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swissgirl Group Title

How to find the complex roots of \(x^4+x^3+3x^2+2x+2 \)

  • 2 years ago
  • 2 years ago

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  1. dumbcow Group Title
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    you have to factor out any real roots easiest to graph the function to obtain any real roots

    • 2 years ago
  2. swissgirl Group Title
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    There are no real roots though

    • 2 years ago
  3. swissgirl Group Title
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    its above the y axis

    • 2 years ago
  4. experimentX Group Title
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    let (x^2+bx+1)(x^2+cx+2) = that expression and find the value of b and c

    • 2 years ago
  5. dumbcow Group Title
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    ok then yeah what @experimentX said if two of the complex roots is in the form: \[x = a \pm bi\] then \[x-a = \pm bi\] \[(x-a)^{2} = -b^{2}\] \[x^{2}+ (-2a)x +(a^{2}+b^{2}) = 0\] anyway , my point is that a pair of complex roots come from a quadratic equation :|

    • 2 years ago
  6. KingGeorge Group Title
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    I imagine you could factor this by grouping and then use the quadratic formula again.

    • 2 years ago
  7. swissgirl Group Title
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    i tried doing the method u taught me yesterday but it wldnt go

    • 2 years ago
  8. KingGeorge Group Title
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    \[\Large x^4+x^3+3x^2+2x+2 \\ \Large =x^4+2x^2+x^3+x^2+2(x+1)\\ \Large =x^4+2x^2+x^2(x+1)+2(x+1)\\ \Large =x^2(x^2+2)+(x^2+2)(x+1)\\ \Large =(x^2+2)(x^2+x+1)\]

    • 2 years ago
  9. KingGeorge Group Title
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    I think the trick is seeing to split up the \(3x^2\) as \(2x^2+x^2\).

    • 2 years ago
  10. swissgirl Group Title
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    omg y is it always sooooo simple and I can never see it lol

    • 2 years ago
  11. swissgirl Group Title
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    Thanksssss guyssss for helping me :))))

    • 2 years ago
  12. KingGeorge Group Title
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    You're welcome.

    • 2 years ago
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