Hello! I'm having trouble with this equation 
because I don't know how to solve for "k". 
Here is the equation. 
If 1+2i is a root of x^2-kx+6=0, find k. 
Can you help me out? 
Sincerely,
Leanne Spence


Substitute 1+2i for x and solve for k:

                    x² - kx + 6 = 0

          (1+2i)² - k(1+2i) + 6 = 0

     (1+2i)(1+2i) - k - 2ik + 6 = 0

1 + 2i + 2i + 4i² - k - 2ik + 6 = 0

         7 + 4i + 4i² - k - 2ik = 0

Since i² = -1 replace i² by -1

       7 + 4i + 4(-1) - k - 2ik = 0

           7 + 4i - 4 - k - 2ik = 0

               3 + 4i - k - 2ik = 0

Get all the k terms on the right

                         3 + 4i = k + 2ik

Factor k out on the right

                         3 + 4i = k(1 + 2i)

Divide both sides by 1 + 2i

                         3 + 4i  
                        -------- = k
                         1 + 2i
     
To divide out the two complex numbers, 
multiply top and bottom by conjugate
of bottom, 1 - 2i

                (3 + 4i)(1 - 2i)
               ------------------ = k
                (1 + 2i)(1 - 2i)
                           
               3 - 6i + 4i - 8i²
             -------------------- = k
               1 - 2i + 2i - 4i²

                    3 - 2i - 8i²
                   -------------- = k
                      1 - 4i²              

Replace i² by -1

                  3 - 2i - 8(-1)
                 ---------------- = k
                    1 - 4(-1) 

                      3 - 2i + 8
                     ------------ = k
                         1 + 4

                         11 - 2i 
                        --------- = k
                            5 
or in A + Bi form in fractions

       k = (11/5) - (2/5)i

or in A + Bi form in decimals

       k = 2.2 - 0.4i 

Edwin