Here's the question you clicked on:
radh
Find the Complex Roots: 4x3 + 5x2 – x + 2. Thank you!
try using the quadratic formula
alright. let me try this out.
how would it work considering the first term is cubed, not squared?
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
how would that help? i don't understand.
i know it simplifies to 4y3 - 37y/12 + 2.99.
do you have an easier way of explaining this?
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
lol.. wow thats neat work... @pathosdebater
Is the problem 4x3 + 5x2 – x + 2 or 4*3 + 5*2 – x + 2? (Just making sure)
It's exponential, yeah. 4x^3 + 5x^2 - x + 2.
What? Oh... I didn't know that. I thought it was 4x*3 + 5x*2 - x+2
This is so strange. I keep getting irrational answers. Is there anything wrong in my calculations which could be causing this?
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
@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. :\
What textbook is it?
It's not a textbook, it's a practice given by the teacher. (I searched online for duplications but it looks original.)
Hmm... well there are definitely only 3 roots, so that narrows it down to C and D.
Thanks for the help though, I'm just going to guess. I appreciate the effort.
I got it! It is definitely C
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...
That was my initial guess, but can you explain how?
2 is not a complex root
Let me see if I can try and explain...
its got to have two complex roots ..if not zero
@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.
have u done complex numbers @pathosdebater
Only briefly. I'm not a wizard at them... yet
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\[\sum_{?}^{?} a=-5/4...\sum_{?}^{?} ab=1/4....abc=-2/4=-1/2\]