write the quadratic equation having the following as roots: a.) 6-2i and 6+2i b.) 3 and 5

There are two ways to construct a quadratic equation if  we are
given the roots x1 and x2.

First method: (x-x1)(x-x2) = 0 and expand
the left to get the equation whose roots are x1 and x2.

Second
method : We assume  ax^2+bx+c = 0 is the equation and then by the relations between the roots and
corfficients x1+x2 = -b/c and x1x2 = c/a, Now solve for b and c , choosing a =
1.

a) By first method.

x1 = 6-2i and x2
= 6+2i.

Therefore (x-1)(x-x2) =
0.

(x-(6-2i))(x+(x+(6+2i)) = 0.

{(x-6)
+2i}{x-6-2i)} = 0.

(x-6)^2 - (2i)^2 =
0.

x^2 -12x+36 +4 = 0.

x^2 -12x +40 = 0
is the equation.

b) By second
method:

Let ax^2+bx+c= 0 be the
equation:

x1=3 and x2 = 5.

x1+x2 = -b/a
and x1x2 = c/a.

3+5 = -b/a and  3*5 =
c/a.

b= -8a and c= 15a.

Put a= 1. Then 
b= -8a= -8, c = 15a = 15*1 = 15.

Therefore ax^2+bx+c = 0 becomes 
1x^2-8x+15 = 0. Or

x^2-8x+15 = 0 is the equation which has the given
roots 3 and 5.