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\[2x^2-ix-2+i=0\]

:/ Sorry... I can't help..

:'(

tried using the quadratic formula yet?

Phew! GO JDOE!

ok..got that

well.. that's.... as far as I can see it going...
what does your answer say?

x=1 and x=-1+0.5i

hmmm I don't see it simplifying ... to that

:(

|dw:1389574709566:dw|

http://en.wikipedia.org/wiki/Ruffini's_rule

how did u do that

that is an algorithm

please explain...never learned that

|dw:1389575164382:dw|

wait...is this like synthetic division

|dw:1389575421091:dw|

this is the briot-ruffini algorithmm, i don't what is synthetic division

so in my question the root is 2?

no the root is one, the first coef in your polynomial is 2

how do i know what the root is?

one, was a easy root to see in that problem

oh. so if it had all 4...then it will be 2?

could u do it step by step and explain what you were doing at each step

|dw:1389575930097:dw|

it doesn't work for another question

post it

\[-ix^2-(3+i)x+2\]

this root is not so easy to see without any calculation, let me think

ok

i think you will have to use the quatratic formula

ok...so my question is how will i know when to use the rule u showed me and quadratic

you know one root already?

it's hard to see in this equation

to use ruffini you must know first one root

For
\[ -ix^2-(3+i)x+2 \]
how far did you get ?

Yes, I saw that... that works, but I would use the above algorithm

ok...umm how would i find that roots from that after i get 4-i

do i just put it into the quadratic form

ok. so after i get \[z=i-(4-1)/-2\] and \[z=1+(4-i)\]

which problem are you doing ?

finding the roots of the second problem i needed

could u stay on..i will try it and see if i get the answers

sorry..i made a mistake

\[\frac{ (3+i)\pm \sqrt{(8+14i)} }{ 2i }\]

i got this so far

I think it should be -2i in the bottom

sorry

no need to be sorry, just careful.

what do i do about the square root

wait y did 8 become negative

could u show me both...bc there r ides in both i don't get that well

Re(z)=-8

Im(z)=-14