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    Could soembody answer these two questions for me please so I can learn the method of how they are done.

    1. Given that 2+i is a root of the equation z^3 -11z +20 = 0, find the other roots of the equation.

    I know that one root is 2-i as it is the complex conjugate but don't know how to find the real root.


    2. z= a+3i
    . . . . 2+ai

    Show that there is only one value of a for which arg z = pi/4, and find this value
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    1. Have you learnt that if you have a cubic ax^3 + bx^2 + cx + d=0 then the sum of the roots is -b/a. In this case, -b/a=0 and you already have two of the roots.
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    (Original post by ttoby)
    1. Have you learnt that if you have a cubic ax^3 + bx^2 + cx + d=0 then the sum of the roots is -b/a. In this case, -b/a=0 and you already have two of the roots.
    Nope. Is it possible you could answer the question?

    Thanks
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    Do you know the factor theorem?

    If so, you should be able to express z^3 -11z + 20 in the form of three linear factors (one will be unknown), or one unknown linear factor and one quadratic factor.

    (are sums and products of roots in the edexcel FP1 spec?)
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    Sorry, I just realised that sums and products of roots don't come up in edexcel FP1. Please ignore my last post.
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    (Original post by Macceroot)
    Nope. Is it possible you could answer the question?

    Thanks
    I think you need to read the posting guidelines. We don't like to give out full solutions in here, but we'll be perfectly happy to give you hints. Do you have any thoughts of your own on the problem?
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    (Original post by generalebriety)
    I think you need to read the posting guidelines. We don't like to give out full solutions in here, but we'll be perfectly happy to give you hints. Do you have any thoughts of your own on the problem?
    Well these are the only two questions I am unable to solve in the first chapter so really want to know how before progressing!

    In a cubic, I just don't understand how to find the real root having been given only the complex roots...
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    You know what two of the roots are, so you know that (x-(2+i)) and (x-(2-i)) are factors of the cubic.
    Therefore, (x-(2+i))(x-(2-i)) is a factor of the cubic. Multiply out the brackets and you will end up with a quadratic.
    Divide your cubic by the quadratic and you will be left with a linear term of the form (x-c).
    Then the final root is x=c.
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    (Original post by Macceroot)
    Well these are the only two questions I am unable to solve in the first chapter so really want to know how before progressing!

    In a cubic, I just don't understand how to find the real root having been given only the complex roots...
    private message me with an email address and i will send you some powerpoints to explain the factor theorem and factorising polynomials with complex roots.
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    (Original post by DFranklin)
    You know what two of the roots are, so you know that (x-(2+i)) and (x-(2-i)) are factors of the cubic.
    Therefore, (x-(2+i))(x-(2-i)) is a factor of the cubic. Multiply out the brackets and you will end up with a quadratic.
    Divide your cubic by the quadratic and you will be left with a linear term of the form (x-c).
    Then the final root is x=c.
    Thanks! I fully udnerstand that now

    Last question for me to round off the complex numbers chapter is:

    2. z= a+3i
    . . . . 2+ai

    Show that there is only one value of a for which arg z = pi/4, and find this value

    Any help very much appreciated
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    (Original post by Macceroot)
    Thanks! I fully udnerstand that now

    Last question for me to round off the complex numbers chapter is:

    2. z= a+3i
    . . . . 2+ai

    Show that there is only one value of a for which arg z = pi/4, and find this value

    Any help very much appreciated
    Well, evaluate arg(z) (remember that arg(p/q) = arg(p) - arg(q)) and set it equal to pi/4, then solve for a.
 
 
 
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