SOLUTION: Okay this is my last question. on this it says... {{{(14-2i)/(3+i)}}} I don't know if that makes sense to you But I think I learned something about taking the conjucate of

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: Okay this is my last question. on this it says... {{{(14-2i)/(3+i)}}} I don't know if that makes sense to you But I think I learned something about taking the conjucate of      Log On


   



Question 181600: Okay this is my last question.
on this it says...
%2814-2i%29%2F%283%2Bi%29
I don't know if that makes sense to you
But I think I learned something about taking the conjucate of 3+i which is 3-1, but I dont know where to go from there, or if im even doing it right.

Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!

%2814-2i%29%2F%283%2Bi%29

Form the conjugate of the denominator, 3%2Bi, which is 3-i.

Place 3-i over itself, like this %283-i%29%2F%283-i%29.

Anything divided by itself is just 1, so %283-i%29%2F%283-i%29
equals 1.

When you multiply somthing by 1, it does not change the value.

Therefore we can multiply %2814-2i%29%2F%283%2Bi%29 by %283-i%29%2F%283-i%29
without changing the value.  So let's perform 
that multiplication:

matrix%281%2C3%2C%2814-2i%29%2F%283%2Bi%29%2C+%22%D7%22%2C+%283-i%29%2F%283-i%29%29

Indicate the multiplication of numerators and 
denominator, all as one fraction:

%28%2814-2i%29%283-i%29%29%2F%28%283%2Bi%29%283-i%29%29

Use FOIL to multiply out the numerator and denominators:

++%2842-14i-6i%2B2i%5E2%29%2F%289-3i%2B3i-i%5E2%29+++ 

Combine the like terms in the top.  In the bottom 
the middle two terms cancel, and we have

++%2842-20i%2B2i%5E2%29++%2F++%289-i%5E2%29++

Now we replace both i%5E2's by -1:

%2842-20i%2B2%28-1%29%29%2F%289-%28-1%29%29

%2842-20i-2%29%2F%289%2B1%29

%2840-20i%29%2F10

Make two fractions:

40%2F10-20i%2F10   

4-+2i%29

Edwin