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Homework Help: Quadratic with alpha and beta

  1. May 22, 2010 #1
    1. The problem statement, all variables and given/known data

    The quadratic equation [tex]x^2 + kx + 2k = 0[/tex] where k is a non-zero constant, has roots [tex]\alpha[/tex] and [tex]\beta[/tex].

    Find a quadratic equation with roots [tex]\frac{\alpha}{\beta}[/tex] and [tex]\frac{\beta}{\alpha}[/tex]. {one is meant to be inverted - the code isn't working properly :( }

    2. Relevant equations

    3. The attempt at a solution

    A question like this, I would normal attempt by using 'u' as a variable and get [tex]\alpha[/tex] on it's own, e.g:
    [tex]\alpha + 1 = u[/tex]
    [tex]\alpha = u - 1[/tex]

    I would then subst. the u - 1 part into the equation and get the answer. I'm unsure where to go as alpha and beta are in a fraction together - help please :)
    Last edited: May 22, 2010
  2. jcsd
  3. May 22, 2010 #2


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    Just a question -- are those both intended to be the same or is one of them supposed to be inverted?
  4. May 22, 2010 #3
    Ahh thank you - meant to be inverted, I'll edit it now :)
  5. May 22, 2010 #4


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    I would have started by writing down the equation of the quadratic with those roots, and only then worry about trying to rewrite everything in terms of k.

    What relationships do you know exist between alpha, beta, and k?
  6. May 22, 2010 #5
    Do you mean;

    [tex](x - \alpha)(x - \beta)[/tex]

    [tex]x2 - \alpha x - \beta x + \alpha\bet = 0[/tex]

    Therefore [tex]\alpha\beta = 2k...?[/tex] :S
  7. May 22, 2010 #6


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    Don't use the sup in combination with tex tags. I think the equation you are trying to write is

    [tex]x^2 -(\alpha + \beta)x + \alpha\beta = x^2 + kx + 2k[/tex]

    There is currently a bug in the "preview" feature for tex; the workaround is to hit the refresh F5 button on your browser after trying to preview.

    Anyway, yes, that gives you a couple of equations to work with for α and β.

    Note to Hurkyl: I have tried to get someone's attention about the preview bug in the HH forum and with a message to Greg. Do you know if anything is being looked at?
  8. May 22, 2010 #7
    Yes it is - okay thank you, I shall give it ago :)
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