Question

Does anyone understand imaginary numbers enough to explain them to me??
I really need to pass but I don't have a clue :(

Answer 1

Well let's find out :)
Do you have something specific that you're having a hard time understanding?

Answer 2

Well, might as well cover the basics :D
The root of all imaginary numbers is the constant i, defined as such...
\[\Large i^2 = -1\]or
\[\Large i = \sqrt{-1}\]

Answer 3

Well for example (3+7x)(2i-19) and to factor it is really hard to understand

Answer 4

that's x? not i?

Answer 5

sorry meant i

Answer 6

Really stressed out

Answer 7

Okay. Don't forget that the constant i behaves like any normal variable. So you can FOIL it as you normally do :)

Answer 8

Thank you :)

Answer 9

That was it?
Could you post your answer? :)

Answer 10

Not exactly :/

Answer 11

I understand it but yet don't if that makes any sense.

Answer 12

FOIL method.
could you multiply these binomials?
\[\Large (7x+3)(2x-19)\]

Answer 13

Remember:\[\Large (ax+b)(cx+d)= acx^2 + (bc+da)x + bd\]

Answer 14

I can barely factor or do any algebraic thing...I was just passed to the next grade without any help and now im getting help with it and it is so confusing

Answer 15

Well you need to practice :)
What say we take an alternative way. Distributing.
(3+7i)(2i-19)

Let's distribute (3+7i) over (2i - 19)

You get
2i(3+7i) - 19(3+7i)

yes?

Answer 16

Wait how exactly did you get the 2i and the 19 outside of the parenthese

Answer 17

:)
If it were
a(2i - 19)
You wouldn't hesitate to distribute the a over the (2i - 19) right? :)
2i(a) - 19(a)

Answer 18

yes

Answer 19

Well, you can treat the (3+7i) as the "a" and distribute it over the
2i - 19 all the same :)

Answer 20

That sounds complicated

Answer 21

but it isn't :)
Just replace (a) with (3+7i)

Answer 22

okay so if i were given an equation equal to zero and told to finnd the intercepts how would I if they involve i?

Answer 23

example?

Answer 24

Thank you for everything :)

Answer 25

O_o