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    Post all your P5 and P6 questions here, I know I've got some!!!

    For a kick off:
    (Solomon, edexcel P6 paper C q2b)

    The points A, B and C are (2, 1, -1), (-2, 4, -2) and (a, -5, 1) respectively relative to the origin O, where a isn't 10.
    a. Find ABxAC
    [I got (10-a)j + 3(10-a)k]

    b. The area of triangle ABC is 4.root(10) square units
    Find the possible values of a

    ????????? Help please!
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    Area of a triangle is (1/2)|ABxAC|
    Try using that, and see how you do...
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    See attachment.
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    (Original post by Gaz031)
    See attachment.
    Cheers gaz, i was being retarded and didn't square the area when i squared the determinant. careless ay, blimey.
    thanks again
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    Here's another:

    solve 5z^4 - 11z^3 + 16z^2 -11z + 5 = 0
    using z^n + z^(-n) = 2cos(nθ)

    I've got as far as 5(z^2 + z^-2) - 11(z + z^-1) +16 = 0

    so 10cos2θ - 22cosθ + 16 = 0

    do you just solve that or what?????
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    solve for [email protected]
    Now substitute for [email protected] into z + z^-1 = 2cos @
    create a quadratic in z.
    solve for z.
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    (Original post by Fermat)
    solve for [email protected]
    Now substitute for [email protected] into z + z^-1 = 2cos @
    create a quadratic in z.
    solve for z.
    Makes sense! Cheers guv
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    Solve the quadratic equation in cosθ, noting that cos2θ = 2cos^2θ + 1.
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    (Original post by JamesF)
    Solve the quadratic equation in cosθ, noting that cos2θ = 2cos^2θ + 1.
    Sorry, I forgot to say it says write the answer in the form a + ib.

    so i guess i write z = a + ib and solve the quadratic in that form to get
    z = 1/2 + [root(3)/2]i
    z = 3/5 + (4/5)i

    think that's right, correct me if I'm wrong
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    Sorry, but ...

    cos @ = 1/2 or [email protected] = 3/5.

    Now substitute for [email protected] into z + 1/z = 2cos @
    You should get two values of z ( = a + ib) for each value of [email protected] - four values in all.
    It is a quartic eqn after all, so four roots.

    Edit:
    what you have is correct - just add in the +/- signs.
 
 
 
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