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
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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 ?
anonymous
  • anonymous
finding the roots of the second problem i needed
anonymous
  • anonymous
could u stay on..i will try it and see if i get the answers
anonymous
  • anonymous
sorry..i made a mistake
phi
  • 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.
anonymous
  • anonymous
\[\frac{ (3+i)\pm \sqrt{(8+14i)} }{ 2i }\]
anonymous
  • anonymous
i got this so far
phi
  • phi
I think it should be -2i in the bottom
anonymous
  • anonymous
sorry
phi
  • phi
no need to be sorry, just careful.
phi
  • phi
if you multiply top and bottom by i you can get \[ \frac{ (3+i)\pm \sqrt{(8+14i)} }{ -2i } \frac{i}{i} = \frac{ (-1+3i)\pm \sqrt{(-8-14i)} }{ 2 } \]
anonymous
  • anonymous
what do i do about the square root
anonymous
  • anonymous
wait y did 8 become negative
phi
  • phi
We could use the algorithm I posted (or use the idea in the video) but both lead to square roots of square roots...
anonymous
  • anonymous
could u show me both...bc there r ides in both i don't get that well
phi
  • phi
OK, Let's do the algorithm \[ \sqrt{z} = \sqrt{\frac{|z| + Re(z)}{2}} + i sgn(Im(z))\sqrt{\frac{|z| - Re(z)}{2}} \] z= -8 - 14i can you list the Re(z) ?
anonymous
  • anonymous
Re(z)=-8
anonymous
  • anonymous
Im(z)=-14
phi
  • phi
and we only really need the sign of Im(z), the -1 last , we need |z| = sqr(a^2 + b^2) where z= a+ bi
anonymous
  • anonymous
why only the sign...why not the number it self?
anonymous
  • anonymous
so will it always be + or - 1 then
phi
  • phi
yes, the formula uses the sign(Im(z)) to determine the sign of the imaginary part of the root.
anonymous
  • anonymous
oh ok..go on....
phi
  • phi
we need |z| = sqr(a^2 + b^2) where z= a+ bi
anonymous
  • anonymous
16.12
anonymous
  • anonymous
or sqr260
phi
  • phi
If we just use decimals, it will be easier (but not exact) Let's use the radicals (so you can compare to the other approach) sqr260= 2sqr65 now plug what we know into the formula
anonymous
  • anonymous
i get 2.015 -2.015i
anonymous
  • anonymous
sorry, 3.47i
phi
  • phi
ok, so now the roots are \[ \frac{ (-1+3i)\pm (2.0155 - 3.4731 i) }{ 2 } \] now it's time to find the 2 separate roots (-1+2.0155)/2 + (3-3.4731)/2 i and (-1 - 2.0155)/2 + (3+3.4731)/2 i
phi
  • phi
after simplifying compare to wolfram's results http://www.wolframalpha.com/input/?i=-ix%5E2-%283%2Bi%29x%2B2%3D0
phi
  • phi
In the box "Complex Solutions" click on Approximate Form to get the decimal answer.
anonymous
  • anonymous
yep. right...how do i do it the other way
phi
  • phi
The idea is that x + i y = \( \sqrt{ -8 - 14i}\) (We require that both x and y are real numbers) square both sides x^2 -y^2 + 2xy i = -8 - 14i equate real to real, and imaginary to imaginary: x^2 -y^2=-8 2xy = -14
phi
  • phi
to solve, let y = -7/x and plug into the 1st equation x^2 - 49/x^2 = -8 multiply by x^2 and re-arrange x^4 +8x^2 -49 =0 now solve for x^2. i.e. let u=x^2 and solve for u in u^2 +8u -49=0
anonymous
  • anonymous
ok. so x=4.06 and x=-12.1
phi
  • phi
you mean u = 4.06 or -12.1 x will be the square root of u because we want x to be real, we use 4.06 (the square root of -12.1 will be imaginary)
anonymous
  • anonymous
yea...sorry so use to writing x
phi
  • phi
you should use a few more decimals
anonymous
  • anonymous
ok. so after i get u ...what do i do?
phi
  • phi
we want x and y (x + i y is the square root we are looking for)
phi
  • phi
x= sqr(u) y= -7/x
anonymous
  • anonymous
ok. so (2.015, -3.474i)
phi
  • phi
which should match what we got before.
anonymous
  • anonymous
yep. thank you phi....so so much. You are a big help.
phi
  • phi
yw
anonymous
  • anonymous
@phi
anonymous
  • anonymous
i need ur help with something
anonymous
  • anonymous
@zepdrix
anonymous
  • anonymous
could i ask u a simple question
anonymous
  • anonymous
@phi could you just answer a simple question that i didn't ask yesterday
anonymous
  • anonymous
hi. you know for this step were the 8+14i become -8-14i \[\frac{ (3+i)\pm \sqrt{(8+14i)} }{ -2i } \frac{i}{i} = \frac{ (-1+3i)\pm \sqrt{(-8-14i)} }{ 2 } \] how does that happen
phi
  • phi
\[ \frac{ (3+i)\pm \sqrt{(8+14i)} }{ -2i } \frac{i}{i} = \frac{ (-1+3i)\pm \sqrt{(-8-14i)} }{ 2 } \] just looking at the square root: \[ i \sqrt{(8+14i)} = \sqrt{i^2} \sqrt{(8+14i)} = \sqrt{i^2(8+14i)} \]
phi
  • phi
of course i^2 = -1 so we get \[ \sqrt{-8 -14i}\]
anonymous
  • anonymous
sorry about this but im gonna ask a stupid question. umm for the sqr i^2....i get the sqr bc im multiplying it it with one..right?
phi
  • phi
I don't understand the question. But look at this example \[ \sqrt{12} = \sqrt{4 \cdot 3} = \sqrt{4}\cdot \sqrt{3} = 2 \sqrt{3} \] now go backwards: \[ 2 \sqrt{3} = \sqrt{2^2}\cdot \sqrt{3}= \sqrt{2^2 \cdot 3}= \sqrt{4\cdot 3}= \sqrt{12}\] I used the same idea with \( i \sqrt{stuff} \)
anonymous
  • anonymous
I don't think I ever was shown how to have a number like 2sqr3 and go backwards....thats what got me off bc i didn't know why you were squaring and taking the square root

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