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    (Original post by StrangeBanana)
    That 1 should be negative. :oops: Fixed it now.

    You were given the polynomial
    2x^3 + px^2 + qx + r

    And some information: the sum of the roots is 6, the product of the roots is -10, and one of the roots is 4. You had to find p, q and r.
    How did you go about doing it??? I know I did it in the most convoluted way going.
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    (Original post by Aph)
    How did you go about doing it??? I know I did it in the most convoluted way going.
    You could get p and r immediately from the sum and product of the roots, and then I factorised the cubic (because (x - 4) was a factor) and deduced q from that.
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    (Original post by StrangeBanana)
    You could get p and r immediately from the sum and product of the roots, and then I factorised the cubic (because (x - 4) was a factor) and deduced q from that.
    I did the same as that except rather than factorising, i subbed in x=4 and made it equal to zero..

    Also, 7ii is still incorrect as the second part of the inequality should be 0<=x<2 rather than 3<=x<2 !
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    (Original post by jpetersgill)
    I did the same as that except rather than factorising, i subbed in x=4 and made it equal to zero..

    Also, 7ii is still incorrect as the second part of the inequality should be 0<=x<2 rather than 3<=x<2 !
    Sounds good.

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    (Original post by StrangeBanana)
    You could get p and r immediately from the sum and product of the roots, and then I factorised the cubic (because (x - 4) was a factor) and deduced q from that.
    I got those 2 yes but what I then did

    \sum\alpha+\beta+\gamma= 6, \alpha\beta\gamma= -10

\Rightarrow \beta=2-\gamma

\Rightarrow 4\gamma(2-\gamma)= -10 and went from there... Do you think that's alright?
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    (Original post by Aph)
    I got those 2 yes but what I then did

    \sum\alpha+\beta+\gamma= 6, \alpha\beta\gamma= -10

=&gt; \beta=2-\gamma

=&gt; 4\gamma(2-\gamma)= -10 and went from there... Do you think that's alright?
    Well I'm not sure about those upside down question marks.

    So that gives you the third root, then what?
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    Anyone know how many marks each question was worth?


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    (Original post by StrangeBanana)
    Well I'm not sure about those upside down question marks.

    So that gives you the third root, then what?
    Well it gives you the 2 roots then you add the 4 and multiply out. You have to put something in about beta and gamma being bi-variable and if one takes one value the other takes its complex congigate.
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    (Original post by Aph)
    Well it gives you the 2 roots then you add the 4 and multiply out. You have to put something in about beta and gamma being bi-variable and if one takes one value the other takes its complex congigate.
    Ah, fair enough. It's a valid method, so as long as you got the right answer I'm sure it'll be accepted!
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    (Original post by StrangeBanana)
    Ah, fair enough. It's a valid method, so as long as you got the right answer I'm sure it'll be accepted!
    Good, that was scary.the answer is 3 right?
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    (Original post by Aph)
    Good, that was scary.the answer is 3 right?
    Nope, p = -12, q = 11, r = 20. Bad luck.
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    (Original post by StrangeBanana)
    Nope, p = -12, q = 11, r = 20. Bad luck.
    Oh... Well I got the other 2 correct
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    (Original post by henrygriff28)
    I did the exact same as you, do you think we'll still get the one mark? I think that would be all I dropped and I was hoping for 100 UMS so will be close.
    Yeah can't lose 2 marks for getting it entirely right! Bullsh*t question, I didn't think I'd have to repeat the obvious lol
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    How many marks would one lose for not shading correctly in the Argand diagram? i.e didn't shade all the way down to minus one


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    (Original post by StrangeBanana)
    Nope, p = -12, q = 11, r = 20. Bad luck.
    NUUUUUUU I got r=123 T.T
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    (Original post by acelle)
    NUUUUUUU I got r=123 T.T
    There were 2 r's in the paper.
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    (Original post by Aph)
    There were 2 r's in the paper.
    O. Thanks that calmed my heart ..

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    (Original post by acelle)
    NUUUUUUU I got r=123 T.T
    :P yeah the other r was 123, don't worry
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    (Original post by StrangeBanana)
    Unofficial Markscheme

    1. x = \dfrac{5}{18}, y = \dfrac{1}{27}


    2. z_1 = 2 + 3j, z_2 = 2 - 3j

    |z_1| = |z_2| = \sqrt{13}

    arg(z_1) = 0.983, arg(z_2) = -0.983


    3. p = -12, q = 11, r = 20


    4.


    5i. Use standard summations.

    ii. k = \dfrac{1}{3}


    6. Run-of-the-mill proof by induction.


    7i. Sketch:


    ii. Values of x for which y \geq 3 are:

    x &lt; -1, 0 \leq x &lt; 2


    8i. \alpha^2 = 9 + 40j, \alpha^3 = -115 + 236j

    ii. q = -7, r = 123

    iii. z = 5 + 4j, 5 - 4j, or -3

    iv. z = 5 + 4j, 5 - 4j, -3 or 1


    9i. A'(0, 0), B'(-4, 0), C'(2, 12)

    ii. Horizontal enlargement with s.f. 4, vertical enlargement with s.f. 2

    iii. The matrix required was:
    \dfrac{1}{48}\begin{pmatrix}

0 & 8 \\

-6 & 12

\end{pmatrix}

    iv. 192 sq. units
    Please marry me

    BUT will i lose some marks for the argand diagram question because for the part where you had do draw the argument boundries i used dotted lines instead of solid?
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    (Original post by Darcy1)
    Please marry me

    BUT will i lose some marks for the argand diagram question because for the part where you had do draw the argument boundries i used dotted lines instead of solid?
    probs 1 or 2 marks because it shows that the line wasn't inclusive of that inequality, and it clearly was. So it sort of shows you didn't understand the condition fully.
 
 
 
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