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Double root

  1. Jan 28, 2012 #1
    r,q are constants. I need to factor this equation such that there is a double root.

    [tex]-\frac{r}{q}u^3+ru^2-\left(\frac{r}{q}+1\right)u+r=0[/tex]

    Are there any tricks for this because this just a nasty equation.

    I don't know if that is a wise approach but:

    [tex](au+b)(cu+d)^2 = ac^2u^3+(2acd+c^2b)u^2+(ad^2+2bcd)u+bd^2[/tex]

    Then

    [tex]ac^2 = -\frac{r}{q}[/tex]

    [tex]2acd+c^2b=bd^2=r[/tex]

    [tex]ad^2+2bcd = \frac{r}{q}+1[/tex]
     
    Last edited: Jan 28, 2012
  2. jcsd
  3. Jan 28, 2012 #2

    Curious3141

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    Hmm...I haven't fully explored this approach, but it might simplify your work. Remember that if a polynomial [itex]p(x)[/itex] has a repeated (double) root [itex]\alpha[/itex], then [itex]\alpha[/itex] is also a root of the derivative [itex]p'(x)[/itex]. You reduce the problem to dealing with a quadratic.
     
  4. Jan 28, 2012 #3

    SammyS

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    Curious_π has a very good idea. Before I read his post, I played around with this for a while.

    What I came up with is the following:
    Let 1/v = u. Substitute that for u, then multiply by v3/r. That gives:
    [itex]\displaystyle v^3-\left(\frac{1}{q}+\frac{1}{r}\right)v^2+v-\frac{1}{q}=0 [/itex]​
    Then notice that a cubic function with leading coefficient of 1, and a repeated root can be written as:
    [itex](v-a)^3(v-b)\quad \to\quad v^3-(2a+b)v^2+(a^2+2ab)v-a^2b[/itex]​
    That's as far as I have taken it. You can try equating coefficients, and/or combining this with Curious3141's suggestion.


    .
     
  5. Jan 28, 2012 #4
    Just to verify that I even took the right approach.

    The question wanted we to show using conditions for a double root that the curve in r-q space is given parametrically by

    [tex]r=\frac{2a^3}{(1+a^2)^2}, \ q=\frac{2a^3}{a^2-1}[/tex]

    and the given equations were

    [tex]r\left(1-\frac{u}{q}\right), \ \frac{u}{1+u^2}[/tex]

    I set them equal to each other and then re-arranged the equation to set it equal to zero.

    That was the right idea though, right?

    Which is where I obtained:

    [tex]-\frac{r}{q}u^3+ru^2-\left(\frac{r}{q}+1\right)u+r=0[/tex]
     
  6. Jan 28, 2012 #5

    SammyS

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    What do you mean by:
    "... and the given equations were

    [itex]\displaystyle r\left(1-\frac{u}{q}\right), \ \frac{u}{1+u^2}[/itex]​
    Those are not equations. No equal signs.
     
    Last edited: Jan 28, 2012
  7. Jan 28, 2012 #6
    Just say the first is U = and the second is V =.
     
  8. Jun 16, 2012 #7
    It looks like you are trying to solve exercise 1 from the book "Mathematical Biology" by Murray. I am also stuck on that same problem, essential they are trying to make the reader derive the parametric equations from:

    r(1-u/q)=u/(1+u^2)

    to get the two parametric equations you gave. i.e. r = 2a^3/(1+a^2)^2 etc

    Does anyone else know how to solve this?
     
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