Can somebody solve in C?

#(ix-5)(6+2ix)(5-2i)=0#

1 Answer
Jan 10, 2018

A product is equal to #0# if and only if at least one of the factors is #0#. The left side is already factored for us.

Explanation:

Solve #(ix-5)=0# or #(6+2ix) = 0# (It is clear that #(5-2i)# cannot be #0#)

#x = 5/i = -5i# or #x = (-6)/(2i) = (-3)/i = 3i#

Check the answers by substitution.