Question

How do you solve the quadratic with complex numbers given `x^2-4x+5=0`?

Answer 1

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

Explanation:

For the equation `ax^2+bx+c=0`, the roots are given by `x=(-b+-sqrt(b^2-4ac))/(2a)`.

It is apparent that if the discriminant `b^2-4ac<0`, we have complex roots.

In the equation `x^2-4x+5=0`, the discriminant is `(-4)^2-4×1×5=16-20=-4<0` and hence roots are complex.

These are `x=(-(-4)+-sqrt(-4))/(2×1)`

= `(4+-2i)/2` i.e.

`x=2+i` and `x=2-i`