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Math Help - Roots of polynomial equation

  1. #1
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    Roots of polynomial equation

    Find the value of \sum\alpha^2, \sum\alpha^2\beta^2, \sum\alpha\beta(\alpha+\beta), and \sum\frac{1}{\alpha} for the following case:
    x^4-x^3+2x+3=0

    I started for the first one:
    \sum\alpha^2=\alpha^2+\beta^2+\gamma^2+\delta^2
    =(\alpha+\beta+\gamma+\delta)^2-2(\alpha\beta+\beta\gamma+\gamma\delta+\alpha\delt  a+\alpha\gamma+\beta\delta)
    =(\sum\alpha)^2-2(\sum\alpha\beta)-2(\alpha\gamma+\beta\delta)
    This is as far as I got, the last term is what stumps me.
    Thanks!
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  2. #2
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    Quote Originally Posted by arze View Post
    Find the value of \sum\alpha^2, \sum\alpha^2\beta^2, \sum\alpha\beta(\alpha+\beta), and \sum\frac{1}{\alpha} for the following case:
    x^4-x^3+2x+3=0

    I started for the first one:
    \sum\alpha^2=\alpha^2+\beta^2+\gamma^2+\delta^2
    =(\alpha+\beta+\gamma+\delta)^2-2(\alpha\beta+\beta\gamma+\gamma\delta+\alpha\delt  a+\alpha\gamma+\beta\delta)
    =(\sum\alpha)^2-2(\sum\alpha\beta)-2(\alpha\gamma+\beta\delta)
    This is as far as I got, the last term is what stumps me.
    Thanks!
    In problems of this type, the usual convention is that \sum\alpha\beta means the sum of all products of pairs of roots (not just products taken in cyclic order, but all pairs of products). In other words, \sum\alpha\beta = \alpha\beta+\beta\gamma+\gamma\delta+\alpha\delta+  \alpha\gamma+\beta\delta, and \sum\alpha^2 = \bigl(\sum\alpha\bigr)^2 - 2\sum\alpha\beta. With this convention, \sum\alpha\beta is equal to the coefficient of x^2 in the quartic equation, so it is equal to 0.
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  3. #3
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    For the second, I worked out the equivalent for cubic equations
    \sum\alpha^2\beta^2=(\sum\alpha\beta)^2-2\alpha\beta\gamma(\sum\alpha)
    Problem is it doesn't work.
    (0)^2-2(3)(-1)=6 answer is 10.
    Thanks!
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  4. #4
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    Hello, arze!

    The problem is not stated clearly.
    And those Greek letters are too hard to type.

    I'll do a few of these . . .


    \text}The roots of: }\,x^4 - x^3 + 2x + 3 \:=\:0\,\text{ are }\,a,b,c,d.

    \text{Find the value of:}

    . . (A)\;\;\sum\alpha^2

    . . (B)\;\;\sum\alpha^2\beta^2

    . . (C)\;\;\sum\alpha\beta(\alpha+\beta)

    . . (D)\;\;\sum\frac{1}{\alpha}


    From Vieta's Theorem, we have these equations:

    . . \begin{array}{cccccc}a + b + c + d \;=\; 1 & [1] \\<br />
ab + ac + ad + bc + bc + cd \;=\; 0 & [2] \\<br />
abc + abd + acd + bcd \;=\; -2 & [3] \\<br />
abcd \;=\; 3 & [4] \end{array}



    (A)\;\text{ We want: }\:a^2 + b^2 + c^2 + d^2

    Square [1]: . (a+b+c+d)^2 \:=\:1^2

    . . a^2 + 2ab + 2ac + 2ad + b^2 + 2bc + 2bd + c^2 + 2cd + d^2 \;=\;1

    . . a^2 + b^2 + c^2 + d^2 + 2\underbrace{(ab + ac + ad + bc + bd + cd)}_{\text{This is 0}} \;=\;1

    Therefore: . a^2+b^2+c^2+d^2 \;=\;1



    (D)\;\text{ We want: }\:\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} + \dfrac{1}{d}

    . . \displaystyle \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} \;=\;\frac{\overbrace{bcd + acd + abd + abc}^{\text{This is }-2}}{\underbrace{abcd}_{\text{This is 3}}}

    Therefore: . \displaystyle \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} \;=\;-\frac{2}{3}


    Get the idea?
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  5. #5
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    Quote Originally Posted by arze View Post
    For the second, I worked out the equivalent for cubic equations
    \sum\alpha^2\beta^2=(\sum\alpha\beta)^2-2\alpha\beta\gamma(\sum\alpha)
    Problem is it doesn't work.
    (0)^2-2(3)(-1)=6 answer is 10.
    Thanks!
    You haven't got the formula for \sum\alpha^2\beta^2 quite right. It should be \sum\alpha^2\beta^2=(\sum\alpha\beta)^2-2(\sum\alpha\beta\gamma)(\sum\alpha) +2\alpha\beta\gamma\delta = 0^2 -2(-2)(1) +2(3) = 10.
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  6. #6
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    Quote Originally Posted by Opalg View Post
    You haven't got the formula for \sum\alpha^2\beta^2 quite right. It should be \sum\alpha^2\beta^2=(\sum\alpha\beta)^2-2(\sum\alpha\beta\gamma)(\sum\alpha) +2\alpha\beta\gamma\delta = 0^2 -2(-2)(1) +2(3) = 10.
    So that formula would do for all quartic equations? Because I have another one, finding \sum\alpha^2\beta^2 for the equation x^4+x-1=0.
    I should get
    (\sum\alpha\beta)^2-2(\sum\alpha\beta\gamma)(\sum\alpha)+2\alpha\btea\  gamma\delta=(o)^2-2(-1)(0)+2(-1)=-2
    answer is 0.
    Thanks!
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