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radh

  • 3 years ago

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

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  1. NickR
    • 3 years ago
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    try using the quadratic formula

  2. radh
    • 3 years ago
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    alright. let me try this out.

  3. NickR
    • 3 years ago
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    ok take your time

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

  5. NickR
    • 3 years ago
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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

  6. radh
    • 3 years ago
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    how would that help? i don't understand.

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

  8. NickR
    • 3 years ago
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    @hartnn ?

  9. NickR
    • 3 years ago
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    do you have an easier way of explaining this?

  10. hartnn
    • 3 years ago
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    let me think...

  11. pathosdebater
    • 3 years ago
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  12. radh
    • 3 years ago
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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

  13. campbell_st
    • 3 years ago
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    lol.. wow thats neat work... @pathosdebater

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

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

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

  17. pathosdebater
    • 3 years ago
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    One second

  18. radh
    • 3 years ago
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    Sorry for the confusion.

  19. pathosdebater
    • 3 years ago
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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?

  20. amriju
    • 3 years ago
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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

  21. radh
    • 3 years ago
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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. :\

  22. pathosdebater
    • 3 years ago
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    What textbook is it?

  23. radh
    • 3 years ago
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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.)

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

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

  26. pathosdebater
    • 3 years ago
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    I got it! It is definitely C

  27. amriju
    • 3 years ago
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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...

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

  29. pathosdebater
    • 3 years ago
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    2 is not a complex root

  30. pathosdebater
    • 3 years ago
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    Let me see if I can try and explain...

  31. amriju
    • 3 years ago
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    its got to have two complex roots ..if not zero

  32. radh
    • 3 years ago
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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.

  33. amriju
    • 3 years ago
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    have u done complex numbers @pathosdebater

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

  35. amriju
    • 3 years ago
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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\[\sum_{?}^{?} a=-5/4...\sum_{?}^{?} ab=1/4....abc=-2/4=-1/2\]

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