## anonymous 3 years ago Find the Complex Roots: 4x3 + 5x2 – x + 2. Thank you!

1. anonymous

try using the quadratic formula

2. anonymous

alright. let me try this out.

3. anonymous

ok take your time

4. anonymous

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

5. anonymous

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. anonymous

how would that help? i don't understand.

7. anonymous

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

8. anonymous

@hartnn ?

9. anonymous

do you have an easier way of explaining this?

10. hartnn

let me think...

11. anonymous

12. anonymous

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

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

14. anonymous

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

15. anonymous

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

16. anonymous

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

17. anonymous

One second

18. anonymous

Sorry for the confusion.

19. anonymous

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

20. amriju

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. anonymous

@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. anonymous

What textbook is it?

23. anonymous

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

24. anonymous

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

25. anonymous

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

26. anonymous

I got it! It is definitely C

27. amriju

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. anonymous

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

29. anonymous

2 is not a complex root

30. anonymous

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

31. amriju

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

32. anonymous

@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

have u done complex numbers @pathosdebater

34. anonymous

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

35. amriju

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