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nilankshi

obtain all the zeroes of 3x^4+6x^2-2x^2-10x-5,if two zeroes are root5/3 and -root 5/3

  • one year ago
  • one year ago

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  1. Lumenaire
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    I assume you mean: 3x^4+6x^3-2x^2-10x-5

    • one year ago
  2. lgbasallote
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    the roots aree \(\large \sqrt{\frac{5}{3}}\) and \(\large -\sqrt{\frac{5}{3}}\) right? or is the root applied on 5 only?

    • one year ago
  3. Lumenaire
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    is there a square root sign at all? I understood that 5/3 and -5/3 are the roots

    • one year ago
  4. apoorvk
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    divide the whole system by its two factors that are given (or try synthetic division) After two whole divisions by each factor, you 'll be left with a quadratic as the quotient, factorising which you'll get you the other two roots.

    • one year ago
  5. lgbasallote
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    he/she said root 5/3 and root -5/3..she already said zeroes so i assume root->radical

    • one year ago
  6. nilankshi
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    yes that is right..

    • one year ago
  7. nilankshi
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    root is on both..

    • one year ago
  8. apoorvk
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    \[\mathbb{yes}\]

    • one year ago
  9. Lumenaire
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    and is it an x^3 in the equation?

    • one year ago
  10. Lumenaire
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    (6x^2) should be 6x^3

    • one year ago
  11. nilankshi
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    yes it is in equion

    • one year ago
  12. lgbasallote
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    \(\large x = \sqrt{\frac{5}{3}}\) \(\large x^2 = \frac{5}{3}\) \(\large 3x^2 = 5\) \(\large 3x^2 - 5 = 0\) as for the other zero... \(\large x = -\sqrt{\frac{5}{3}}\) \(\large x^2 = \frac{5}{3}\) \(\large 3x^2 = 5\) \(\large 3x^2 - 5 = 0\) so two of the factors would be (3x^2 - 5)^2

    • one year ago
  13. lgbasallote
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    something doesnt feel right...

    • one year ago
  14. nilankshi
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    yaa..

    • one year ago
  15. lgbasallote
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    hmm idk how to do this sorry... @apoorvk may know though

    • one year ago
  16. nilankshi
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    its ok..

    • one year ago
  17. apoorvk
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    Sorry, Trance. :) and Nilakshi. I 'll work on the solution now

    • one year ago
  18. TranceNova
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    Thanks :) @Rohangrr if you want to talk about it, please make a new question.

    • one year ago
  19. phi
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    I would multiply out \[ (x-\sqrt{5/3})(x+\sqrt{5/3}) = x^2-\frac{5}{3}\] we can write this as \(3x^2-5\) (why? see below) divide this into the original equation to get \(x^2+2x+1 \) now factor this to get the remaining roots. * we are looking for \((x^2-\frac{5}{3}) (x^2+2x+1) =0\) if we multiply both sides by 3 we get \( 3(x^2-\frac{5}{3})(x^2+2x+1) =0\) \( (3x^2-5) (x^2+2x+1)=0\)

    • one year ago
  20. apoorvk
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    Hmm, I got so bored, Phi already posted it. :/ Good job, Phi

    • one year ago
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