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Homework Help: Trouble with very basic algebra question

  1. Sep 2, 2008 #1
    1. The problem statement, all variables and given/known data

    Determine all x in R such that the following hold:

    1) (x+4/x-2) < x
    2) |x+4/x-2| < x
    3) |2x| > |5-2x|


    2. Relevant equations

    We have the triangle inequality, |a + b| [tex]\leq[/tex] |a| + |b|, which also implies the relation ||a| - |b|| [tex]\leq[/tex] |a-b|.

    Also, relations such as: |x| < a implies -a < x < a, etc.
    |ab| = |a|*|b|

    3. The attempt at a solution

    I feel very bad asking about these, I'm sure I should have learned this sort of thing earlier than analysis I (call it my weak spot). Anyway, here is what is bothering me:

    1) (x+4/x-2) < x --> 0 < x - (x+4/x-2) --> 0 < [x(x-2) - (x+4)]/(x-2)
    --> 0 < (x^2 - 3x + 4)/2. Now if we use the quadratic formula to find the roots of the numerator, we get (3 +/- [tex]\sqrt{9-16}[/tex] )/2, so the roots are complex. So am I to assume that there are no x in R such that this inequality is true? I can accept that, but then the next one is the same, except the absolute value. So for that one:

    2) |x+4/x-2| < x --> -x < (x+4/x-2) < x --> -x(x-2) < x + 4 < x(x-2)
    --> -x^2 + x -4 < 0 < x^2 - 3x - 4. Now the lhs doesn't factor, but the quadratic formula results in complex roots again, and the rhs roots, which would be x=4 and x=-1, pose problems: if you substitute x = 4 or x = -1 into the inequality you get 4 < 4 and -1 < -1 which is obviously never true! So - do I have to assume again that there are no real x in R such that this inequality holds? I'm getting worried that all my problems (there are more very similar to these) seem to have no real x possible! It makes me think I am doing something wrong...

    3) |2x| > |5-2x| --> 2|x| > |5-2x| >= ||5| + |-2x|| --> I'm not sure what to do next! There is obviously some very simple answer that I should know. Maybe I am just having a slow night (lack of sleep does that!). I thought I'd be using the triangle inequality, but then, it's a difference and not a sum... so I'm just not sure what I'm supposed to do.

    Any tips for those of us who are SLOW SLOW SLOW with this type of inequality? I feel like I can handle proofs pretty well, with sup/inf concepts, proving other similar sorts of things, and have always been fine at abstract and linear algebra. But concrete math problems with numbers, whoa! No good at that... :grumpy:
     
  2. jcsd
  3. Sep 2, 2008 #2
    Are you trying to graph them on a number line and put them into inequalities?
     
  4. Sep 2, 2008 #3
    I'm just supposed to determine which real x's would satisfy them. I'm not sure what level of rigor is expected - I'm not sure if he just wanted to review a few computational examples of the basic real number theorems or if we are supposed to be very rigorous in showing how we make each step. I guess we are trying to put them into inequalities. I just haven't seen this sort of problem for about 7 years!
     
  5. Sep 2, 2008 #4
    I am trying to help out lol, let me read up a little more and I will get back to you.
     
  6. Sep 2, 2008 #5
    thanks! :-D
     
  7. Sep 3, 2008 #6
    The first one is pretty interesting. first u simply subtract "x" from each side to get
    (x+4)/(x-2)-x<0
    now, to find a common denominator
    ((x+4)-x(x-2))/(x-2)<0
    now, you CANT multiply each side by "x-2" because u dont know if x>2 or x<2 yet.
    Instead, just simplify some more
    (x+4-x^2+2x)/(x-2)<0
    (-x^2+3x+4)/(x-2)<0
    now, it is easier (but not necessary) to multiply each side by -1 (and flipping the < to a >)
    (x^2-3x-4)/(x-2)>0
    and, now comes the interesting part!
    (x+1)(x-4)/(x-2)>0
    alright, notice how the left hand side is only true, when the product of (x+1) and (x-4) is divided by (x-2).
    Notice, however, that the same "greater than zero" rules apply for products as they do for quotients. what i mean by this is: the previous equation is also true when
    (x+1)(x-4)(x-2)>0
    Now, all you have to do is find "for what values of x is (x+1) and (x-4) and (x-2) all positive?"
    Once you find that, you have to go through the different cases when 2 of them are negative, and the last other one is positive.
    The solution ends up being:

    -1<x<2 or x>4

    this solution can be easily verified, if you check out each case of "x" from x=-2 all the way to x=5
     
  8. Sep 3, 2008 #7
    You were very close in this one! when u plugged in x=4 and x=-1, you found the bounds of what x can or cannot be. All you need to do is try 3 situations. x<-1 , -1<x<4 , 4<x.
    If you try the numbers x=-2,1,5 you find the the statement becomes False,False,True
    that means that the correct answer is simply

    x>4

    you can also prove this one very easily by using any sort of graphing device.
     
  9. Sep 3, 2008 #8
    For the 3rd one, an educated guess and check will suffice.
    |2x|>|5-2x|
    2|x|>2|(5-2x)/2|

    |x|>|5/2-x|

    If we guess x<0, you can see the equation becomes
    -x>5/2-x
    now add x to each side
    0>5/2

    This is NEVER true, which means that x must be greater than 0.

    Now if we try x>5/2 our equation simplifies to:
    x>-(5/2-x)
    x>x-5/2
    subtracting x from each side gives
    0>-5/2

    Now this is ALWAYS true, so that means that x could be greater than 5/2.

    Now for our last attempt, we will have 0<x<5/2
    This leaves us
    x>5/2-x
    adding x to each side
    2x>5/2
    x>5/4

    Now combining out previous results, we obtain
    This means that (x>5/4 AND x>0) OR x>5/2
    the "(x>5/4 AND x>0)" simplifies to "x>5/4". Thus:
    (x>5/4) OR x>5/2
    and this is only true when x>5/4. Thus, the final answer to this simple yet hard problem is just

    x>5/4
     
  10. Sep 3, 2008 #9

    HallsofIvy

    User Avatar
    Science Advisor

    The simplest way to solve complicated inequalities is to first solve the corresponding equality.

    For example, to solve |2x|> |5-2x| first solve |2x|= |x-2x| which then can be writen as either 2x= 5- 2x or 2x= -(5- 2x).

    If 2x= 5- 2x then 4x= 5 so x= 5/4. If 2x= -(x- 2x)= -5+ 2x then 0= -5 which is not true. The equality is true only for x= 5/4

    Those two points separate intervals where the inequality is true from intervals whereit is false. It is only necessary to look at a single point in each interval.

    For example, 0< 5/4 and |2(0)|= 0 which is NOT greater than |5- 2(0)|= 5.
    The inequality is false for all x< 5/4.

    5/4< 2 and |2(2)|= 4 which is greater than |4- 2(2)|= 1. The inequality is true for all x> 5/4.
     
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