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Math Help - Find value of expression involving roots of a quadratic

  1. #1
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    Angry Find value of expression involving roots of a quadratic

    The roots of the equation x^2 - 2x + 7=0 are alpha and beta. Without finding the value of alpha and beta explicitly, determine 2B^3 - 3a^2 + 6 (a being alpha and B being beta)

    i post it here as i feel that its a complex number question but i found this question from grade 9 where complex numbers have yet to be introduced (singapore)

    really desperate to solve this question. the answer seem to be a clear cut integer instead of multiple answers in complex form.

    by the way, this has to do with sum and product of roots
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  2. #2
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    Re: Find value of expression involving roots of a quadratic

    I haven't solved the whole problem by hand, but here's the idea: you aply the division algorithm to the polynomials 2x^3 with divisor x^2-2x+7 and the polynomial -3x^2 with the same divisor as the anterior. Next you substitute the B and a in the quotient and it is equal to 0. you will then have only operations on numbers.
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  3. #3
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    Re: Find value of expression involving roots of a quadratic

    Hello, perrification!

    Well, I've made some progress, but . . .


    \text{The roots of the equation }x^2 - 2x + 7\:=\:0\,\text{ are }\alpha\text{ and }\beta.
    \text{Without finding the value of }\alpha\text{ and }\beta\text{ explicitly, determine: }\:2\beta^3 - 3\alpha^2 + 6

    I post it here as i feel that its a complex number question, but i found this question
    from grade 9 where complex numbers have yet to be introduced (Singapore).

    Really desperate to solve this question. the answer seem to be a clear-cut integer
    instead of multiple answers in complex form.

    By the way, this has to do with sum and product of roots.

    Since \alpha and \beta are roots of the first equation: . \begin{Bmatrix}\alpha + \beta &=& 2 & [1]\\ \alpha\!\cdot\!\beta &=& 7 & [2]\end{Bmatrix}

    Square [1]: . (\alpha + \beta)^2 \:=\:2^2 \quad\Rightarrow\quad \alpha^2 +2(\alpha\beta) + \beta^2\:=\:4

    Substitute [2]: . \alpha^2 + 2(7) + \beta^2 \:=\:4 \quad\Rightarrow\quad \alpha^2 + \beta^2 \:=\:\text{-}10

    It seems that we are dealing with complex numbers.


    We have: . \alpha^3 + \beta^3 \;=\;(\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2) \;=\;(2)(\text{-}10 - 7) \;=\;-34


    So we have: . \begin{Bmatrix} \alpha\cdot\beta &=& 7 \\ \alpha + \beta &=& 2 \\ \alpha^2 + \beta^2 &=& \text{-}10 \\ \alpha^3 + \beta^3 &=& \text{-}34 \end{Bmatrix} \;\;\hdots\:\text{ all real numbers!}

    Does this help?


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


    I did it "the long way" and got a surprising answer.


    The roots of the quadratic are: . x \:=\:1 \pm i\sqrt{6}

    Let: . \begin{Bmatrix} \alpha &=& 1 + i\sqrt{6} \\ \beta &=& 1 - i\sqrt{6} \end{Bmatrix}

    Then: . \begin{Bmatrix}\alpha^2 &=& -5 + 2i\sqrt{6} \\ \beta^3 &=& -17 + 3i\sqrt{6} \end{Bmatrix}


    Therefore: . 2\beta^3-3\alpha^2 + 6 \;\;=\;\;2(-17 + 3i\sqrt{6}) - 3(-5 + 2i\sqrt{6}) + 6

    . . . . . . . . . . . =\;\; -34 + 6i\sqrt{6} + 15 - 6y\sqrt{6} + 6 \;\;=\;\;\boxed{-13} \quad ha!

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  4. #4
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    Re: Find value of expression involving roots of a quadratic

    but if we were restricted with not using complex number it is most likely more on the manipulation of the alpha and betas.
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  5. #5
    Member ModusPonens's Avatar
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    Re: Find value of expression involving roots of a quadratic

    Have you understood my answer or do you want me to explain it more explicitly?
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