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Homework Help: Get in terms of d

  1. Feb 24, 2006 #1
    I dont know why I cant do this, all I need to do is to get the variable d on one side and everything else on the other side:

    constants: the f's and s
    variables: l and d

    [tex]\frac{1}{\frac{1}{\frac{1}{l}-\frac{1}{f_2}}+d}-\frac{1}{d+l-s}=\frac{1}{f_1}[/tex]

    [tex]\frac{f_2-l}{df_2-dl+lf_2}-\frac{1}{d+l-s}=\frac{1}{f_1}[/tex]

    [tex]\frac{d_f2+f_2l-f_2s-f_2s-dl-l^2+ls-df_2+dl-lf_2}{d^2f-d^2l+dlf_2+df_2l-dl^2+f_2l^2-sdf_2+dls-slf_2}=\frac{1}{f_1}[/tex]

    when I try to solve for d, i get d^2(...)+d(....)=constants
    I cant solve this because both sides will have a d in it, is there something I'm missing?
     
    Last edited: Feb 24, 2006
  2. jcsd
  3. Feb 24, 2006 #2

    Hurkyl

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    What kind of equation is d^2(...)+d(....)=constants?

    I bet you know how to solve (...) x^2 + (...) x + (...) = 0.
     
  4. Feb 24, 2006 #3
    yes but the (...) has the variable l in it (l is not a constant), so I cant solve for d. I need to solve for d in terms of l
     
  5. Feb 24, 2006 #4

    Hurkyl

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    So? How does that change anything?
     
  6. Feb 24, 2006 #5

    quasar987

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    What matters is that l is not a function of d.

    What I'm saying is, if x and t are two independant variables, and we have

    [tex]f(t)x^2+g(t)x+h(t)=0[/tex]

    then the solutions for x are a function of t:

    [tex]x_{1,2}(t) = \frac{-g(t) \pm \sqrt{g^2(t) - 4f(t)h(t)}}{2f(t)}[/tex]
     
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