SOLUTION: I've never been good at math at all. I am in a college algebra class and need help doing quadratic equations and inequalitites. The question says to find each of the following quot

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Question 345278: I've never been good at math at all. I am in a college algebra class and need help doing quadratic equations and inequalitites. The question says to find each of the following quotients and then express the answers in the standard form of a complex number. The problem is:
(5+i)/(2+9i)
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!

%285%2Bi%29%2F%282%2B9i%29

The denominator is 2%2B9i.  Form the conjugate by
changing the sign only the term that contains the letter
i.

So the conjugate of red%282%2B9i%29 is red%282-9i%29

Now put that conjugate over itself: red%28%282-9i%29%2F%282-9i%29%29 

That just equals 1 because any number (other than 0) divided by
itself is 1.  So we multiply the original problem by that:

expr%28%285%2Bi%29%2F%282%2B9i%29%29%2Ared%28expr%28%282-9i%29%2F%282-9i%29%29%29

%28%285%2Bi%29red%28%282-9i%29%29%29%2F%28%282%2B9i%29red%28%282-9i%29%29%29

Use FOIL on the top and the bottom:

%2810-45i%2B2i-9i%5E2%29%2F%284-18i%2B18i-81i%5E2%29

Simplify:

%2810-43i-9i%5E2%29%2F%284-81i%5E2%29

Now use the fact that since i=sqrt%28-1%29, the i%5E2=-1
So we replace i%5E2 by -1

%2810-43i-9%28-1%29%29%2F%284-81%28-1%29%29

%2810-43i%2B9%29%2F%284%2B81%29

%2819-43i%29%2F85

Make two fractions

19%2F85-expr%2843%2F85%29i

Edwin