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  1. #1
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    quadratic roots question

    Write a quadratic equation that has 5 + 2i as one of its roots.

    Can someone show me a step by step method on solving the above problem.
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  2. #2
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    Quote Originally Posted by cityismine View Post
    Write a quadratic equation that has 5 + 2i as one of its roots.

    Can someone show me a step by step method on solving the above problem.
    Take your pick:

    0 = (x - [5 + 2i])(ax + b) where a and b are any numbers you please ......

    However, if you want your equation to have real coefficients, then

    0 = (x - [5 + 2i])(x - [5 - 2i]) = ([x - 5] - 2i)([x - 5] + 2i) = [x - 5]^2 + 4 = ........
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    Mr. fantastic can you explain the second method of "real coefficients" a bit more thoroughly. I'm still a little confused how you got the initial equation.
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    Quote Originally Posted by cityismine View Post
    Mr. fantastic can you explain the second method of "real coefficients" a bit more thoroughly. I'm still a little confused how you got the initial equation.
    The only way to make disappear such a complex number is to multiply by its conjugate.

    Let \bar{z} be the conjugate of z.

    (x-z)(x-\bar{z})=x^2-x(\underbrace{z+\bar{z}}_{\text{real number}})+\underbrace{z \cdot \bar{z}}_{\text{real number}}

    See why these are real numbers by noting that z=a+ib (therefore, \bar{z}=a-ib)
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