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Math Help - Finding roots a+bi form

  1. #1
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    Finding roots a+bi form

    Find the roots in simplest a+bi form y=2x^2-4x+5

    [Math]\frac{-b+-\sqrt{ b^2-4ac}}{2a}[/tex]

    [Math]\frac{4+-\sqrt{16-4(2)(5)}}{2(2)}[/tex]

    [Math]\frac{4+-\sqrt{24}}{4}[/tex]

    to put in a+bi form I believe that that it has to be a perfect square and that I can not have a radical in the answer? Please help with what I am missing Thanks
    Last edited by IDontunderstand; December 15th 2008 at 06:10 PM.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by IDontunderstand View Post
    Find the roots in simplest a+bi form y=2x^2-4x+5

    \frac{-b+-{{Sqrt}b^2-4ac}}{2a}
    As you (kind of) put down the roots of ax^2+bx+c are x=\frac{b\pm\sqrt{b^2-4ac}}{2a}

    So we can see that the solutions of \overbrace{2}^{a}x^2-\overbrace{4}^{b}x+\overbrace{5}^{c} are x=\frac{4\pm\sqrt{(-4)^2-4(2)(5)}}{2(2)}=\frac{4\pm\sqrt{16-40}}{4}=\frac{4\pm2\sqrt{6}i}{4}
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  3. #3
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    Radical

    So am I able to have a radical in simplest a+bi form? I have never seen a radical in simplest form or a fraction which I dont think that should be a problem.

    So are we able to have a radical and a fraction in the answer in simple a+bi form?
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  4. #4
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    So are we able to have a radical and a fraction in the answer in simple a+bi form?
    Technically speaking we should keep going to get
    \frac{4\pm2\sqrt{6}i}{4} = \frac{4}{4}\pm \frac{2\sqrt{6}}{4}i

    then cancel common factors with the denominator, but in reality I would expect the answer given by Mathstud to be worth full marks. Radicals and fractions are perfectly appropriate.
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