We'll recall the fact that a quadratic equation has two zeroes. We notice that the given zero is a complex number and we know that the complex roots come in pairs. Therefore, we'll determine the other zero of the quadratic to be found as being the conjugate of the given zero, namely 2 + 2i.
We'll write the factor form of a quadratic:
(x - x1)(x - x2) = 0, where x1 and x2 represent the zeroes of the quadratic.
We'll substitute x1 and x2 by the given roots:
x1 = 2 -2i and x2 = 2 + 2i
(x - 2 + 2i)(x - 2 - 2i) = 0
We notice that the product above represents a special product that returns a difference of two squares:
`(x - 2 + 2i)(x - 2 - 2i) = (x-2)^2 - (2i)^2`
We'll expand the binomial:
`(x - 2 + 2i)(x - 2 - 2i) = x^2 - 4x + 4 - 4i^2`
But `i^2 = -1`
`(x - 2 + 2i)(x - 2 - 2i) = x^2 - 4x + 4 + 4`
`(x - 2 + 2i)(x - 2 - 2i) = x^2 - 4x + 8`
The requested quadratic, whose roots are 2 - 2i and 2 + 2i, is `x^2 - 4x + 8 = 0.`