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Inequalities help

  1. Jul 1, 2007 #1
    stupid error in the "tex" thing, see first post.
    Last edited: Jul 1, 2007
  2. jcsd
  3. Jul 1, 2007 #2
    damn latex keeps screwing up see next post, soz
    Last edited: Jul 1, 2007
  4. Jul 1, 2007 #3
    just starting to learn some stuff on inequalities and need some help. Not sure where to put this topic, so move it if necessary. thnx

    Heres one problem I think I've solved but need clarification from you it's right or where i've gone wrong. I also may have missed an obvious solution but im going off a very short and "concise" introduction to inequalities so I havn't really got a full understanding of how to solve a given problem. Thnx

    Find the set of real numbers [tex] x \neq 0[/tex] such that [tex]2x + 1/x < 3[/tex]:

    I then manipulated 2x + 1/x < 3 to get to

    [tex]\frac{(2x-1)(x-1)}{x} < 0[/tex]

    Im pretty sure up till there is right, but it's the next stage that I think my working in dubious. I'm pretty sure I can't times both sides by x because it could be negative, could be positive.

    I know [tex]\frac{(2x-1)(x-1)}{x}[/tex] is negative. Because its less than O. This means that either 2x-1, x-1, or x is negative, or all 3. I kinda think all that makes sense.

    (this is where is gets real dubious lol).

    So, the following are the options.

    2x-1>0, x-1<0, x>0
    2x-1>0, x-1>0, x<0
    2x-1<0, x-1<0, x<0
    2x-1<0, x-1>0, x>0

    which cancels too

    x>1/2, x<1, x>0
    x>1/2, x>1, x<0
    x<1/2, x<1, x<0
    x<1/2, x>1, x>0

    the 2nd and 4th possiblilities are impossible so that leaves


    the first doesn't work either because it says find real numbers and subbing in "1" give 3<3, which is false.

    so then i conclude the answer has to be [tex]x \leq -1[/tex]

    is this right? i know i probably did it wrong, but need to know where I went wrong, thnx.
  5. Jul 1, 2007 #4
    should be fine now, fixed the latex.
  6. Jul 1, 2007 #5


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    Science Advisor

    The easiest thing to do is multiply both sides by x^2, which you know will be non-negative. This gets you to,

    [tex] (2x-1)(x-1)(x) < 0[/tex]

    This a cubic with three real roots at x=0, x=1/2 and x=1. Make a quick sketch of it if you like, it's easy to see that it is negative for x < 0 and for 1/2 < x < 1.

    BTW: You don't really need to sketch it, that was just a suggestion in case you where not already familar with what a cubic looks like. If you are familar with a cubic you'll know that a positive cubic (x^3 term positive) will go from -ive to +ive through the first zero, from +ive to -ive through the second zero and so on. So it's not hard to figure out where it is negative.
    Last edited: Jul 1, 2007
  7. Jul 1, 2007 #6
    o rite i see

    was my answer correct?
  8. Jul 1, 2007 #7


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    Homework Helper

    No, it's not 1/2 > x < 1, in fact, it should read:

    1/2 < x < 1

    No, you don't substitute "1" in. 1/2 < x < 1, x = 1 does not satisfy the requirement. The solution is 1/2 < x < 1, so if you want to check, well take any x that is greater than 1/2, and less than 1, you can choose .999, but definitely not 1. Can you see why? :)

    Btw, how did you get x <= -1? You should note that x is a real number, not an integer.

    There's a more common method, are you familiar with drawing tables, and testing the sign on each interval?
  9. Jul 1, 2007 #8
    o crap haha, misread the question lol. stupid me.

    yeah thnx for that, yeah i think the 1/2 > x < 1 was a typo.
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