Question

How do you find the quadratic equation with roots at (2-i) and its complex conjugate?

Answer 1

Multiply out `(x-(2-i))(x-(2+i))` and simplify to find equation:

`x^2-4x+5 = 0`

Explanation:

The Complex conjugate of `2-i` is `2+i`. You get this by inverting the sign of the imaginary part.

Multiply out `(x-(2-i))(x-(2+i))` using the difference of squares identity:

`a^2-b^2 = (a-b)(a+b)`

with `a=(x-2)` and `b=i` as follows:

`(x-(2-i))(x-(2+i))`

`=((x-2)-i)((x-2)+i)`

`=(x-2)^2-i^2`

`=x^2-4x+4+1`

`=x^2-4x+5`