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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

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

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Other answers:

tried using the quadratic formula yet?
Phew! GO JDOE!
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
\(\large \begin{array}{llll} {\color{red}{ 2}}x^2&{\color{blue}{ -i}}x&{\color{green}{ -2+i}}=0\\ \uparrow &\ \uparrow &\quad \uparrow \\ {\color{red}{ a}}&\ {\color{blue}{ b}}&\quad {\color{green}{ c}} \end{array}\\ \quad \\ \bf x=\cfrac{i\pm\sqrt{i^2-4(2)(-2+i)}}{2(2)}\implies x=\cfrac{i\pm\sqrt{-1-8(-2+i)}}{2(2)}\\ \quad \\ x=\cfrac{i\pm\sqrt{-1+16-8i}}{2(2)}\implies x=\cfrac{i\pm\sqrt{15-8i}}{4}\) 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
how did u do that
that is an algorithm
please explain...never learned that
|dw:1389575164382:dw| this like synthetic division
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
it doesn't work for another question
post it
this root is not so easy to see without any calculation, let me think
i think you will have to use the quatratic formula 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
  • phi
see 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 ?
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....
  • phi
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
  • 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...
ok. so after i get \[z=i-(4-1)/-2\] and \[z=1+(4-i)\]
  • phi
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
  • phi
If it is this one \[ -ix^2-(3+i)x+2 = 0 \] then you have a= -i, b= -(3+i) and c=2 but this is going to lead to some *very* ugly numbers.
\[\frac{ (3+i)\pm \sqrt{(8+14i)} }{ 2i }\]
i got this so far
  • phi
I think it should be -2i in the bottom
  • phi
no need to be sorry, just careful.
  • phi
if you multiply top and bottom by i you can get \[ \frac{ (3+i)\pm \sqrt{(8+14i)} }{ -2i } \frac{i}{i} = \frac{ (-1+3i)\pm \sqrt{(-8-14i)} }{ 2 } \]
what do i do about the square root