anonymous
  • anonymous
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
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
\[2x^2-ix-2+i=0\]
anonymous
  • anonymous
:/ Sorry... I can't help..
anonymous
  • anonymous
:'(

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anonymous
  • anonymous
@jdoe0001
jdoe0001
  • jdoe0001
tried using the quadratic formula yet?
anonymous
  • anonymous
Phew! GO JDOE!
anonymous
  • 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
  • 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
  • anonymous
ok..got that
jdoe0001
  • jdoe0001
well.. that's.... as far as I can see it going... what does your answer say?
anonymous
  • anonymous
x=1 and x=-1+0.5i
jdoe0001
  • jdoe0001
hmmm I don't see it simplifying ... to that
anonymous
  • anonymous
:(
anonymous
  • anonymous
|dw:1389574709566:dw|
anonymous
  • anonymous
http://en.wikipedia.org/wiki/Ruffini's_rule
anonymous
  • anonymous
how did u do that
anonymous
  • anonymous
that is an algorithm
anonymous
  • anonymous
please explain...never learned that
anonymous
  • anonymous
|dw:1389575164382:dw|
anonymous
  • anonymous
wait...is this like synthetic division
anonymous
  • anonymous
|dw:1389575421091:dw|
anonymous
  • anonymous
this is the briot-ruffini algorithmm, i don't what is synthetic division
anonymous
  • anonymous
so in my question the root is 2?
anonymous
  • anonymous
no the root is one, the first coef in your polynomial is 2
anonymous
  • anonymous
how do i know what the root is?
anonymous
  • anonymous
one, was a easy root to see in that problem
anonymous
  • anonymous
oh. so if it had all 4...then it will be 2?
anonymous
  • anonymous
could u do it step by step and explain what you were doing at each step
anonymous
  • anonymous
|dw:1389575930097:dw|
anonymous
  • anonymous
it doesn't work for another question
anonymous
  • anonymous
@RaphaelFilgueiras
anonymous
  • anonymous
post it
anonymous
  • anonymous
\[-ix^2-(3+i)x+2\]
anonymous
  • anonymous
this root is not so easy to see without any calculation, let me think
anonymous
  • anonymous
ok
anonymous
  • anonymous
i think you will have to use the quatratic formula
anonymous
  • anonymous
ok...so my question is how will i know when to use the rule u showed me and quadratic
anonymous
  • anonymous
you know one root already?
anonymous
  • anonymous
it's hard to see in this equation
anonymous
  • anonymous
to use ruffini you must know first one root
phi
  • 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
  • phi
For \[ -ix^2-(3+i)x+2 \] how far did you get ?
anonymous
  • 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
  • phi
Yes, I saw that... that works, but I would use the above algorithm
anonymous
  • anonymous
ok...umm how would i find that roots from that after i get 4-i
anonymous
  • anonymous
do i just put it into the quadratic form
phi
  • 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
  • anonymous
ok. so after i get \[z=i-(4-1)/-2\] and \[z=1+(4-i)\]
phi
  • phi
which problem are you doing ?