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    If we had an equation to find, and we are given one complex root, say  z= 1+5i would the following be acceptable for finding the equation/quadratic factor

     z = 1+5i

     (z-1)^2 =-25

     z^2-2z +26 =0

    \therefore equation/quadratic factor is  z^2-2z+26

    It seems like the least laborious way to me.
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    (Original post by NotNotBatman)
    If we had an equation to find, and we are given one complex root, say  z= 1+5i would the following be acceptable for finding the equation/quadratic factor

     z = 1+5i

     (z-1)^2 =-25

     z^2-2z +26 =0

    \therefore equation/quadratic factor is  z^2-2z+26

    It seems like the least laborious way to me.
    Yeah. Although I'd think you'd want to start off from z = 1 \pm 5i so the implication follows.
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    (Original post by Zacken)
    Yeah. Although I'd think you'd want to start off from z = 1 \pm 5i so the implication follows.
    ah, good idea. Thanks.
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    (Original post by alfmeister)
    Attachment 531011

    I drew a diagram and found the displacement of point B from A. Then I knew that point C will be displayed the same distance from the origin as it is the opposite side of BA so I was able to deduce the coordinate of C.
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    Thank you how would we work out the area, would we divide it into 2 triangles?
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    Does anyone know how we would do inductions with inequalities like in this question? Not seen one like this before! Thanks Name:  image.jpg
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    Can anyone point me in the direction of any 2016 papers + mark schemes?
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    (Original post by economicss)
    Does anyone know how we would do inductions with inequalities like in this question? Not seen one like this before! Thanks Name:  image.jpg
Views: 183
Size:  447.9 KB
    which paper and qn? i can't view the pic (viewing on my phone)
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    (Original post by economicss)
    Does anyone know how we would do inductions with inequalities like in this question? Not seen one like this before! Thanks Name:  image.jpg
Views: 183
Size:  447.9 KB
    1. Show true for n=1.

    2. Assume true for n=k: \sum_{r=1}^k r > \frac{1}{2}k^2

    3. \displaystyle 

\begin{align*}\sum_{r=1}^{k+1} r &= \sum_{r=1}^{k} r + (k+1) \\ & > \frac{1}{2}k^2 + (k+1) \\ &= \frac{1}{2}(k^2 + 2k + 2) \\ &= \frac{1}{2}(k+1)^2 + \frac{1}{2} \\ & > \frac{1}{2}(k+1)^2 \end{align*}

    It's much the same as normal induction and it won't up in a normal edexcel paper.
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    I'm doing FP1 IAL Edexcel,
    Does anyone have notes for roots of quadratic equations?
    Or topic questions that i could use to practice?
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    (Original post by Ayman!)
    Good luck - it's quite a nice module to get 100 in, which I unfortunately missed out on. Edexcel are loving their conic sections in FP1 from what it seems, so try getting really good at those!
    you got any good conics qns w/ answers?
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    (Original post by Zacken)
    1. Show true for n=1.

    2. Assume true for n=k: \sum_{r=1}^k r > \frac{1}{2}k^2

    3. \displaystyle 

\begin{align*}\sum_{r=1}^{k+1} r &= \sum_{r=1}^{k} r + (k+1) \\ & > \frac{1}{2}k^2 + (k+1) \\ &= \frac{1}{2}(k^2 + 2k + 2) \\ &= \frac{1}{2}(k+1)^2 + \frac{1}{2} \\ & > \frac{1}{2}(k+1)^2 \end{align*}

    It's much the same as normal induction and it won't up in a normal edexcel paper.
    Great, thanks
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    (Original post by tazza ma razza)
    you got any good conics qns w/ answers?
    If you haven't done the review exercises try out the links iMackJack posted previously in this thread. I think they'd be good practice (I haven't done them myself)
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    Links to conics qns - they are killing me atm
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    does anyone have the 2016 ial jan paper link?
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    (Original post by tazza ma razza)
    does anyone have the 2016 ial jan paper link?
    This was all I could find, it has worked solutions on it though https://0025309b76bc88a0e4c3444acd03...%20kprime2.pdf
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    (Original post by economicss)
    This was all I could find, it has worked solutions on it though https://0025309b76bc88a0e4c3444acd03...%20kprime2.pdf
    many thanks ma G
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    How would you do part d?
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    (Original post by Reeyap)
    How would you do part d?
    If you know ON is perpendicular to PQ, how can you connect the gradient of ON to the gradient of PQ? Use this and you should find that you're able to isolate a p^2q^2 term
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    (Original post by kingaaran)
    If you know ON is perpendicular to PQ, how can you connect the gradient of ON to the gradient of PQ? Use this and you should find that you're able to isolate a p^2q^2 term
    Okay so (gradient of ON) x (gradient of PQ) = -1 ?
    How would you simplify the terms though? :ashamed2:
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    (Original post by Reeyap)
    Okay so (gradient of ON) x (gradient of PQ) = -1 ?
    How would you simplify the terms though? :ashamed2:
    What's your gradient of ON and what is your gradient for PQ first of all?
 
 
 
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