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    (Original post by TWP1992)
    but you can find a quite easily, did it ask you to solve in terms of a?
    I must be being extremely thick here - enlighten me?
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    (Original post by Mr M)
    I may regret asking this but where did you get 3.2 from?

    You will clearly lose 3 accuracy marks (1 in the first part and 2 in the second part). It is possible you could all get the method marks but it depends what you did.
    I can't remember exactly. I wasn't sure what to do, so I multiplied the roots out (x-imag number times x-conjugate times x-a) and compared the x coefficient. I ended up with ai-ai+5a+5a-26=6. So 10a=32. a=3.2

    Second part I just compared the x^2 coefficient and the number.
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    (Original post by destined-monkey)
    I can't remember exactly. I wasn't sure what to do, so I multiplied the roots out (x-imag number times x-conjugate times x-a) and compared the x coefficient. I ended up with ai-ai+5a+5a-26=6. So 10a=32. a=3.2

    Second part I just compared the x^2 coefficient and the number.
    That method sounds OK but surely you knew 3.2 couldn't be correct?
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    (Original post by Mr M)
    That method sounds OK but surely you knew 3.2 couldn't be correct?
    Why? I put it all back in and it seemed ok. I was pretty sure I got it wrong, yeah, as those questions usually gave nice numbers, but I couldn't see where I went wrong.
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    How can you find the real root of the cubic equation without knowing p and q? I think it was question 6. I managed to figure out p and q by putting (5+i) and (5-i) back in to the cubic then using simultaneous equations, from there I could work out the real root, but I don't understand how you could get it without knowing the full equation?

    I hope that made sense!
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    (Original post by chrishalliday7)
    How can you find the real root of the cubic equation without knowing p and q? I think it was question 6. I managed to figure out p and q by putting (5+i) and (5-i) back in to the cubic then using simultaneous equations, from there I could work out the real root, but I don't understand how you could get it without knowing the full equation?

    I hope that made sense!
    (x - 5 + i)(x - 5 - i) = x^2 - 10x + 26

    (x - a) (x^2 - 10x + 26) = x^3 + px^2 + 6x + q
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    (Original post by Mr M)
    (x - 5 + i)(x - 5 - i) = x^2 - 10x + 26

    (x - a) (x^2 - 10x + 26) = x^3 + px^2 + 6x + q
    That makes sense! Would my method of doing the question the wrong way round gain many marks?
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    (Original post by Mr M)
    (x - 5 + i)(x - 5 - i) = x^2 - 10x + 26

    (x - a) (x^2 - 10x + 26) = x^3 + px^2 + 6x + q
    Ahhhh! Now you've put that up, i've just realised i had 10a-26=6 instead of 10a+26=6. That would explain it! Stupid!!!
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    (Original post by chrishalliday7)
    That makes sense! Would my method of doing the question the wrong way round gain many marks?
    I'm not totally sure what you did. Did you get p and q correct by using the factor theorem?
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    (Original post by Mr M)
    I'm not totally sure what you did. Did you get p and q correct by using the factor theorem?
    I put x=5-i and x=5+i back into the original cubic (thus eliminating x), which left me with 2 equations with 2 unknows (p and q). I did some tidying up and found p=-8 and q=52.

    In essence I answered part ii) first (albeit using a strange method) to leave me with x^3 - 8x^2 + 6x + 52 = 0.

    I then cheated slightly by putting the equation into my calculator, it told me x = -2.

    So obviously no method marks for part 1) but is my part ii) method acceptable?
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    (Original post by chrishalliday7)
    I put x=5-i and x=5+i back into the original cubic (thus eliminating x), which left me with 2 equations with 2 unknows (p and q). I did some tidying up and found p=-8 and q=52.

    In essence I answered part ii) first (albeit using a strange method) to leave me with x^3 - 8x^2 + 6x + 52 = 0.

    I then cheated slightly by putting the equation into my calculator, it told me x = -2.

    So obviously no method marks for part 1) but is my part ii) method acceptable?
    I think you will get full marks on both bits. You answered the question.
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    Also, for the last part of question 10, did we have to write down the image of (1,0)?
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    (Original post by chrishalliday7)
    Also, for the last part of question 10, did we have to write down the image of (1,0)?
    You should say where the image of one point is. It didn't have to be that point.
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    Does anyone have a copy of the paper they can upload?
    I came out thinking I'd done OK, now I'm not so sure!
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    I'll upload mine today.
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    Yeah I wouldn't mind checking out the paper again.
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    Is anyone else sick of matrices after this exam? They were in almost half of the questions
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    (Original post by mcp2)
    I'll upload mine today.
    Has anyone got a copy of this OCR Jan 2010 FP1 paper. Would really like to see a copy now.
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    guys im taking fp1 in a few days i understand all of this paper except question 3
    the answer is z = 2 + 5i but i dont have a clue on the method to do this
    help anyone?
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    btw the question itself was
    find the complex number z which satisfies
    z + 2iz* = 12 + 9i
 
 
 
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