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Math Help - Further Pure help! quadradic equation

  1. #1
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    Further Pure help! quadradic equation

    We are given that:

    x^2+4x+7=0

    Find the equation with roots \alpha+2\beta and 2\alpha+\beta

    The method which you have to do it in is the simplest method where u=?.

    We are NOT allowed to use a brute force method

    Can anyone please help?
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  2. #2
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    Hello, anthmoo!

    We are given that: . x^2+4x+7\:=\:0

    Find the equation with roots \alpha+2\beta and 2\alpha+\beta

    I assume that \alpha and \beta are the roots of the given equation.

    The roots of the quadratic are: . \begin{array}{cc}\alpha \:=\:\text{-}2 + i\sqrt{3} \\ \beta \:=\:\text{-}2 - i\sqrt{3}\end{array}

    Let P\:=\:\alpha + 2\beta\:=\:(\text{-}2 + i\sqrt{3}) + 2(\text{-}2 - i\sqrt{3}) \:=\:\text{-}6 - i\sqrt{3}

    Let Q\:=\:2\alpha + \beta\:=\:2(\text{-}2 + i\sqrt{3}) + (\text{-}2 - i\sqrt{3}) \:=\:\text{-}6 + i\sqrt{3}


    Then: . P + Q\:=\:(\text{-}6-i\sqrt{3}) + (\text{-}6 + i\sqrt{3}) \:=\:\text{-}12
    . . This is the negative of the x-coefficient.

    And: . PQ\:=\:(\text{-}6-i\sqrt{3})(\text{-}6+i\sqrt{3}) \:=\:39
    . . This is the constant term.


    Therefore, the equation is: . \boxed{x^2 + 12x + 39\:=\:0}

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  3. #3
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    Quote Originally Posted by Soroban View Post
    Hello, anthmoo!


    I assume that \alpha and \beta are the roots of the given equation.

    The roots of the quadratic are: . \begin{array}{cc}\alpha \:=\:\text{-}2 + i\sqrt{3} \\ \beta \:=\:\text{-}2 - i\sqrt{3}\end{array}

    Let \text{-}2 + i\sqrt{3}) + 2(\text{-}2 - i\sqrt{3}) \:=\:\text{-}6 - i\sqrt{3}" alt="P\:=\:\alpha + 2\beta\:=\\text{-}2 + i\sqrt{3}) + 2(\text{-}2 - i\sqrt{3}) \:=\:\text{-}6 - i\sqrt{3}" />

    Let Q\:=\:2\alpha + \beta\:=\:2(\text{-}2 + i\sqrt{3}) + (\text{-}2 - i\sqrt{3}) \:=\:\text{-}6 + i\sqrt{3}


    Then: . \text{-}6-i\sqrt{3}) + (\text{-}6 + i\sqrt{3}) \:=\:\text{-}12" alt="P + Q\:=\\text{-}6-i\sqrt{3}) + (\text{-}6 + i\sqrt{3}) \:=\:\text{-}12" />
    . . This is the negative of the x-coefficient.

    And: . \text{-}6-i\sqrt{3})(\text{-}6+i\sqrt{3}) \:=\:39" alt="PQ\:=\\text{-}6-i\sqrt{3})(\text{-}6+i\sqrt{3}) \:=\:39" />
    . . This is the constant term.


    Therefore, the equation is: . \boxed{x^2 + 12x + 39\:=\:0}


    Thankyou Soroban but it says that I am not allowed to know or work out the roots of the equation. I am only supposed to work it out without knowing the roots.




    In another question, it was asked

    The equation x^2+4x+7=0 has roots  \alpha \ and \ \beta . By not using a brute force method, find the equations with roots  3\alpha, 3\beta


    I completed this one successfully by using this method simply (the one i need to use):

     Let \ u=3\alpha
    Therefore \ (1/3)u=\alpha

    This gives me the new equation:

    (\frac{1}{3}u^2)+4(\frac{1}{3}u)+7=0
    \frac{u^2}{9}+ \frac{4u}{3}+7=0
    u^2+12u+63=0

    which is the new equation

    x^2+12x+63=0

    Although you got the right answer Soroban, I was strictly not allowed to use such a method! Can you get the same answer using my method?
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  4. #4
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    By the way, I still need help with the question I stated in the first post! The second doesn't mean I know how to do the question in the first post
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  5. #5
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    Hello again, anthmoo!

    I was afraid that used Brute Force . . . I'll try again.


    We are given that: x^2+4x+7=0

    Find the equation with roots \alpha+2\beta and 2\alpha+\beta

    Since \alpha and \beta are roots of the quadratic,
    . . we know that: . \alpha + \beta \:=\:-4 and \alpha\beta\:=\:7

    The roots of the new equation are: . P \:=\:\alpha + 2\beta and Q \:=\:2\alpha + \beta


    The x-coefficient of the new equation is:
    . . -(P + Q) \:=\:-\left[(\alpha + 2\beta) + (2\alpha + \beta)\right] \:=\:-3(\alpha + \beta)

    Since \alpha + \beta\:=\:-4, then: . -(P+Q)\:=\:-3(-4) \:=\:12
    . . Hence, the x-coefficient is: 12


    The constant term of the new equation is:
    . . PQ\:=\:(\alpha + 2\beta)(2\alpha+\beta)
    . . . . . =\:2\alpha^2 + 5\alpha\beta + 2\beta^2
    . . . . . = \:2\alpha^2 + 4\alpha\beta + 2\beta^2 + \alpha\beta
    . . . . . =\:2(\alpha^2 + 2\alpha\beta + \beta^2) + \alpha\beta
    . . . . . =\:2(\alpha + \beta)^2 + \alpha\beta

    Since \alpha + \beta\:=\:-4 and \alpha\beta\:=\:7, then: . PQ\:=\:2(-4)^2 + 7 \:=\:39
    . . Hence, the constant term is 39.


    Therefore, the equation is: . x^2 + 12x + 39\:=\:0

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  6. #6
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    Ah thankyou VERY much Soroban!

    This is due in tomorrow too! Thanks again man!
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  7. #7
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    Quote Originally Posted by anthmoo View Post
    We are given that:

    x^2+4x+7=0

    Find the equation with roots \alpha+2\beta and 2\alpha+\beta

    The method which you have to do it in is the simplest method where u=?.

    We are NOT allowed to use a brute force method

    Can anyone please help?
    The quadratic with roots \alpha+2\beta and 2\alpha+\beta and leading coefficient 1 is:

    x^2-[(\alpha+2\beta)+(2\alpha+\beta)]x + (\alpha+2\beta)(\alpha+2\beta),

    or after simplifying:

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

    But \alpha and \beta are the roots of:

    x^2+4x+7,

    so \alpha+\beta=-4, and \alpha \beta = 7, so
    the quadratic with the required roots is:


    x^2+3\times 4x + 2\times 16+7,

    or:

    x^2+12x + 39.

    RonL

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  8. #8
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    Thankyou captainblack!

    Hmmm it sounds like CB's way is the way that the book probably wanted it but both ways get the right answer so thanks to both of you.

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