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

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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 ?

- anonymous

finding the roots of the second problem i needed

- anonymous

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

- anonymous

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.

- anonymous

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

- anonymous

i got this so far

- phi

I think it should be -2i in the bottom

- anonymous

sorry

- phi

no need to be sorry, just careful.

- phi

if you multiply top and bottom by i you can get
\[ \frac{ (3+i)\pm \sqrt{(8+14i)} }{ -2i } \frac{i}{i} =
\frac{ (-1+3i)\pm \sqrt{(-8-14i)} }{ 2 }
\]

- anonymous

what do i do about the square root

- anonymous

wait y did 8 become negative

- 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

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

- 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

Re(z)=-8

- anonymous

Im(z)=-14

- 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

why only the sign...why not the number it self?

- anonymous

so will it always be + or - 1 then

- phi

yes, the formula uses the sign(Im(z)) to determine the sign of the imaginary part of the root.

- anonymous

oh ok..go on....

- phi

we need |z| = sqr(a^2 + b^2) where z= a+ bi

- anonymous

16.12

- anonymous

or sqr260

- 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

i get 2.015 -2.015i

- anonymous

sorry, 3.47i

- 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

after simplifying compare to wolfram's results
http://www.wolframalpha.com/input/?i=-ix%5E2-%283%2Bi%29x%2B2%3D0

- phi

In the box "Complex Solutions" click on Approximate Form to get the decimal answer.

- anonymous

yep. right...how do i do it the other way

- 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

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

ok. so x=4.06 and x=-12.1

- 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

yea...sorry so use to writing x

- phi

you should use a few more decimals

- anonymous

ok. so after i get u ...what do i do?

- phi

we want x and y (x + i y is the square root we are looking for)

- phi

x= sqr(u)
y= -7/x

- anonymous

ok. so (2.015, -3.474i)

- phi

which should match what we got before.

- anonymous

yep. thank you phi....so so much. You are a big help.

- phi

yw

- anonymous

@phi

- anonymous

i need ur help with something

- anonymous

@zepdrix

- anonymous

could i ask u a simple question

- anonymous

@phi could you just answer a simple question that i didn't ask yesterday

- 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

\[ \frac{ (3+i)\pm \sqrt{(8+14i)} }{ -2i } \frac{i}{i} =
\frac{ (-1+3i)\pm \sqrt{(-8-14i)} }{ 2 } \]
just looking at the square root:
\[ i \sqrt{(8+14i)} = \sqrt{i^2} \sqrt{(8+14i)} = \sqrt{i^2(8+14i)} \]

- phi

of course i^2 = -1
so we get
\[ \sqrt{-8 -14i}\]

- 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

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

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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