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Mathematics 23 Online
OpenStudy (anonymous):

1. For i = −1, if 3i(2 + 5i) = x + 6i, then x = ? A. –15 B. 5 C. 5i D. 15i E. 27i please explain. :D

jimthompson5910 (jim_thompson5910):

What do you get when you distribute 3i(2 + 5i)

OpenStudy (anonymous):

5i+15i?

OpenStudy (anonymous):

6i* i ment

jimthompson5910 (jim_thompson5910):

6i is correct, but it should be 15i^2 since i*i = i^2

jimthompson5910 (jim_thompson5910):

so it's really 6i + 15i^2

jimthompson5910 (jim_thompson5910):

but what is i^2 really?

OpenStudy (anonymous):

1?

jimthompson5910 (jim_thompson5910):

close

jimthompson5910 (jim_thompson5910):

\[\Large i = \sqrt{-1}\] square both sides

OpenStudy (anonymous):

i do not know.

jimthompson5910 (jim_thompson5910):

squaring a square root effectively undoes it

jimthompson5910 (jim_thompson5910):

and they cancel

jimthompson5910 (jim_thompson5910):

\[\Large i = \sqrt{-1}\] \[\Large i^2 = \left(\sqrt{-1}\right)^2\] \[\Large i^2 = -1\]

OpenStudy (anonymous):

then?

jimthompson5910 (jim_thompson5910):

so 15i^2 is really 15(-1) = -15

OpenStudy (anonymous):

ohh

jimthompson5910 (jim_thompson5910):

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

jimthompson5910 (jim_thompson5910):

3i(2 + 5i) = x + 6i -15 + 6i = x + 6i I'm sure you can see what x is

OpenStudy (anonymous):

Thank you.

jimthompson5910 (jim_thompson5910):

yw

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