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Math Help - Polynomial help again ..

  1. #1
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    Polynomial help again ..

    The roots of the quadratic equation x^2+px+q=0 , where q is not 0 are \alpha and \beta and the
    roots of the quadratic equation x^2+p'x+q=0 is k\alpha and \frac{\beta}{k}. For any p ,p' and q , show that the sum of the four possible values of k is \frac{pp'}{q}.
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    Quote Originally Posted by mathaddict View Post

    The roots of the quadratic equation x^2+px+q=0 , where q is not 0 are \alpha and \beta and the
    roots of the quadratic equation x^2+p'x+q=0 is k\alpha and \frac{\beta}{k}. For any p ,p' and q , show that the sum of the four possible values of k is \frac{pp'}{q}.
    so you have: \begin{cases} \alpha + \beta=-p \\ \alpha \beta = q \\ k\alpha + \frac{\beta}{k}=-p' \end{cases}. fro the last equation you get \alpha k^2 + \beta=-p'k, which with the first equation give us: \alpha=\frac{p-p'k}{k^2-1}, \ \beta=\frac{(p'-pk)k}{k^2 - 1}. [here i'm assuming that k \neq \pm 1.]

    putting what we got for \alpha, \beta in the second equation, i.e. \alpha \beta = q, gives us: (p-p'k)(p'-pk)k=q(k^2-1)^2. thus: qk^4 -pp'k^3 + (p^2 + p'^2 - 2q)k^2 - pp'k + q=0 and the result follows.
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    Re :

    Quote Originally Posted by NonCommAlg View Post
    so you have: \begin{cases} \alpha + \beta=-p \\ \alpha \beta = q \\ k\alpha + \frac{\beta}{k}=-p' \end{cases}. fro the last equation you get \alpha k^2 + \beta=-p'k, which with the first equation give us: \alpha=\frac{p-p'k}{k^2-1}, \ \beta=\frac{(p'-pk)k}{k^2 - 1}. [here i'm assuming that k \neq \pm 1.]

    putting what we got for \alpha, \beta in the second equation, i.e. \alpha \beta = q, gives us: (p-p'k)(p'-pk)k=q(k^2-1)^2. thus: qk^4 -pp'k^3 + (p^2 + p'^2 - 2q)k^2 - pp'k + q=0 and the result follows.

    Thanks NonCommAlg , I understood what you did , but i don seem to get the result . The sum of all possible values of k ..
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    Quote Originally Posted by mathaddict View Post
    Thanks NonCommAlg , I understood what you did , but i don seem to get the result . The sum of all possible values of k ..
    Hint: The four possible values are the roots of that quartic equation....
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    RE :

    This is what i did so far , trying to get the roots of this equation .

    qk^2-pp'k+(p^2+p'^2-2q)-\frac{pp'}{k}+\frac{q}{k^2}=0

    Let w=k+\frac{1}{k^2} and substituting into the equation , i got

    q(w-2)-pp'w+p^2+p'^2-2q=0

    Till here , i have no idea how to get the roots .
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  6. #6
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    You dont need to find the roots. All you need is the sum of the roots. Looking at the polynomial equation, cant you tell the sum of the roots in terms of its coefficients?
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    Re :

    Oh alright , my mistake . By the way , i understand for quadratic equation

    ax^2+bx+c=0

    The sum of roots is -b/a

    But for quartic equation , i am not sure which is the sum of roots .
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    Quote Originally Posted by mathaddict View Post
    Oh alright , my mistake . By the way , i understand for quadratic equation

    ax^2+bx+c=0

    The sum of roots is -b/a

    But for quartic equation , i am not sure which is the sum of roots .
    Oh... sorry then

    For a quartic ax^4 + bx^3 + cx^2 + dx + e = 0, the sum of the roots is -\frac{b}{a}. Use this idea and you will immediately get the answer.
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    Re :

    Thanks a lot iso , for clearing all my doubts ..
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  10. #10
    Lord of certain Rings
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    I think it is a good thing to know Vieta's formulae to solve problems similar to the ones you have posed. You can read it from here.
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    Display problem

    Quote Originally Posted by Isomorphism View Post
    I think it is a good thing to know Vieta's formulae to solve problems similar to the ones you have posed. You can read it from here.
    I couldn't get this page to display correctly in my browser (Firefox). Any suggestions?

    Grandad

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  12. #12
    Lord of certain Rings
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    Quote Originally Posted by Grandad View Post
    I couldn't get this page to display correctly in my browser (Firefox). Any suggestions?

    Grandad

    I use Firefox version 3.0.5 and it loads perfectly for me. Are you having problems with the TeX symbols?
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  13. #13
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    Display problem

    Quote Originally Posted by Isomorphism View Post
    I use Firefox version 3.0.5 and it loads perfectly for me. Are you having problems with the TeX symbols?
    Just tried again - works fine now! (I also have Firefox 3.0.5.)

    Grandad

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