Math question

Indeed you did, you had answered before i had even clicked on the question

The quadratic expression x2 - 2x + 15 can be factored to give x2 - 2x - 15 = (x - 5)(x + 3). Thus the quadratic equation x2 - 2x + 15 = 0 has roots x = 5 and x = -3. You can perform this from the end to the beginning.

One quadratic equation with roots 5 and -3 is (x - 5)(x + 3) = 0. Expanding this gives x2 - 2x + 15 = 0.

What I am saying is a quadratic equation with roots a and b is (x - a)(x - b) = 0. This fact is true, even ifa and b are complex numbers. Hence one quadratic equation with roots 4 - i and 4 + i is

[x - (4 - i)][x - (4 + i)] = 0.

Expand the left side.

When you are finished you can use the quadratic formula to verify that 4 - i and 4 + 1i are roots.

Man, I typed fast!!

Is that right?

Let's expand upon Hayley's answer [it is correct  !!!]

[x - (4 - i)][x - (4 + i)] = 0

x^2 - x (4 - i) - x (4 + i) +  [4 - i] [ 4 + i]  =

x^2 - 4x + ix - 4x - ix  + 16 -4i + 4i - i^2  =

x^2 - 8x + 16  + 1  =

x^2 - 8x + 17