Could someone walk me through this question and explain to me what happens. I have been working on it for 7 hours. I don't understand my prof and I might fail the class and I want to cry.
Its solving a quadratic equation that had complex numbers

- anonymous

- schrodinger

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- anonymous

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

- anonymous

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

- anonymous

:'(

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## More answers

- anonymous

@jdoe0001

- jdoe0001

tried using the quadratic formula yet?

- anonymous

Phew! GO JDOE!

- anonymous

I tried it, i solved it but i got the wrong answers. I tried it over and over again. I looked up videos, still wrong answer

- jdoe0001

\(\large \begin{array}{llll}
{\color{red}{ 2}}x^2&{\color{blue}{ -i}}x&{\color{green}{ -2+i}}=0\\
\uparrow &\ \uparrow &\quad \uparrow \\
{\color{red}{ a}}&\ {\color{blue}{ b}}&\quad {\color{green}{ c}}
\end{array}\\ \quad \\ \bf
x=\cfrac{i\pm\sqrt{i^2-4(2)(-2+i)}}{2(2)}\implies x=\cfrac{i\pm\sqrt{-1-8(-2+i)}}{2(2)}\\ \quad \\
x=\cfrac{i\pm\sqrt{-1+16-8i}}{2(2)}\implies x=\cfrac{i\pm\sqrt{15-8i}}{4}\)

- anonymous

ok..got that

- jdoe0001

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

- anonymous

x=1 and x=-1+0.5i

- jdoe0001

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

- anonymous

:(

- anonymous

|dw:1389574709566:dw|

- anonymous

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

- anonymous

how did u do that

- anonymous

that is an algorithm

- anonymous

please explain...never learned that

- anonymous

|dw:1389575164382:dw|

- anonymous

wait...is this like synthetic division

- anonymous

|dw:1389575421091:dw|

- anonymous

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

- anonymous

so in my question the root is 2?

- anonymous

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

- anonymous

how do i know what the root is?

- anonymous

one, was a easy root to see in that problem

- anonymous

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

- anonymous

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

- anonymous

|dw:1389575930097:dw|

- anonymous

it doesn't work for another question

- anonymous

@RaphaelFilgueiras

- anonymous

post it

- anonymous

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

- anonymous

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

- anonymous

ok

- anonymous

i think you will have to use the quatratic formula

- anonymous

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

- anonymous

you know one root already?

- anonymous

it's hard to see in this equation

- anonymous

to use ruffini you must know first one root

- phi

see https://en.wikipedia.org/wiki/Complex_square_root#Algebraic_formula
for how to take the square root of a complex number in rectangular form (a + bi)
\[ \sqrt{z} = \sqrt{\frac{|z| + Re(z)}{2}} + i sgn(Im(z))\sqrt{\frac{|z| - Re(z)}{2}} \]
Here is an example:
\[ \sqrt{15-8i} \]
the magnitude = |z|= \( \sqrt{15^2+8^2}= \sqrt{225+64} = 17 \)
The Real part of z = Re(z)= 15
The Imaginary part of z= Im(z)= -8. The sign of Im(z)= sgn(Im(z))= -1
putting in the numbers we get
\[ \sqrt{z} = \sqrt{\frac{17+ 15}{2}} + -1\cdot i \sqrt{\frac{17 - 15}{2}} \]
simplifying, we get
4 - i

- phi

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

- anonymous

well, i didn't try this one bc i spent 5 hours on the first one....i do it a different way that i saw in a math video..idk if that will work....http://www.youtube.com/watch?v=kn37LC8MfBU

- phi

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

- anonymous

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

- anonymous

do i just put it into the quadratic form

- phi

I posted how to do the square root of a complex number, because that is the tricky part when you use the quadratic formula.
For your problem, use the quadratic formula and simplify until you get to the square root of a complex number...

- anonymous

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

- phi

which problem are you doing ?