Show that a=3 and find the value of b.

Question

az²+bz+10i=0 where a and b are real, has a root of 3-i. Show that a=3 and find the value.

I know that 3+i is also a root and I would be able to find the value of b if I could show that a is equal to 3.

Please help.

Answer 1

Since 3-i is a root, a(3-i)²+b(3-i)+10i = 0

 

So, a(9-6i-1)+3b-bi+10i = 0

 

     (8a+3b) + (-6a-b+10)i = 0

 

Therefore, 8a+3b = 0  and -6a-b+10 = 0

 

We have the system of equations:  8a+3b = 0

                                                  6a+ b = 10

 

Multiply the second equation by -3:  8a+3b = 0

                                                 -18a-3b = -30

 

Adding the equations, we get -10a = -30.  So, a = 3

 

Since a = 3 and 6a+b = 10, we obtain 18+b = 10

                                                           b = -8

 

Solution:  a = 3, b = -8