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Math Help - Relations between Coefficents of Polynomials

  1. #1
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    Relations between Coefficents of Polynomials

    Given ax^3 + bx^2 + cx + d = 0

    and roots are e, 2e, 3e

    Find relation between coefficients.

    I approached this firstly with addition of roots

    6e = -b/a
    e = -b/6a

    Then subbed into equation to attain 5b^3 - 36acb + 216a^2d = 0

    However, the solution proposes 11ad = bc, by solving via sum of roots, sum of roots 2 at a time and product of roots then combining together.

    Which solution is correct or are they the same?
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  2. #2
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    Since the roots are e, 2e, 3e, if you were to substitute them into the polynomial, you would get 0.

    So substituting each of them in gives a set of equations...

    ae^3 + be^2 + ce + d = 0
    a(2e)^3 + b(2e)^2 + c(2e) + d = 0
    a(3e)^3 + b(3e)^2 + c(3e) + d = 0


    ae^3 + be^2 + ce = -d
    8ae^3 + 4be^2 + 2ce = - d
    27ae^3 + 9be^2 + 3ce = -d.


    Apply R_2 - 8R_1 \to R_2 and R_3 - 27R_1 \to R_3 to get

    ae^3 + be^2 + ce = -d
    -4be^2 - 6ce = 7d
    -18be^2 - 24ce = 26d


    Apply R_3 - \frac{9}{2}R_2 \to R_3 to give

    ae^3 + be^2 + ce = -d
    -4be^2 - 6ce = 7d
    3ce = -\frac{11}{2}d.


    Solving the third equation for c:

    c = -\frac{11d}{6e}.


    Substituting into the second equation and solving for b:

    -4be^2 + 11d = 7d

    -4be^2 = -4d

    b = \frac{d}{e^2}.


    Substituting into the first equation and solving for a:

    ae^3 + d - \frac{11d}{6} = -d

    ae^3 - \frac{5d}{6} = -d

    ae^3 = -\frac{d}{6}

    a = -\frac{d}{6e^3}.


    Therefore the solution is:

    a = -\frac{d}{6e^3}, b = \frac{d}{e^2}, c = -\frac{11d}{6e}.
    Last edited by Prove It; September 11th 2010 at 02:54 AM.
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    Sorry it was probably my question being too obscure, but it is the relation between coefficents of polynomials, that is the eradication of e, and linking a,b,c,d. Now ive gotten 2 answers as i stated,
    5b^3 - 36acb + 216a^2d = 0 and 11ad = bc, but I am not sure if the 1st identity is valid as they are not the same.
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  4. #4
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    The relationship is through d and e.
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    Mabey you attacking the question in far greater depth than I am. I merely want the relationship between the coefficents, i.e. a,b,c,d. I think what your syaing is that this relationship between d and e will satisfy the two solutions is stated? Is that correct?
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    Quote Originally Posted by Lukybear View Post
    Sorry it was probably my question being too obscure, but it is the relation between coefficents of polynomials, that is the eradication of e, and linking a,b,c,d. Now ive gotten 2 answers as i stated,
    5b^3 - 36acb + 216a^2d = 0 and 11ad = bc, but I am not sure if the 1st identity is valid as they are not the same.
    Do you know Vieta's formulas? We know that If x_{1}, x_{2}, and x_{3} are the roots of the equation a_{3}x^3+a_{2}x^2+a_{1}x+a_{0}, then -\frac{a_{2}}{a_{3}} = x_{1}+x_{2}+x_{3}, \frac{a_{1}}{a_{3}} = x_{1}x_{2}+x_{1}x_{3}+x_{3}x_{2} , and  -\frac{a_{0}}{a_{3}} = x_{1}x_{2}x_{3}. So if e, 2e, 3e are the roots of f(x) = ax^2+bx^2+cx+d, then the Vieta relations are:   e+2e+3e = 6e = -\frac{b}{a}[/Math], call it (1), [Math] (e)(2e)+(e)(3e)+(2e)(3e) = 11e^2 = \frac{c}{a}, call it (2), and [LaTeX ERROR: Convert failed] , call it (3). Put (1) into (2), and we have \frac{36a^2c}{11b^2} = a. Now put (1) into (3) and we get \frac{36a^3d}{b^3} = a. But a = \frac{36a^2c}{11b^2}, so we have \frac{36a^2c}{11b^2} = \frac{36a^3d}{b^3}. That is the relation between the coefficients.
    Last edited by TheCoffeeMachine; September 12th 2010 at 12:50 AM.
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    Yes, which is 11ad = bc simplified, and if e + 2e + 3e = 6e.

    But since 6e = -d/c
    e=-d/6c

    f(-d/6c) would also render a relationship between coefficents will it not? since e satisfies f(e) = 0, by definition. This renders 5b^3 - 36acb + 216a^2d = 0 .

    Just wondering whether the two relations are equivalent.
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  8. #8
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    Hello, Lukybear!

    Given: . ax^3 + bx^2 + cx + d \:= \:0 .with roots: e, 2e, 3e

    Find relation between coefficients.

    You know about the sum-of-roots.
    Do you know the rest of that theorem?


    The equation is: . x^3 + \dfrac{b}{a}x^2 + \dfrac{c}{a}x + \dfrac{d}{a} \:=\:0 .
    [Leading coefficient must be 1.]


    If the roots are \,p,q,r, then :

    . . \begin{array}{ccccc}p + q + r &=&\text{-}\dfrac{b}{a}  & \text{(Roots "taken one at a time")}\\ \\[-3mm]<br /> <br />
pq + qr + pr &=& \dfrac{c}{a} & \text{(Roots "taken two at a time")}\\ \\[-3mm]<br /> <br />
pqr &=& \text{-}\dfrac{d}{a} & \text{(Roots "taken three at a time")} \end{array}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    \begin{array}{cccccccc}\text{1-at-a-time:} & e + 2e + 3e &=& 6e &=& \text{-}\dfrac{b}{a} & [1] \\ \\[-3mm]<br /> <br />
\text{2-at-a-time:} & (e)(2e) + (2e)(3e) + (e)(3e) &=& 11e^2 &=& \dfrac{c}{a} &[2] \\ \\[-3mm]<br /> <br />
\text{3-at-a-time:} & (e)(2e)(3e) &=& 6e^3 &=& \text{-}\dfrac{d}{a} &[3] \end{array}


    . . \begin{array}{ccccccccc}\text{From [1], we have:} & 6e &=& \text{-}\dfrac{b}{a} & \Rightarrow & a &=& \text{-}\dfrac{b}{6e} & [4] \\ \\[-3mm]<br />
\text{From [2], we have:} & 11e^2 &=& \dfrac{c}{a} & \Rightarrow& a & =& \dfrac{c}{11e^2} & [5] \\ \\[-3mm]<br />
\text{From [3], we have:} & 6e^3 &=& \text{-}\dfrac{d}{a} & \Rightarrow & a &=& \text{-}\dfrac{d}{6e^3}& [6]  \end{array}


    Equate [4], [5], [6]: . a \:=\:\text{-}\dfrac{b}{6e} \:=\:\dfrac{c}{11e^2} \:=\:\text{-}\dfrac{d}{6e^3}


    Multiply by 66e^3\!:\;\;66ae^3 \:=\:\text{-}11be^2 \:=\:6ce \:=\:\text{-}11d


    We have: . \begin{Bmatrix}66ae^3 &=& \text{-}11be^2 & \Rightarrow & e &=& \text{-}\dfrac{b}{6a} & [7] \\ \\[-3mm]<br />
6ce &=& \text{-}11d & \Rightarrow & e &=& \text{-}\dfrac{11d}{6c} & [8] \end{Bmatrix}


    Equate [7] and [8]: . \text{-}\dfrac{b}{6a} \:=\:\text{-}\dfrac{11d}{6c} \quad\Rightarrow\quad \boxed{bc \:=\:11ad}



    Edit: TheCoffeeMachine beat me to it . . . *sigh*
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    a(x-e)(x-2e)(x-3e)
    =ax^3-6eax^2+11ae^2x-6ae^3


    a=a, b=-6ea, c=11ae^2, d=-6ae^2
    be^2-d=0  \Rightarrow e^2=\frac{d}{b}
    c=11a(\frac{d}{b})
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  10. #10
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    Quote Originally Posted by Lukybear View Post
    Just wondering whether the two relations are equivalent.
    I don't think they are equivalent. But that's odd -- they have to be. Let's ask Soroban.
    Quote Originally Posted by Soroban View Post
    TheCoffeeMachine beat me to it . . . *sigh*
    Hey, Soroban, don't sigh. I might have been quicker but I can never match your painstaking clarity.

    What do you think: is the relationship given by f(-\frac{b}{6a}) equavalent to the one that you have found?
    Last edited by TheCoffeeMachine; September 20th 2010 at 07:16 PM.
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  11. #11
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    Well obviously not. But it is still valid, as it is a relationship between coefficents.
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