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    Where a question asks you to find the quadratic equation that has a+bi as one of its roots

    how do we know the first step to solving this will be realising that

    (x-(a+bi))(x-(a-bi)) ??

    why do i sub the pair into the place of the constants?
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    (Original post by timebent)
    Where a question asks you to find the quadratic equation that has a+bi as one of its roots

    how do we know the first step to solving this will be realising that

    (x-(a+bi))(x-(a-bi)) ??

    why do i sub the pair into the place of the constants?
    Assuming the coefficients of the quadratic are real, if a+bi is a root, then its conjugate a-bi is also a root.
    Now a root of p will come from a factor of (x-p), whatever the value of p...
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    (Original post by HapaxOromenon3)
    Assuming the coefficients of the quadratic are real, if a+bi is a root, then its conjugate a-bi is also a root.
    Now a root of p will come from a factor of (x-p), whatever the value of p...
    This bit that's it.

    How do i know it's in the form (x-a)(x-b)? where a and b are the roots of the equation
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    (Original post by timebent)
    This bit that's it.

    How do i know it's in the form (x-a)(x-b)? where a and b are the roots of the equation
    Well if a quadratic has roots  \alpha \text{ and } \beta then it can be written as  (x-\alpha )(x-\beta ) .
    Think about it, when you solve a quadratic, you set it to 0 and then you factorise and find the values of x for when each of the factors is 0. So you know that a quadratic with roots 2 and 3 can be written as (x-2)(x-3) because if you had (x-2)(x-3)=0 you would have no problem telling me the roots.
    From this it follows that if one the roots, \alpha , is  a+bi then the quadratic can be written as  (x-(a+bi))(x-\beta ) .
    If a quadratic (or any degree polynomial) with real coefficients has complex root w=a+bi then its complex conjugate  w^* = a-bi is also a root of the quadratic (this holds true for any degree polynomial with all real coefficients).
    From this we know that a-bi is also a root of the quadratic equation so we can say that  \beta = a-bi so the quadratic can be written as  (x-(a+bi))(x-(a-bi)) .
    But generally if any polynomial of degree n with the form
     \displaystyle x^n + a_1 x^{n-1} + a_2x^{n-2} + \ldots + a_{n-1}x + a_{n} has roots  \alpha _1, \alpha_2, \ldots \alpha _n , then it can be expressed in the form
     \displaystyle (x-\alpha _1 )(x-\alpha _2 )\ldots (x-\alpha _{n-1} )(x-\alpha _n ) .
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    (Original post by B_9710)
    Well if a quadratic has roots  \alpha \text{ and } \beta then it can be written as  (x-\alpha )(x-\beta ) .
    Think about it, when you solve a quadratic, you set it to 0 and then you factorise and find the values of x for when each of the factors is 0. So you know that a quadratic with roots 2 and 3 can be written as (x-2)(x-3) because if you had (x-2)(x-3)=0 you would have no problem telling me the roots.
    From this it follows that if one the roots, \alpha , is  a+bi then the quadratic can be written as  (x-(a+bi))(x-\beta ) .
    If a quadratic (or any degree polynomial) with real coefficients has complex root w=a+bi then its complex conjugate  w^* = a-bi is also a root of the quadratic (this holds true for any degree polynomial with all real coefficients).
    From this we know that a-bi is also a root of the quadratic equation so we can say that  \beta = a-bi so the quadratic can be written as  (x-(a+bi))(x-(a-bi)) .
    OHHH!!1 thanks for that, just needed some clarification on that ^-^
    I really am simple minded and forget stuff sometimes
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    Just an easier way to find the quadratic equation.

    The quadratic equation will be in the form x^2 -(a+b)x + ab...

    Where a = 1 root and in this case, b = it's complex conjugate


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