Math again ....?

Question

find the complex solutions of the following quadratic equations. write the solutions in standard form.

z^2 - 4z + 5 = 0

and
(4)b^2 = 2b - 1

Answer 1

z^2 - 4z + 5 = 0

We can either use the quadratic formula or complete the square. I prefer to complete the square.

z^2 - 4z + 4 + 1 = 0
(z - 2)^2 = -1
z - 2 = +/- i
z = 2 +/- i, so
z = {2 + i, 2 - i}

2) 4b^2 = 2b - 1

Going to complete the square again.

4b^2 - 2b + 1 = 0
4( b^2 - (1/2)b ) + 1 = 0
4( b^2 - (1/2)b + 1/16) + 1 - 4/16 = 0
4(b - 1/4)^2 + 4/4 - 1/4 = 0
4(b - 1/4)^2 + 3/4 = 0
4(b - 1/4)^2 = -3/4
(b - 1/4)^2 = -3/16
b - (1/4) = +/- sqrt(3)/4 i
b = (1/4) +/- sqrt(3)/4 i

Answer 2

z^2 - 4z + 5 = 0
(z-5)(z+1)= 0
z = (5,-1)

(4)b^2 = 2b - 1
4b^2-2b+1 = 0
x= [2 +/- sqrt(2^2 -16)]/8
x = 1/4 +/- 1/4 sqrt(3)i