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Math Help - Roots of cubic equations and new equation

  1. #1
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    Roots of cubic equations and new equation

    The equation x^3 + 4x^2 + 3x + 2 = 0 has roots \alpha , \beta, \gamma.

    Find the equation with roots \beta \gamma , \gamma \alpha , \alpha \beta

    For this I so far have:

    \alpha + \beta + \gamma = -4
    \beta \alpha + \gamma \alpha + \alpha \gamma = 3
    \alpha \beta \gamma = -2

    How would I progress from here?
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  2. #2
    Senior Member Dinkydoe's Avatar
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    work out (x-\alpha\gamma)(x-\alpha\beta)(x-\beta\gamma) = x^3-(\beta\gamma+\alpha\gamma+\alpha\beta)x^2 +(\alpha^2\beta\gamma+\alpha\beta^2\gamma+\alpha\b  eta\gamma^2)x - \alpha^2\beta^2\gamma^2 with the relations you have.

    (Then (x-\alpha\gamma)(x-\alpha\beta)(x-\beta\gamma) = 0 is your desired equation)
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  3. #3
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    Quote Originally Posted by db5vry View Post
    The equation x^3 + 4x^2 + 3x + 2 = 0 has roots \alpha , \beta, \gamma.

    Find the equation with roots \beta \gamma , \gamma \alpha , \alpha \beta
    I would've made x^3 + 4x^2 + 3x + 2 = (x-\alpha)(x-\beta)(x-\gamma) and then factorised the LHS to find \beta \gamma , \alpha


    Quote Originally Posted by db5vry View Post

    \alpha + \beta + \gamma = -4
    \beta \alpha + \gamma \alpha + \alpha \gamma = 3
    \alpha \beta \gamma = -2

    How would I progress from here?
    You should be able to solve it by substitution and elimination

    for example

    \beta \alpha + \gamma \alpha + \alpha \gamma = 3

    Is the same as

    \alpha (\beta + 2\gamma )= 3

    \alpha = \frac{3}{\beta + 2\gamma }
    Last edited by pickslides; January 4th 2010 at 12:12 PM. Reason: bad latex
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  4. #4
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    Quote Originally Posted by Dinkydoe View Post
    work out (x-\alpha\gamma)(x-\alpha\beta)(x-\beta\gamma) = x^3-(\beta\gamma+\alpha\gamma+\alpha\beta)x^2 +(\alpha^2\beta\gamma+\alpha\beta^2\gamma+\alpha\b  eta\gamma^2)x - \alpha^2\beta^2\gamma^2 with the relations you have.
    How does this work? I can see why you substitute \beta \gamma + \gamma \alpha + \alpha \beta into the coeffient space but why does it become \alpha^2\beta^2\gamma^2 at the end? I don't understand that (yet).
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  5. #5
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    Quote Originally Posted by db5vry View Post
    but why does it become \alpha^2\beta^2\gamma^2 at the end?
    It is the product of the last term in each factor

    -\alpha\gamma\times -\alpha\beta\times -\beta\gamma = -\alpha^2\beta^2\gamma^2
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  6. #6
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    Right, I found an answer for this question in detail now but I'm now sure how this works either:

    It says consider \beta\gamma + \gamma\alpha + \alpha\beta = 3

    then

    \beta\gamma^2\alpha + \gamma\alpha^2\beta\gamma + \alpha\beta^2\gamma = \alpha\beta\gamma(\alpha + \beta + \gamma)

    = -2 multiplied by -4 = 8

    How does the equation with the squares originate? What do you calculate in order to get to that?
    Last edited by db5vry; January 4th 2010 at 12:42 PM.
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  7. #7
    Senior Member Dinkydoe's Avatar
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    How does this work? I can see why you substitute into the coeffient space but why does it become at the end? I don't understand that (yet).
    There was asked for an equation with roots  \alpha\gamma,\beta\gamma, \alpha\beta .

    (x-\alpha\gamma)(x-\beta\gamma)(x-\alpha\beta) = 0 is ofcourse such an equation.

    If you multiply all the factors you see the result is what I've written out.

    the last coefficient is -\alpha\gamma\cdot - \alpha\beta\cdot - \beta\gamma = -\alpha^2\beta^2\gamma^2
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  8. #8
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    Quote Originally Posted by Dinkydoe View Post
    There was asked for an equation with roots  \alpha\gamma,\beta\gamma, \alpha\beta .

    (x-\alpha\gamma)(x-\beta\gamma)(x-\alpha\beta) = 0 is ofcourse such an equation.

    If you multiply all the factors you see the result is what I've written out.

    the last coefficient is -\alpha\gamma\cdot - \alpha\beta\cdot - \beta\gamma = -\alpha^2\beta^2\gamma^2
    I now understand the last coefficient but I'm not sure about the one that I described in another post. Is there some way you can help me with the other one please
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  9. #9
    Senior Member Dinkydoe's Avatar
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    Have you actually tried to write out the expression (x-\alpha\beta)(x-\beta\gamma)(x-\alpha\gamma) ?

    You can multiply do you? Then collect the coefficients for the powers of x and decide what their values are on the basis of the relations you have.
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  10. #10
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    Quote Originally Posted by Dinkydoe View Post
    Have you actually tried to write out the expression (x-\alpha\beta)(x-\beta\gamma)(x-\alpha\gamma) ?

    You can multiply do you? Then collect the coefficients for the powers of x and decide what their values are on the basis of the relations you have.
    I have multiplied that out but it doesn't make anything easier or make absolutely any sense to me!

    My mark scheme says:

    = -4
    = 3
    = -2

    then consider = 3 [I understand this all up to here]



    = -2 multiplied by -4 = 8 [I do not know how this works - where is the above line coming from?]

    \beta\gamma . \gamma\alpha . \alpha\beta = \alpha^2\beta^2\gamma^2

    = 4 [I understand this]

    then the required equation is

    x^3 - 3x^2 + 8x - 4 = 0 [I can see how it got here considering the previous working]

    All I really want to know is how the mark scheme has come to this! I understand most of it, barring the line which results in "=8".
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  11. #11
    Senior Member Dinkydoe's Avatar
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    Then I wonder how you came to these relations

    = -4
    = 3
    = -2
    The only way you could have found these relations is by considering (x-\alpha)(x-\beta)(x-\gamma) = x^3-(\alpha+\beta+\gamma)x^2+(\alpha\beta+\alpha\gamma  +\beta\gamma)x - \alpha\beta\gamma = x^3 + 4x^2 + 3x + 2

    Then the relations you start out with follow.

    After that you want an equation with roots \alpha\gamma,\beta\gamma,\alpha\beta
    An equation that satisfies this property is: (x-\alpha\gamma)(x-\beta\gamma)(x-\alpha\beta) = 0

    But we like to know explicitly how this polynomial looks like from the relations we derived.

    Thus we write out this factored form: This can be tenacious work but it's not that hard actually. Mathematics is 10 % inspiration, 90% perspiration.



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  12. #12
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    Quote Originally Posted by Dinkydoe View Post
    Then I wonder how you came to these relations

    The only way you could have found these relations is by considering (x-\alpha)(x-\beta)(x-\gamma) = x^3-(\alpha+\beta+\gamma)x^2+(\alpha\beta+\alpha\gamma  +\beta\gamma)x - \alpha\beta\gamma = x^3 + 4x^2 + 3x + 2

    Then the relations you start out with follow.

    After that you want an equation with roots \alpha\gamma,\beta\gamma,\alpha\beta
    An equation that satisfies this property is: (x-\alpha\gamma)(x-\beta\gamma)(x-\alpha\beta) = 0

    But we like to know explicitly how this polynomial looks like from the relations we derived.

    Thus we write out this factored form: This can be tenacious work but it's not that hard actually. Mathematics is 10 % inspiration, 90% perspiration.


    I came to the relations that you quoted by using: -b/a to get -4, c/a to get 3 and -d/a to get -2 which was simple enough.

    I agree with what you said about mathematics, except with me much of the 90% perspiration is lost in confusion.

    Please could you write out a solution? I have tortured myself too long over this question and have tried what you said - expanding those brackets but I can't do that without obtaining x^2 and x in a number of terms and it doesn't add up to what my mark scheme details.

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