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Thread: 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
    $\displaystyle x^2 + 4x + 7$
    Find an equation with roots $\displaystyle \alpha$ + $\displaystyle 2\beta$ and $\displaystyle \beta$ + $\displaystyle 2\alpha$

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

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

    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 $\displaystyle \alpha$ and $\displaystyle \beta$ are roots of the equation $\displaystyle x^2 + 4x + 7 = 0$. Find an equation with roots $\displaystyle \alpha+2\beta$ and $\displaystyle \beta + 2\alpha$
    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. $\displaystyle x^2 + 4x + 7 = (x -\alpha)(x-\beta)$

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

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

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

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

    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. $\displaystyle x^2 + 4x + 7 = (x -\alpha)(x-\beta)$

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

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

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

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

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

    Bobak
    Thankyou very much .
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