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Solve Inequalities algebraically

  1. Sep 29, 2013 #1
    1. The problem statement, all variables and given/known data
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    3. The attempt at a solution
    So when I started this question, I knew that 2r +3 > 0 in order for this whole thing to work so r > -3/2, and I know that if the middle term has to be more than 0, than only one more of the others is going to be negative, for the first term i found that r < 4/3 and the third was r < 2, so I know that the answer should be (4/3, 2), but how do I prove that algebraically?
     
    Last edited: Sep 29, 2013
  2. jcsd
  3. Sep 29, 2013 #2

    eumyang

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    Homework Helper

    I wouldn't solve it "algebraically." I would solve it by making a two-column table. First, find values of r that would make the left side 0 or undefined (I call these critical values), which appears you have already done. Then in the first column list all critical values and all intervals bounded by these critical values.
    Code (Text):
    Test interval|Value
    -------------+----------
    (-inf, -3/2) |undefined
    r = -3/2     |0
    (-3/2, 4/3)  |
    r = 4/3      |0
    (4/3, 2)     |
    r = 2        |0
    (2, inf)     |
     
    In the second column, you fill in 0, undefined, positive, or negative. For the remaining intervals, pick a test value to plug into the inequality and you will eventually determine the solution set.
     
  4. Sep 29, 2013 #3
    ohhhh ok, that makes sense. thank you.
     
  5. Sep 29, 2013 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Since the square root, [itex]\sqrt{2r+ 3}[/itex], is, by definition, nonnegative, you can (almost) ignore it. The product of the two numbers, 3r- 4 and [itex]\sqrt[3]{r- 2}[/itex], will be negative if and only if the two factors are of opposite order. In other words, either

    1) 3r- 4> 0 and [itex]\sqrt[3]{r- 2}< 0[/itex] or
    2) 3r- 4< 0 and [itex]\sqrt[3]{r- 2}> 0[/itex]

    Similarly, a positive number, cubed, is positive and a negative number, cubed, is negative. So that can be reduced to
    1) 3r- 4> 0 and r- 2< 0 or
    2) 3r- 4< 0 and r- 2> 0.

    3r- 4> 0 for r> 4/3 and r- 2< 0 for r< 3. Those will both be true for 4/3< r< 3.
    [itex]\sqrt{2r+ 3}= 0[/itex] only for r= -3/2 and that does not lie between 4/3 and 3.

    (1) is true for 4/3< r< 3.

    3r- r< 0 for r< 4/3 and r- 2> 0 for r> 3. (2) is never true.
     
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