SOLUTION: Okay this is my last question. on this itsays... 14-2i Over 3+i. I dont know if that makes sense to you But i think i learned something about taking the conjucate of 3+i w

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: Okay this is my last question. on this itsays... 14-2i Over 3+i. I dont know if that makes sense to you But i think i learned something about taking the conjucate of 3+i w      Log On


   

Question 181601: Okay this is my last question.
on this itsays...
14-2i Over 3+i.
I dont 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.
Found 2 solutions by stanbon, ankor@dixie-net.com:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
14-2i Over 3+i.
-------------
= [(14-2i)(3-i)] / [(3+i)(3-i)]
(42-14i-6i-2) / (9+1}
= (40-20i)/10
= (20 - 10i)/ 5
= 4 - 2i
=============
Cheers,
Stan H.

Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
%28%2814-2i%29%29%2F%28%283%2Bi%29%29
you multiply it by the conjugate of the denominator over itself
%28%2814-2i%29%29%2F%28%283%2Bi%29%29 * %28%283-i%29%29%2F%28%283-i%29%29 = %28%2842-14i-6i%2B2i%5E2%29%29%2F%28%289%2B3i-3i-i%5E2%29%29 = %28%2842-20i%2B2i%5E2%29%29%2F%28%289-i%5E2%29%29
i^2 = -1, therefore
%28%2842-20i%2B2%28-1%29%29%29%2F%28%289-%28-1%29%29%29 = %28%2842-20i-2%29%29%2F%28%289%2B1%29%29 = %28%2840-20i%29%29%2F%2810%29
Divide 10 into both terms
4 - 2i