anonymous
  • anonymous
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!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
i like your picture :D
anonymous
  • anonymous
Haha, thanks! :D Yours is rather adorable, as well!
zzr0ck3r
  • zzr0ck3r
3i(2+5i) = 6i+15i^2 = 6i+15(-1) = -15+6i

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zzr0ck3r
  • zzr0ck3r
so if you are talking about the form x+yi then x = -15
zzr0ck3r
  • zzr0ck3r
understand?
anonymous
  • anonymous
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.
anonymous
  • anonymous
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
anonymous
  • anonymous
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.
anonymous
  • anonymous
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...
anonymous
  • anonymous
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.
anonymous
  • anonymous
follow @zzr0ck3r steps...they show how to group the i terms and replace them
zzr0ck3r
  • zzr0ck3r
i=i i^2 = -1 i^3 = -i i^4 = 1 i^5 = i . . .
anonymous
  • anonymous
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
anonymous
  • anonymous
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!

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