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Really Simple I just forgot

  1. Jan 17, 2005 #1
    If I have a problem such as x^2 + 3x <10 and I got -5 and 2 as my answers, would it be x< -5 and x<2? or -5<x<2?? and when there is an absolute value symbol in the problem, when finding the other answer, do I flip the inequality sign? Please help me remember the rules on inequality!!
     
  2. jcsd
  3. Jan 17, 2005 #2

    dextercioby

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    In this case,it's better if u do a graph.
    First of all,write your inequation under the form
    [tex] x^{2}+3x-10<0 [/tex]

    Plot the parabola and see for which "x",the values of "y" (of the function whose graph is the parabola) are less than 0.

    For the second part,please be more specific.Give an example,maybe...

    Daniel.
     
  4. Jan 17, 2005 #3
    Thanks. For the second part, something like l 3x-4 l >5
     
  5. Jan 17, 2005 #4

    dextercioby

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    Okay,explicitate the modulus...According to its definition,of course...

    Daniel.
     
  6. Jan 18, 2005 #5

    HallsofIvy

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    Dexter, behave yourself!

    A general technique for solving inequalities is to solve the equation first (just what you did) then check one point in each interval.

    For example, to solve x2+ 3x< 10, first solve x2+ 3x= 10.
    x2+ 3x- 10= (x+ 5)(x- 2)= 0 so x= -5 or 2. Those points divide the number line into 3 intervals: x< -5, -5< x< 2, and x> 2. Since a continuous function can change from positive to negative and vice versa (or < 10 to > 10) where it is equal to 0 (or = 10) x2+3x- 10 must have the same sign throughout each of those intervals.
    It's easy to calculate that (-6)2+ 3(-6)= 36- 18= 18> 10 so x2+ 3x> 10 for all x< -5.
    It's easy to calculate that 02+ 3(0)= 0< 10 so x2+ 3x< 10 for all x between -5 and 2.
    It's easy to calculate that 32+ 3(3)= 12> 10 so x2+ 3x> 10 for a x> 2.

    Likewise to solve |3x-4|> 5, first solve |3x- 4|= 5. Since absolute value "loses" the sign, either 3x-4= 5 or 3x-4= -5. In the first case, 3x= 9 so x= 3. In the second,
    3x= -1 so x= -1/3.
    The two points, x= -1/3 and x= 3 divide the number line into 3 intervals: x< -1/3,
    -1/3< x< 3, and x> 3.

    x= -1 is in x< -1/3. |3(-1)- 4|= |-7|= 7> 5. |3x-4|> 5 for all x< -1/3.

    x= 0 is in -1/3< x< 3. |3(0)-4|= |-4|= 4< 5. |3x-4|< 5 for all x in -1/3< x< 3.

    x= 4 is in x> 3. |3(4)-4|= |8|= 8> 5. |3x- 4|> 5 for all x> 3.
     
  7. Jan 18, 2005 #6

    dextercioby

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    What's the "catch",Halls?? :confused:

    Daniel.

    P.S.Did i say somethin' stupid,again??? :yuck:
     
  8. Jan 18, 2005 #7
    in this case, since both since are positive
    you may squares both side so that the abs sign will no longer exist
     
  9. Jan 18, 2005 #8

    dextercioby

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    Why complicate the problem uselessly???He could just apply the definition of modulus as i had advised earlier...

    Mathematics is better comprehendable when put in the simpest terms... :smile:

    Daniel.
     
  10. Jan 18, 2005 #9

    HallsofIvy

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    "explicate the modulus"??

    I would think that most basic algebra students would not recognize the word "modulus" as meaning absolute value.
     
  11. Jan 18, 2005 #10

    dextercioby

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    "Explicitate",so when wording is concerned,it's one-a-piece... :tongue2: Check out the "erf area" in the math thread... :wink:

    Daniel.

    P.S.I'm not used to spelling English words for math names... :redface:
     
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