Question

Pre-Calculus // Complex Zeros and the FTA
http://prntscr.com/aahswg
Help please? Where am I doing this wrong?

Answer 1

hehhe

Answer 2

so.. I assume you used the quadratic formula to get the roots?

Answer 3

well, gimme a sec to post something

Answer 4

ok I updated the Q now delete those comments lol

Answer 5

eheh

Answer 6

>_>

Answer 7

deleteeeeeee qq

Answer 8

\(\textit{quadratic formula}\\
{\color{blue}{ 1}}x^2{\color{red}{ +1}}x{\color{green}{ +1}}=0
\qquad \qquad
x= \cfrac{ - {\color{red}{ 1}} \pm \sqrt { {\color{red}{ 1}}^2 -4{\color{blue}{ (1)}}{\color{green}{ (1)}}}}{2{\color{blue}{ (1)}}}
\\ \quad \\
x=\cfrac{-1\pm\sqrt{1-4}}{2}\implies x=\cfrac{-1\pm\sqrt{-3}}{2}
\\ \quad \\
x=\cfrac{-1\pm\sqrt{3}\cdot \sqrt{-1}}{2}\implies x=\cfrac{-1\pm\sqrt{3}\cdot \ {\color{brown}{ i}}}{2}
\\ \quad \\
x=
\begin{cases}
-\cfrac{1}{2}+\cfrac{\sqrt{3}\ i}{2}\\
-\cfrac{1}{2}-\cfrac{\sqrt{3}\ i}{2}
\end{cases}\)

Answer 9

; A; wah so much information

Answer 10

...WAIT THAT IS FOR THE OLD PROBLEM WHAT THE HECK >_>

Answer 11

hmmm ok.. thought it was the same.

Answer 12

for the new one
well, notice above the quadratic formula
you ended up with a complex result BUT notice the +/- signs
complex values do not come alone by themselves
the come in pairs

Answer 13

ayy... delete all the old ones i am so confused now lol

Answer 14

anyhow, for your new "edited" posting
if P has a solution of a + bi
a + bi doesn't come alone by herself, she comes with her sister, the conjugate
a - bi

Answer 15

oh, so dual answer? \[\pm(a+bi)\]

Answer 16

oops, mistake... i meant\[a\pm bi\]

Answer 17

:|

Answer 18

@hartnn

Answer 19

\(\color{#0cbb34}{\text{Originally Posted by}}\) @jdoe0001
anyhow, for your new "edited" posting
if P has a solution of a + bi
a + bi doesn't come alone by herself, she comes with her sister, the conjugate
a - bi
\(\color{#0cbb34}{\text{End of Quote}}\)


if a+bi is one root, then a-bi is also one root!
given the co-efficients are real.

Answer 20

hmm... I think I tried that and got "wrong answer" but I will try it again

Answer 21

ya, still wrong :-\

Answer 22

http://prntscr.com/aaikcs

Answer 23

why are you putting a+bi?

Answer 24

a+bi is already there
just put
`a-bi`

Answer 25

I did... see link? :-(

Answer 26

i see :
`a+bi,a-bi`

we are asking you to input
`a-bi`

Answer 27

just a-bi?

Answer 28

yes

Answer 29

... now it is right... i am so confused lol

Answer 30

so is the normal format, "a-bi" instead of "a+bi" for complex numbers ??

Answer 31

its `a+bi`
only.

ex: if `3+4i` is one of the zeros, then `3-4i` is also a zero.

Answer 32

mmm... is it because (x-(a+bi)) becomes (x-a-bi)??

Answer 33

(x-(a+bi)) does becomes (x-a-bi)
but thats not the reason

Answer 34

oh... ok... then idk lol

Answer 35

thats because when we multiply the 2 factors, a+bi, a-bi
only then the co-efficients become real!

\((x- (a+bi)) (x-(a-bi))\)

if you try multiplying a+bi, -a+bi then the co-efficients will not be real.

Answer 36

???
are you saying that one of the values for | a + bi | doesn't work? o_o

Answer 37

where does |a+bi| come into picture?

Answer 38

oh i meant \[\pm(a+bi)\]sorry about that

Answer 39

only changing the sign of imaginary part works.

a+bi, a-bi

-a+bi, -a-bi

not any other combination

Answer 40

oh... ok I am still confused lol
but it's ok I will ask my teacher. I got the gist/most of it, it's fine

Answer 41

thank you :-D !!

Answer 42

good luck :)
welcome ^_^

Answer 43

well.. I have more Q... can I post them on the other one .. ^-^;

Answer 44

sure...there are many others who will help you :)
i gotta go, have a nice day.

Answer 45

D-: but you are the best at helping...

Answer 46

already tagged you... lol sorry

Answer 47

i hihly doubt that, but thanks for the compliment :P

Answer 48

noo.... never seen a problem where you did not help me when i asked... lol