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radh

Find the Complex Roots: 4x3 + 5x2 – x + 2. Thank you!

  • one year ago
  • one year ago

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  1. NickR
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    try using the quadratic formula

    • one year ago
  2. radh
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    alright. let me try this out.

    • one year ago
  3. NickR
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    ok take your time

    • one year ago
  4. radh
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    how would it work considering the first term is cubed, not squared?

    • one year ago
  5. NickR
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    Look for a simple substitution that eliminates the quadratic term of 4 x^3+5 x^2-x+2. Eliminate the quadratic term by substituting y = x+5/12: 29/12+5 (y-5/12)^2+4 (y-5/12)^3-y = 0

    • one year ago
  6. radh
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    how would that help? i don't understand.

    • one year ago
  7. radh
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    i know it simplifies to 4y3 - 37y/12 + 2.99.

    • one year ago
  8. NickR
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    @hartnn ?

    • one year ago
  9. NickR
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    do you have an easier way of explaining this?

    • one year ago
  10. hartnn
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    let me think...

    • one year ago
  11. pathosdebater
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    • one year ago
  12. radh
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    That is definitely not one of the options in this multiple choice question. Let me give the possible answers: (all negative and positive of this number) a. 1, 1/2, 1/4, 2 b. 1, 1/2, 2, 4 c. 1, 1/2, 1/4 d. 1, 1/4, 2

    • one year ago
  13. campbell_st
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    lol.. wow thats neat work... @pathosdebater

    • one year ago
  14. pathosdebater
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    Is the problem 4x3 + 5x2 – x + 2 or 4*3 + 5*2 – x + 2? (Just making sure)

    • one year ago
  15. radh
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    It's exponential, yeah. 4x^3 + 5x^2 - x + 2.

    • one year ago
  16. pathosdebater
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    What? Oh... I didn't know that. I thought it was 4x*3 + 5x*2 - x+2

    • one year ago
  17. pathosdebater
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    One second

    • one year ago
  18. radh
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    Sorry for the confusion.

    • one year ago
  19. pathosdebater
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    This is so strange. I keep getting irrational answers. Is there anything wrong in my calculations which could be causing this?

    • one year ago
  20. amriju
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    u have an easy way of solving....firstly...note that a cubic eqn has atmost two complex sol..and atleast 1 real soln....so go for the real soln by trial and error...then u hav a factor...divide the polynomial by the factor...u have a quad eqn now as the quotnt....solve it to get the complex roots

    • one year ago
  21. radh
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    @pathosdebater I'm not sure, I got similar answers (which is why I came here). There may be a typo in the question (not what I typed; the actual question itself), but no one I know has said anything about it. :\

    • one year ago
  22. pathosdebater
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    What textbook is it?

    • one year ago
  23. radh
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    It's not a textbook, it's a practice given by the teacher. (I searched online for duplications but it looks original.)

    • one year ago
  24. pathosdebater
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    Hmm... well there are definitely only 3 roots, so that narrows it down to C and D.

    • one year ago
  25. radh
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    Thanks for the help though, I'm just going to guess. I appreciate the effort.

    • one year ago
  26. pathosdebater
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    I got it! It is definitely C

    • one year ago
  27. amriju
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    do u mind if i say that the options, all of them are wrong..? its so coz...u cant have three complex roots for a REAL cubic eqn...complex roots come in pairs...so...

    • one year ago
  28. radh
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    That was my initial guess, but can you explain how?

    • one year ago
  29. pathosdebater
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    2 is not a complex root

    • one year ago
  30. pathosdebater
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    Let me see if I can try and explain...

    • one year ago
  31. amriju
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    its got to have two complex roots ..if not zero

    • one year ago
  32. radh
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    @amriju I don't mind, I found it to be that way too when I substituted it in. I don't know. Not my practice, it was the teachers. I got all of the other questions on finding zeros, this one was just confusing.

    • one year ago
  33. amriju
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    have u done complex numbers @pathosdebater

    • one year ago
  34. pathosdebater
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    Only briefly. I'm not a wizard at them... yet

    • one year ago
  35. amriju
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    u really cant have an option with more than two values...that settles it...as for the solution u got to try this way...if a,b,c be the roots then\[\sum_{?}^{?} a=-5/4...\sum_{?}^{?} ab=1/4....abc=-2/4=-1/2\]

    • one year ago
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