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    Question:

    When  (1-3x)^p is expanded the coefficient of  x^2 = 54

    Given that  p > 0, use this information to find the value of the constant p.

    So far, I worked out (although I'm most likely wrong) that -

    the coefficient in terms of p is equal to -3 \dfrac{p(p-1)}{2!}
    and therefore:

     54 = -3 \dfrac{p(p-1)}{2!}



p^2-p+9 = 0

    As far as I know the quadratic isn't solveable by simple factorising (C2), so I'm presuming I've gone wrong somewhere - does anyone know where?
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    (Original post by Wilko94)
    So far, I worked out (although I'm most likely wrong) that -

    the coefficient in terms of p is equal to -3 \dfrac{p(p-1)}{2!}
    and therefore:
    Close, but don't forget it's "-3x" that you're squaring, which doesn't give -3x^2.


    Hope that wasn't too cryptic.

    Edit: And as Ben121 observed, you've not multplied and divided into the 54 correctly.
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    I'm not entirely sure you should have that 9 there. You have made a mistake when writing the quadratic.
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    (Original post by ghostwalker)
    Close, but don't forget it's "-3x" that you're squaring, which doesn't give -3x^2.


    Hope that wasn't too cryptic.
    Completely forgot about that.

    So would this be correct?

     54 = -3^2 \dfrac{p(p-1)}{2!}





(\div 9)       6= \dfrac{p(p-1)}{2!}





(\times 2)        12= p^2-p





p^2-p-12=0



(p-4)(p+3)



p = 4
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    (Original post by Wilko94)

     p = 4
    Yep.
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    (Original post by ghostwalker)
    Yep.
    Thank you very much to the both of you.

    Ben121 you have been repped, ghostwalker when I try to rep you, I get the message -

    Please rate some other members before rating this member again.
    You've probably helped me already with my Maths as I always rep the person that helps me, and I tend to be asking for help frequently. Thank you very much anyway!
 
 
 
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Updated: April 2, 2011

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