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    (Original post by Zacken)
    A quadratic equations with those roots are:

    x^2 - x(\alpha + 1 + \beta - 1) + (\alpha +1)(\beta - 1)
    Yh got it.

    For 9ii. Why is 1/(2n-3) included surely it cancels out with a term before.
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    (Original post by Super199)
    Yh got it.

    For 9ii. Why is 1/(2n-3) included surely it cancels out with a term before.
    Nopes. What term would it cancel with?
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    (Original post by Zacken)
    Nopes. What term would it cancel with?
    Two terms before it. Since n=4, the 1/5 cancels with the 1/5 from n=2
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    Hey Can Someone Help Me With Matrix Transformations Please? How Do You Know Whether It Is A Reflection Etc?
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    (Original post by Super199)
    Two terms before it. Since n=4, the 1/5 cancels with the 1/5 from n=2
    From a quick glance, looks like you're right. (Can't do the question myself right now) does the markscheme say it's included? Have you looked at the direction of the cancelling? Probably best just to put your working up tbh.
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    (Original post by Zacken)
    From a quick glance, looks like you're right. (Can't do the question myself right now) does the markscheme say it's included? Have you looked at the direction of the cancelling? Probably best just to put your working up tbh.
    Bit messy but this is what I tried to do.
    http://m.imgur.com/0FB2bJb
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    (Original post by Super199)
    Bit messy but this is what I tried to do.
    http://m.imgur.com/0FB2bJb
    Yeah, that's (almost) correct. The \frac{1}{2n-3} does cancel but the \frac{1}{2n-1} doesn't.
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    (Original post by Zacken)
    Yeah, that's (almost) correct. The \frac{1}{2n-3} does cancel but the \frac{1}{2n-1} doesn't.
    ah yes I see it now. How do I do q10ii same paper :/
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    (Original post by Super199)
    ah yes I see it now. How do I do q10ii same paper :/
    It's a disguised quadratic, make the sub u = z^2
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    (Original post by Zacken)
    It's a disguised quadratic, make the sub u = z^2
    So it's literally the same answers from the previous part?

    u^2-4u+9=0

    Completing the square gives: u= 2+- square root of 5i
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    (Original post by Super199)
    So it's literally the same answers from the previous part?

    u^2-4u+9=0

    Completing the square gives: u= 2+- square root of 5i
    Yep.
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    ffs dont worry
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    (Original post by Zacken)
    Yep.
    Yo I think im going mad but can you explain how to work out the centre of the circle from loci's. e.g q6ii
    https://3ed6c47c6c1ead854571a830aaba...0FP1%20OCR.pdf
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    (Original post by Super199)
    Yo I think im going mad but can you explain how to work out the centre of the circle from loci's. e.g q6ii
    https://3ed6c47c6c1ead854571a830aaba...0FP1%20OCR.pdf
    Locus |z-a| = r has radius r and centre a.
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    (Original post by Zacken)
    Locus |z-a| = r has radius r and centre a.
    How do you work the matrix for a shear? q8ii this is?
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    (Original post by Super199)
    How do you work the matrix for a shear? q8ii this is?
    Set up a matrix with unknown components and make matrix * (1, 1) = (1,2) or whatever points they've given and use that to find the matrix components.
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    (Original post by Zacken)
    Set up a matrix with unknown components and make matrix * (1, 1) = (1,2) or whatever points they've given and use that to find the matrix components.
    How though

    let the matrix be
    a b
    c d

    a +b =1
    c+d = 2
    not sure what to do from then.
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    (Original post by Super199)
    How though

    let the matrix be
    a b
    c d

    a +b =1
    c+d = 2
    not sure what to do from then.
    Yeah but you also know any point on the y-axis is invariant

    So (matrix) * (0 1) = (something - draw a diagram and think it through, what does a shear to to units vectors?)
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    (Original post by Zacken)
    Yeah but you also know any point on the y-axis is invariant

    So (matrix) * (0 1) = (something - draw a diagram and think it through, what does a shear to to units vectors?)
    a b
    c d * (0 1) = 0 1 ?
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    (Original post by Super199)
    a b
    c d * (0 1) = 0 1 ?
    I think so, I dunno. Try it.
 
 
 
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