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Maths 1998 Q 6b

  • 05-06-2010 5:30pm
    #1
    Alice10
    Closed Accounts Posts: 70 ✭✭


    Can anyone tell me how to do this question

    1998 Q6b Paper one


Comments

  • #2
    Conor108
    Registered Users, Registered Users 2 Posts: 3,871 ✭✭✭Conor108


    Higher level?


  • #3
    Alice10
    Closed Accounts Posts: 70 ✭✭Alice10


    yeah


  • #4
    Trogdor
    Registered Users, Registered Users 2 Posts: 2,310 ✭✭✭Trogdor


    Well dy/dx=2xk

    input this value of dy/dx into the formula you're given and you'll come out with:
    2(x^2)k+2(x^2)(k^2)+k(x^2)=0

    Which factorises to X^2(3k+2k^2)=0

    So then either x^2=0 (irrelevant to question)

    or 3k+2k^2=0 (from that you get k to be -3/2)

    Hope this helps :)


  • #5
    Alice10
    Closed Accounts Posts: 70 ✭✭Alice10


    yeah it did thanks :)


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