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Homework Help: B less or equal than 8

  1. Sep 30, 2011 #1
    1. The problem statement, all variables and given/known data
    This isn't going to be this complicated that I have to section my problem down to segments, this is just a part of an advanced problem. There is also something wrong with my browser, I can only type into a textbox where I start typing at the beginning. I can't press enter or anything like that.The problem is this: I cannot understand why this happens: http://j.mp/q2ySH9. I would think that b would be less or equal than BOTH of the numbers... yet it becomes an interval. EDIT: sorry, in the second step it's supposed to be b, not b squared.


    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited by a moderator: Apr 26, 2017
  2. jcsd
  3. Sep 30, 2011 #2

    SammyS

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    (Now your image is visible.)

    [itex]b^2\le8[/itex][itex]b^2-(\sqrt{8})^2\le0[/itex][itex](b-\sqrt{8})(b+\sqrt{8})\le0[/itex]

    Standard methods give the interval you questioned: [itex]-\sqrt{8}\le b\le\sqrt{8}\,.[/itex]
     
    Last edited by a moderator: Apr 26, 2017
  4. Sep 30, 2011 #3

    HallsofIvy

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    The "standard methods" are: pq< 0 if and only if p and q are of different sign: either "p< 0 and q> 0" or "p> 0 and q< 0".

    Since [itex]b^2- 8= (b-\sqrt{8})(b+\sqrt{8})[/itex], for that to be less than 0 we must have one of
    (a) [itex]b- \sqrt{8}> 0[/itex] and [itex]b+ \sqrt{8}< 0[/itex] or
    (b) [itex]b- \sqrt{8}< 0[/itex] and [itex]b+ \sqrt{8}> 0[/itex]

    (a) gives [itex]b> \sqrt{8}> 0[/itex] and [itex]b< -\sqrt{8}< 0[/itex] but those cannot both be true. Therefore, we must have (b) which gives [itex]b< \sqrt{8}[/itex] and [itex]b> -\sqrt{8}[/itex] or [itex]-\sqrt{8}< b< \sqrt{8}[/itex].

    Since the initial problem had "[itex]\le 0[/itex]" rather than just "<", we must have
    [tex]-\sqrt{8}\le b\le \sqrt{8}[/tex].

    By the way, since 8= 4(2) and 4 is a "perfect square", you can write [itex]\sqrt{8}= 2\sqrt{2}[/itex] and that is typically preferred:
    [tex]-2\sqrt{2}\le b\le 2\sqrt{2}[/tex]
    might be preferred as an answer.
     
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