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Polynomial inequality

  1. Aug 13, 2004 #1
    show that for the equation y^3 - 3y + 4 = 0 ,
    one of the root lies between -3 and -2

    i don't know how to show that one of the root lies between -3 and -2, but i can show that one of the root ( or more ) is smaller than - 3^(1/2), pay in mind that -3 and -2 are also smaller than -3^(1/2).
    Here is my method, but it doesn't solve the question, i wrote it just for your reference so you got more idea to solve it. U help is meaningful to me, thanks you !

    y^3 - 3y + 4 = (y)(y^2 - 3) + 4 = 0
    so this mean in order to make the equation becomes zero,
    the term (y)( y^2 - 3) must equal -4, in order word, it must less than zero...
    hence it is right to write (y)(y^2 - 3) < 0
    by number line method or graph method, we know that the range of y for this inequality is y < -3^(1/2) or 0 < y < 3^(1/2)
    so this imply that one of the root ( or more ) is smaller than -3^(1/2).

    But, this is not the correct answer, we need to show it lies between -3 and -2 , not the negative squate root of 3 !!!!
    Please, any expert, if u know the method, please show me as soon as possible, u help is meaningful to me!
  2. jcsd
  3. Aug 13, 2004 #2

    matt grime

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    what happens when you let y=-3? and when you let y=-2? so at some point in between -3 and -2 the function must have been zero because...?
  4. Aug 13, 2004 #3
    when matt says let y=-3,he wants u to consider the function
    f(y)=y^3 - 3y + 4
    and evaluate f(-3).
    (Just thought i would post this cuz many ppl who have asked me such question don't understand what we are trying to do here, they find the zero crossing argument quite difficult to understand)

    -- AI
  5. Aug 14, 2004 #4


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    Or, more in general, what can you say about f(x) in the immediate neighborhood, but on opposite sides of a root (ie : f(x'+a) and f(x'-a), when x' is a root of f) ?
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