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Need the domain of integral values that satisfy

  1. Mar 13, 2006 #1
    I have this inequality:

    [tex]4x^2 - 160x + 1500 \le 900[/tex]

    I brought the 900 to the LHS and found the roots (35.81, 4.19). Now I just need the domain of integral values that satisfy. Can I say [tex]5 \le x \le 35[/tex]?
     
    Last edited: Mar 14, 2006
  2. jcsd
  3. Mar 13, 2006 #2
    The range ([itex]5\le x\le 35[/itex]) you have is not correct. Thinking of like this might help get you a range: [itex](x-35.81)(x-4.91)\le 0[/itex]. From the solution you can see what range you should have.
     
  4. Mar 13, 2006 #3
    Sorry, I meant domain not range. Or is that what you meant too?
     
  5. Mar 13, 2006 #4
    [itex](x-35.81)(x-4.91)\le 0[/itex] gives you two inequalities, right? What can you conclude from them?
     
  6. Mar 13, 2006 #5
    [tex]x \le 35.81[/tex]
    [tex]x \le 4.91[/tex]

    The second one can't be right...
     
  7. Mar 13, 2006 #6
    Yep. And what can you say from this?
     
  8. Mar 13, 2006 #7
    Well, the first would be redundant.

    How can this be correct, because if I substitute x = 3 in my orginal inequality, I get [itex]1056 \le 900[/itex]
     
  9. Mar 14, 2006 #8

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    No, that's not right. Think about the graph of a quadratic. If
    [tex]y= 4x^2 - 160x + 1500[/tex] that's a parabola opening upward.
    Where is
    [tex]y= 4x^2 - 160x + 1500= 0[/tex]
    y will be negative between them.
     
  10. Mar 14, 2006 #9

    VietDao29

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    Homework Helper

    Hmm, okay, say you have the a quadratic function:
    f(x) := ax2 + bx + c, which has 2 solutions:
    x1, and x2. (And x1 < x2)
    Then if x0 is in ]x1, x2[
    Then af(x0) < 0, that means if a > 0, then f(x0) < 0, and vice versa, if a < 0, then f(x0) > 0.
    If x0 is not in ]x1, x2[, then af(x0) > 0.
    ---------------
    Example:
    Solve:
    x2 - 3x - 1 > -3
    <=> x2 - 3x + 2 > 0
    <=> (x - 2) (x - 1) > 0
    <=> x < 1, or x > 2 (since a = 1 > 0, x1 = 1, and x2 = 2).
    So can you apply it to your problem? :)
     
    Last edited: Mar 14, 2006
  11. Mar 14, 2006 #10
    So if I'm interpreting you guys correctly, then the domain in my OP is correct? (5 <= x <= 35)
     
  12. Mar 16, 2006 #11

    VietDao29

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    Homework Helper

    Yes, it's correct. :smile:
     
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