Question

Problem: For i=√-1, if 3i(2+5i) = x+6i, then x=?
Answer: -15

My work using -1:
3(-1)(2+5(-1)) = x+6(-1)
-3(2-5) = x-6
-3(-3) = x-6
9 = x-6
[+6]
x=15

My work using +1:
3(1)(2+5(1)) = x+6(1)
3 (2+5) = x+6
3(7) = x+6
21 = x=6
[-6]
x=15



I keep on getting +15. What am I doing wrong?

Also, I would seriously appreciate it if someone could sort of explain this to me? Particularly the (√-1). Thanks in advance!

Answer 1

i like your picture :D

Answer 2

Haha, thanks! :D Yours is rather adorable, as well!

Answer 3

3i(2+5i) = 6i+15i^2 = 6i+15(-1) = -15+6i

Answer 4

so if you are talking about the form x+yi then x = -15

Answer 5

understand?

Answer 6

Sort of, but not really. I understand what you're saying in sections (ex, I now know where you got 6i+15i^2 from, but everything after? And putting it all together? Not really. Why add (-1) to to 15 but not 6? What I mean is, why 15(-1) but not 6(-1). Why do you leave the latter as 6i?

Also, you mentioned the form x+yi, but I'm not really seeing where this fits into everything (aside from the second half of the equation (x+6i). Am I making sense? I seriously apologize if I'm not (or if I'm just asking stupid questions). This is just really confusing.

Answer 7

You're directly subbing in -1 for you i terms. You must wait until you have squared terms to substitute:
\[i = \sqrt{-1}\]
\[i^2 = -1\]

Expand your fractions first and 'keep' the i terms (treat them as any variable)
then when you see a i^2, replace it directly with a -1

Answer 8

Once again I'm in a 'sort of, but not really' situation of understanding, but I think at this point it's more on me than you guys. I'm going to go back and re-read the passage on complex numbers and whatnot before attempting this again/publically. Sorry for the inconvenience.

Answer 9

It's not inconvenient. By showing what steps you attempted and where you got mixed up it's pretty clear what your having a problem with. Let me try and find a table that might clear things up...

Answer 10

Oh, I got it! I was focusing so much on the first half of the equation that I was ignoring the second half. The √-1 was really throwing me off, but in the end you solve if like you'd do any other...ah, I forgot the term for this setup.

Answer 11

follow @zzr0ck3r steps...they show how to group the i terms and replace them

Answer 12

i=i
i^2 = -1
i^3 = -i
i^4 = 1
i^5 = i
.
.
.

Answer 13

Anyway:
6(√-1)+15(√-1) = x+6(√-1)
6*-1+15*-1 = x-6
-6+-15 = x-6
-21 = x-6 = x-6
[+6]
X=-15

Answer 14

I *think* I got it now. I'm going to need a lot more practice, though, but this makes a lot more sense. Thanks so much @Mathmuse and @zzr0ck3r - you were both immensely helpful.

I'm copying everything down now so I can study it further. Once again, thanks so much!