(25 + ki)*(3+2i) = 75+3ki+50i-2k
= 75-2k + (3k+50)i
So, you need 3k+50 = 0, since real numbers have zero imaginary part.
For what real number k does the product (25 + ki)*(3+2i) equal a real number?
6 answers
are you sure? I need the actual answer.
well, geez. You cant solve 3k+50 = 0?
k = -50/3
(25 - 50/3 i)*(3+2i) = 325/3
which is a real number
k = -50/3
(25 - 50/3 i)*(3+2i) = 325/3
which is a real number
Uhh 325/3 isn't right
Um, Lelface the answer isnt 325/3. It's -50/3, like steve said earlier. He was just substituting the answer in to show it worked.
it's -50/3
3 answers