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Math Help - Roots of polynomial equations: Substitution method.

  1. #1
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    Roots of polynomial equations: Substitution method.

    How would you do this one with the substitution method?
    Equation
     x^2 + 4x + 7
    Find an equation with roots \alpha + 2\beta and \beta + 2\alpha

    Example of what i mean:
    x^2 + 5x + 7 = 0 has roots \alpha and \beta
    Find an equation with 2\alpha and 2\beta
    Substitution would be:
    Let u = 2\alpha
    hence \alpha = \frac{u}{2}

    Sub it into equation gives.
    (\frac{u}{2})^2 + 5(\frac{u}{2}) + 7
    When multiplied out gives
    u^2 + 10u + 28

    Uh, i don't mean to sound patronising when i gave an example since i know a lot of you are muuuch better at maths then me . Just wanted to make sure my question was clear, thats all .

    Thankyou very much
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  2. #2
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    Quote Originally Posted by AshleyT View Post
    Given that \alpha and \beta are roots of the equation  x^2 + 4x + 7 = 0. Find an equation with roots \alpha+2\beta and \beta + 2\alpha
    Hello Ashley, A lengthy example is not required you just need to clearly define your question.

    Given the root of your quadratic you can write. x^2 + 4x + 7 = (x -\alpha)(x-\beta)

    Expand and compare coefficients to give.
    \alpha+\beta = -4 \ \ \ (1)
    \alpha \beta = 7 \ \ \ \ \ \ \  \ (2)

    now you require a quadratic such that \alpha+2\beta and \beta + 2\alpha are roots. So your required quadratic can be written as (x -( \alpha+2\beta))(x-(\beta + 2\alpha))

    \Rightarrow x^2 -3(\alpha+\beta)x + \alpha \beta +2 \beta^2 + 4 \alpha \beta + 2 \alpha^2

    \Rightarrow x^2 -3(\alpha+\beta)x + \alpha \beta +2(\beta + \alpha)^2

    Now appropriately substitute (1) and (2) to get your answer.

    Bobak
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  3. #3
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    Quote Originally Posted by bobak View Post
    Hello Ashley, A lengthy example is not required you just need to clearly define your question.

    Given the root of your quadratic you can write. x^2 + 4x + 7 = (x -\alpha)(x-\beta)

    Expand and compare coefficients to give.
    \alpha+\beta = -4 \ \ \ (1)
    \alpha \beta = 7 \ \ \ \ \ \ \  \ (2)

    now you require a quadratic such that \alpha+2\beta and \beta + 2\alpha are roots. So your required quadratic can be written as (x -( \alpha+2\beta))(x-(\beta + 2\alpha))

    \Rightarrow x^2 -3(\alpha+\beta)x + \alpha \beta +2 \beta^2 + 4 \alpha \beta + 2 \alpha^2

    \Rightarrow x^2 -3(\alpha+\beta)x + \alpha \beta +2(\beta + \alpha)^2

    Now appropriately substitute (1) and (2) to get your answer.

    Bobak
    Thankyou very much .
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