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    Q) How do I go about solving this equation?

    I missed a lecture on it and the lecturer hasn't put up any work up
    on it yet.

    https://gyazo.com/cecacb87537926b92213aca1372d52c3

    Could someone explain the method on how to go about solving the question. :puppyeyes:
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    (Original post by XxKingSniprxX)
    Q) How do I go about solving this equation?

    I missed a lecture on it and the lecturer hasn't put up any work up
    on it yet.

    https://gyazo.com/cecacb87537926b92213aca1372d52c3

    Could someone explain the method on how to go about solving the question. :puppyeyes:
    Can your studies in Further Maths not help you?

    Roots of polynomials???
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    (Original post by Maths is Life)
    Can your studies in Further Maths not help you?

    Roots of polynomials???
    I didn't do further maths.

    I only self-taught FP1 over the summer and can't remember much of it.
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    (Original post by XxKingSniprxX)
    I didn't do further maths.

    I only self-taught FP1 over the summer and can't remember much of it.
    Alpha and beta are the same in each expression.
    I cba with the symbols sorry.

    Sum if the roots is -b/a
    Product of the roots is c/a

    It's the morning - go over FP1 roots of polynomials

    Hint: for the first expression a=2

    For the other equation a is the largest denominator...
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    (Original post by XxKingSniprxX)
    Q) How do I go about solving this equation?

    I missed a lecture on it and the lecturer hasn't put up any work up
    on it yet.

    https://gyazo.com/cecacb87537926b92213aca1372d52c3

    Could someone explain the method on how to go about solving the question. :puppyeyes:
    In further maths you cover that when a quadratic in the form ax^2+bx+c=0 \Rightarrow x^2+\frac{b}{a}x+\frac{c}{a}=0 has roots \alpha and \beta then it must be true that x^2+\frac{b}{a}x+\frac{c}{a}=(x-\alpha)(x-\beta) and from expanding RHS you get x^2+\frac{b}{a}x+\frac{c}{a}=x^2-(\alpha+\beta)x+\alpha \beta

    and now by comparison of coefficients you can see that \frac{b}{a}=-(\alpha+\beta) \Rightarrow -\frac{b}{a}=\alpha+\beta and \frac{c}{a}=\alpha \beta

    From here you can get your values for \alpha+\beta and \alpha \beta

    When it comes to getting a quadratic with roots \alpha+\frac{1}{\beta} and \beta+\frac{1}{\alpha} you know that:

    -q=(\alpha+\frac{1}{\beta})+( \beta + \frac{1}{\alpha})
    r=(\alpha+\frac{1}{\beta})(\beta  +\frac{1}{\alpha})

    and you need to manipulate the RHS on each expression to get it into terms of \alpha+\beta and \alpha \beta

    Once you've done that, just substitute the values you found earlier and you got it.
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    I feel so dumb, compared to all of you.
 
 
 
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