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Delta with discriminate

  1. Aug 18, 2011 #1
    1. The problem statement, all variables and given/known data
    With the discriminate, why is delta sometimes used?


    2. Relevant equations
    [itex]\Delta[/itex] = b2 - 4ac


    3. The attempt at a solution
    I get the logic behind what the discriminate is and how and why it works, but I don't understand why delta is used in the equation. What change is occuring?
     
  2. jcsd
  3. Aug 18, 2011 #2

    Ray Vickson

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    Nothing is changing. They need to have some symbol for the discriminant, and they have decided to use Delta (Greek D for Discriminant). I have never seen that notation, but it does make some sense.

    RGV
     
  4. Aug 18, 2011 #3

    NascentOxygen

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    I'd forgotten that delta is sometimes used, but thanks for the reminder. :smile:

    It's probably as good a choice as any, because the discriminant relates directly to the step either side of the peak in the parabola where it crosses the axis.

    i.e., x = -b +/- sqrt(b2 -4ac) ....etc

    So, the offset up (and down) about -b/(2a) is determined by delta. To be exact, delta/(2a)

    If the coefficient of x is unity, then delta actually is the distance between the roots. If delta = 0 then the roots coincide; there is no distance between them.
     
    Last edited: Aug 19, 2011
  5. Aug 19, 2011 #4

    PeterO

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    No change - it is just shorter - sort of like a name for the dicriminant.
     
  6. Aug 19, 2011 #5

    Mark44

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    I believe that delta is used only because the word discriminant starts with "d", the same sound as the letter delta represents.
     
  7. Aug 20, 2011 #6
    Hey, could you elaborate on the "offset up and down" portion? I've never seen that terminology used with parabola's before.
     
  8. Aug 21, 2011 #7

    NascentOxygen

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    Well, this gives me the opportunity to make a correction to what I wrote. (Sharp eyes would have noted that I omitted the essential string sqrt in one or two places.)
    Harking back to your first encounter with graphing the parabola, you found that the parabola's minimum (or maximum) occurs where x=-b/(2a)
    and the parabola crosses the x-axis at two points offset from this by an amount +/- sqrt(b2 - 4ac)/(2a)

    So you can see this offset is directly related to delta. (To the square root of delta, to be more precise.)
     
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