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## radh Group Title Find the Complex Roots: 4x3 + 5x2 – x + 2. Thank you! one year ago one year ago

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1. NickR Group Title

try using the quadratic formula

2. radh Group Title

alright. let me try this out.

3. NickR Group Title

ok take your time

4. radh Group Title

how would it work considering the first term is cubed, not squared?

5. NickR Group Title

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 Group Title

how would that help? i don't understand.

7. radh Group Title

i know it simplifies to 4y3 - 37y/12 + 2.99.

8. NickR Group Title

@hartnn ?

9. NickR Group Title

do you have an easier way of explaining this?

10. hartnn Group Title

let me think...

11. pathosdebater Group Title

12. radh Group Title

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 Group Title

lol.. wow thats neat work... @pathosdebater

14. pathosdebater Group Title

Is the problem 4x3 + 5x2 – x + 2 or 4*3 + 5*2 – x + 2? (Just making sure)

15. radh Group Title

It's exponential, yeah. 4x^3 + 5x^2 - x + 2.

16. pathosdebater Group Title

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

17. pathosdebater Group Title

One second

18. radh Group Title

Sorry for the confusion.

19. pathosdebater Group Title

This is so strange. I keep getting irrational answers. Is there anything wrong in my calculations which could be causing this?

20. amriju Group Title

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 Group Title

@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 Group Title

What textbook is it?

23. radh Group Title

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

24. pathosdebater Group Title

Hmm... well there are definitely only 3 roots, so that narrows it down to C and D.

25. radh Group Title

Thanks for the help though, I'm just going to guess. I appreciate the effort.

26. pathosdebater Group Title

I got it! It is definitely C

27. amriju Group Title

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 Group Title

That was my initial guess, but can you explain how?

29. pathosdebater Group Title

2 is not a complex root

30. pathosdebater Group Title

Let me see if I can try and explain...

31. amriju Group Title

its got to have two complex roots ..if not zero

32. radh Group Title

@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 Group Title

have u done complex numbers @pathosdebater

34. pathosdebater Group Title

Only briefly. I'm not a wizard at them... yet

35. amriju Group Title

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$\sum_{?}^{?} a=-5/4...\sum_{?}^{?} ab=1/4....abc=-2/4=-1/2$