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NickR
 2 years ago
Best ResponseYou've already chosen the best response.0try using the quadratic formula

radh
 2 years ago
Best ResponseYou've already chosen the best response.0alright. let me try this out.

radh
 2 years ago
Best ResponseYou've already chosen the best response.0how would it work considering the first term is cubed, not squared?

NickR
 2 years ago
Best ResponseYou've already chosen the best response.0Look for a simple substitution that eliminates the quadratic term of 4 x^3+5 x^2x+2. Eliminate the quadratic term by substituting y = x+5/12: 29/12+5 (y5/12)^2+4 (y5/12)^3y = 0

radh
 2 years ago
Best ResponseYou've already chosen the best response.0how would that help? i don't understand.

radh
 2 years ago
Best ResponseYou've already chosen the best response.0i know it simplifies to 4y3  37y/12 + 2.99.

NickR
 2 years ago
Best ResponseYou've already chosen the best response.0do you have an easier way of explaining this?

radh
 2 years ago
Best ResponseYou've already chosen the best response.0That 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

campbell_st
 2 years ago
Best ResponseYou've already chosen the best response.0lol.. wow thats neat work... @pathosdebater

pathosdebater
 2 years ago
Best ResponseYou've already chosen the best response.1Is the problem 4x3 + 5x2 – x + 2 or 4*3 + 5*2 – x + 2? (Just making sure)

radh
 2 years ago
Best ResponseYou've already chosen the best response.0It's exponential, yeah. 4x^3 + 5x^2  x + 2.

pathosdebater
 2 years ago
Best ResponseYou've already chosen the best response.1What? Oh... I didn't know that. I thought it was 4x*3 + 5x*2  x+2

pathosdebater
 2 years ago
Best ResponseYou've already chosen the best response.1This is so strange. I keep getting irrational answers. Is there anything wrong in my calculations which could be causing this?

amriju
 2 years ago
Best ResponseYou've already chosen the best response.0u 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

radh
 2 years ago
Best ResponseYou've already chosen the best response.0@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. :\

pathosdebater
 2 years ago
Best ResponseYou've already chosen the best response.1What textbook is it?

radh
 2 years ago
Best ResponseYou've already chosen the best response.0It's not a textbook, it's a practice given by the teacher. (I searched online for duplications but it looks original.)

pathosdebater
 2 years ago
Best ResponseYou've already chosen the best response.1Hmm... well there are definitely only 3 roots, so that narrows it down to C and D.

radh
 2 years ago
Best ResponseYou've already chosen the best response.0Thanks for the help though, I'm just going to guess. I appreciate the effort.

pathosdebater
 2 years ago
Best ResponseYou've already chosen the best response.1I got it! It is definitely C

amriju
 2 years ago
Best ResponseYou've already chosen the best response.0do 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...

radh
 2 years ago
Best ResponseYou've already chosen the best response.0That was my initial guess, but can you explain how?

pathosdebater
 2 years ago
Best ResponseYou've already chosen the best response.12 is not a complex root

pathosdebater
 2 years ago
Best ResponseYou've already chosen the best response.1Let me see if I can try and explain...

amriju
 2 years ago
Best ResponseYou've already chosen the best response.0its got to have two complex roots ..if not zero

radh
 2 years ago
Best ResponseYou've already chosen the best response.0@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.

amriju
 2 years ago
Best ResponseYou've already chosen the best response.0have u done complex numbers @pathosdebater

pathosdebater
 2 years ago
Best ResponseYou've already chosen the best response.1Only briefly. I'm not a wizard at them... yet

amriju
 2 years ago
Best ResponseYou've already chosen the best response.0u 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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