1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
    Staff Emeritus
    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Trouble with very basic algebra question
  1. Basic Algebra Question (Replies: 13)

Loading...