smart you are!
    I'm realy happy to saw your solution.
    Now i have one more problem and I hope you are not feel borred because I always have more and more problem.
    My problem is:
    In ABC rectangle have BO bisects CBA , CO bisects ACB, MN // BC .
    FInd the perimeter of AMN rectangle. Thank you advance.
    my email address is "sophearot@yahoo.ca"
    Last edited by CaptainBlack; May 22nd 2006 at 09:03 PM.
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