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Thread: Quadratic with complex roots

  1. #1
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    Quadratic with complex roots

    Hi there,

    I'm looking for some confirmation that i have done this correctly/the easiest way,

    Rgds.

    Resolve $\displaystyle 9x^2-36x+37$ into factors.

    = 9($\displaystyle x-\alpha$)($\displaystyle x-\beta$)

    $\displaystyle \frac{36+or-\sqrt{36^2-4*9*37}}{18}$

    $\displaystyle \frac{36+or-\sqrt{-36}}{18}$

    $\displaystyle \frac{36+or-j6}{18}$

    $\displaystyle 2+or-j\frac{1}{3}$

    $\displaystyle 9(x-2+j\frac{1}{3})(x-2-j\frac{1}{3})$

    $\displaystyle 3(x-2+j\frac{1}{3})3(x-2-j\frac{1}{3})$

    $\displaystyle (3x-6+j)(3x-6-j)$
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  2. #2
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    Quote Originally Posted by andyw View Post
    Hi there,

    I'm looking for some confirmation that i have done this correctly/the easiest way,

    to confirm it is correct simply expand your answer

    $\displaystyle (3x-6+j)(3x-6-j)$

    does it give?

    $\displaystyle 9x^2-36x+37$
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  3. #3
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    It certainly does!

    Thanks.
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