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    Given that 2 + i is a root of the equation
    z^2 + bz + c = 0, where b and c are real constants,
    (i) write down the other root of the equation,
    (ii) find the value of b and the value of c.


    For part two, after quite a long simulatenous equation i get the right answer. When I look at the Mark Scheme they have done something which makes my method look pathetic, but I do not understand it:

     (2 - i)^2 + b(2 + i) + c = 0
    Imaginary parts b = – 4
    Real parts c + 3 + 2b = 0
    c = 5

    How can they sub both roots into the equation like that? That's all I want to know. I know how to compare coefficients after.

    Thanks

    EDIT: I did it this way:

     (2 - i)^2 + b(2 - i) + c = 0
     (2+ i)^2 + b(2 + i) + c = 0

    And then solved
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    Roots of a polynomial with real coefficients occur in complex conjugate pairs. Standard FP1 result. I can explain if you wish.
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    (Original post by 2710)

     (2 - i)^2 + b(2 + i) + c = 0
    Imaginary parts b = – 4
    Real parts c + 3 + 2b = 0
    c = 5

    How can they sub both roots into the equation like that? That's all I want to know. I know how to compare coefficients after.
    It's at typo!

    If you work through the equation as given in the quote you get b=4, not -4
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    Yes I know that. But I don't understand how you can sub both roots into the equation. Its like saying

     x^2 +ax +b =0 , roots = 5 and 8

    You can't just sub 5 and 8 into the formula and get the values of a and b:

     5^2 + 8a +b =0

    Can you?
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    (Original post by ghostwalker)
    It's at typo!

    If you work through the equation as given in the quote you get b=4, not -4
    Thank you
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    (Original post by 2710)
    Yes I know that. But I don't understand how you can sub both roots into the equation. Its like saying

     x^2 +ax +b =0 , roots = 5 and 8

    You can't just sub 5 and 8 into the formula and get the values of a and b:

     5^2 + 8a +b =0

    Can you?

    Did you get my point? Because b and c are real, any complex roots must be in conjugate pairs, so if 2+i is one root, 2-i must be the other.
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    (Original post by ian.slater)
    Did you get my point? Because b and c are real, any complex roots must be in conjugate pairs, so if 2+i is one root, 2-i must be the other.
    You do not understand me. I know what complex conjugates pairs are. I was just using that as an example, and it was perfectly fine, since in this question, you cannot sub both pairs into the equation to get the answer, since, as ghostwalker has pointed out, it is wrong. My question was not about complex conjugate pairs, but whether you could do what the Mark scheme did, and the answer is no.

    EDIT: I think you may be looking at part i of the question. I am actually doing ii, as I have already stated
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    (Original post by ian.slater)
    Did you get my point? Because b and c are real, any complex roots must be in conjugate pairs, so if 2+i is one root, 2-i must be the other.
    Do you get his point?
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    That's weird, putting both pairs in like that.
    If you can do that, it's new to me.
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    Yeah it must be a typo.
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    (Original post by 2710)
    You do not understand me. I know what complex conjugates pairs are. I was just using that as an example, and it was perfectly fine, since in this question, you cannot sub both pairs into the equation to get the answer, since, as ghostwalker has pointed out, it is wrong. My question was not about complex conjugate pairs, but whether you could do what the Mark scheme did, and the answer is no.

    EDIT: I think you may be looking at part i of the question. I am actually doing ii, as I have already stated
    Ah - sorry - I thought you meant 'how did they know the other root?'

    On working out b and c I'd use b = - sum of roots = -4 and c = product of roots = 5
 
 
 
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