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Inequalities and absolute value

  1. Oct 23, 2009 #1
    1. The problem statement, all variables and given/known data
    1) x^5 > x^2
    2) 7| x + 2 | + 5 > 4
    3) 3 - | 2x + 4 | <= 1

    2. Relevant equations



    3. The attempt at a solution
    1)
    x5 - x2 > 0
    x2(x3 - 1) > 0
    x2(x - 1)(x2 + x + 1) > 0
    Im not too sure what to do next. I cant factor it any further, at least I dont think so. Which leads me to ask how exactly am I suppose to find the numbers to check what the solution is?

    2)

    7| x + 2 | + 5 > 4
    7| x + 2 | > -1
    |x + 2 | > -1/7
    Can this be correct? The absolute value must always equal 0, or a positive number, right? How would I go about solving this? Or should I say the solutions do not exist?

    3)
    3 - | 2x + 4 | <= 1
    - | 2x + 4 | <= -2
    | 2x + 4 | => 2
    2x + 4 => 2
    2x => -2
    x => -1
    or
    2x + 4 <= -2
    2x <= -6
    x <= -3
    is this the right answer?

    Thanks for your time
     
  2. jcsd
  3. Oct 23, 2009 #2
    For 1), now that you've factored it, find where the graph crosses the x-axis to get some intervals between those points. Each interval will be either above or below the x-axis.

    [Edit] I must have thought the the inequality sign for 2) was the other way...
    The absolute value of a real number is always ≥0, so |x + 2| > -1/7 is always true, for any real x.

    3 seems correct.
     
    Last edited: Oct 23, 2009
  4. Oct 23, 2009 #3
    When at |x + 2 | > -1/7, recall that this is an inequality, not an equation, it doesn't say that |x+2| is less than 0, it says that it is greater than -1/7. No value for x would make this untrue, so x can be any real number.
     
  5. Oct 23, 2009 #4
    x5 - x2 > 0
    x2(x3 - 1) > 0
    x2(x - 1)(x2 + x + 1) > 0

    Solving an inequality would mean to express the solution as a union of intervals. In this case, which values of x will result in a value greater than 0 when plugged into the inequality.
     
  6. Oct 24, 2009 #5
    So, I should go about solving the equation then?

    Such as,

    x + 2 > -1/7
    x > -1/7 - 2
    x > -15/7
    or
    x + 2 < 1/7
    x < -13/7

    It seems these answers conflict, though. How can x be greater than -15/7, and less than
    -13/7.

    I'm rather confused about absolute value because they have drilled it into my head that they always must be positive, or 0. So, when I saw an absolute inequality with it saying > -1/7, I assumed that the absolute value, while greater than 1/7, was still a negative. Does this mean in the cases of absolute values and inequalities, it doesn't matter if there is a negative value after one of the <,> signs?

    Thanks Again
     
    Last edited: Oct 24, 2009
  7. Oct 24, 2009 #6
    Think of an absolute value as a distance in that a distance is going to be positive. The statement is true because since you know |x + 2| is always positive, you know |x + 2| is greater than -1/7 no matter what value of x you plug in. Remember it is not an equation, so it even if it said |x + 2| > -100,000 it would still be true.
     
  8. Oct 24, 2009 #7
    I understand. Was the posted solution to that question correct? The answers left me confused.
     
  9. Oct 24, 2009 #8
    Don't think of plugging those values of x into |x + 2 | > -1/7. You're trying to find out which values of x make this statement true: 7| x + 2 | + 5 > 4. Try plugging your solution into the inequality for x and then seeing if that proves true.
     
  10. Oct 24, 2009 #9
    I plugged them, and they work. I was just concerned because the textbook asks for me to solve the question also in a graph form.
     
  11. Oct 24, 2009 #10
    Since you are confused that the answers seem to overlap, think about what that means. It means that all real numbers are included.
     
  12. Oct 24, 2009 #11
    That makes sense!
    Thanks for your help.
     
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