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NickRBest ResponseYou've already chosen the best response.0
try using the quadratic formula
 one year ago

radhBest ResponseYou've already chosen the best response.0
alright. let me try this out.
 one year ago

radhBest ResponseYou've already chosen the best response.0
how would it work considering the first term is cubed, not squared?
 one year ago

NickRBest ResponseYou've already chosen the best response.0
Look 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
 one year ago

radhBest ResponseYou've already chosen the best response.0
how would that help? i don't understand.
 one year ago

radhBest ResponseYou've already chosen the best response.0
i know it simplifies to 4y3  37y/12 + 2.99.
 one year ago

NickRBest ResponseYou've already chosen the best response.0
do you have an easier way of explaining this?
 one year ago

radhBest ResponseYou've already chosen the best response.0
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

campbell_stBest ResponseYou've already chosen the best response.0
lol.. wow thats neat work... @pathosdebater
 one year ago

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

radhBest ResponseYou've already chosen the best response.0
It's exponential, yeah. 4x^3 + 5x^2  x + 2.
 one year ago

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

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

amrijuBest ResponseYou've already chosen the best response.0
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

radhBest 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. :\
 one year ago

pathosdebaterBest ResponseYou've already chosen the best response.1
What textbook is it?
 one year ago

radhBest ResponseYou've already chosen the best response.0
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

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

radhBest ResponseYou've already chosen the best response.0
Thanks for the help though, I'm just going to guess. I appreciate the effort.
 one year ago

pathosdebaterBest ResponseYou've already chosen the best response.1
I got it! It is definitely C
 one year ago

amrijuBest ResponseYou've already chosen the best response.0
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

radhBest ResponseYou've already chosen the best response.0
That was my initial guess, but can you explain how?
 one year ago

pathosdebaterBest ResponseYou've already chosen the best response.1
2 is not a complex root
 one year ago

pathosdebaterBest ResponseYou've already chosen the best response.1
Let me see if I can try and explain...
 one year ago

amrijuBest ResponseYou've already chosen the best response.0
its got to have two complex roots ..if not zero
 one year ago

radhBest 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.
 one year ago

amrijuBest ResponseYou've already chosen the best response.0
have u done complex numbers @pathosdebater
 one year ago

pathosdebaterBest ResponseYou've already chosen the best response.1
Only briefly. I'm not a wizard at them... yet
 one year ago

amrijuBest ResponseYou've already chosen the best response.0
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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