Question

How do I find all real and complex zeros of `x^2-4x+7`?

Answer 1

Use the quadratic formula to find zeros occur for:

`x = 2+-i sqrt(3)`

Explanation:

The quickest way is probably to use the quadratic formula.

`x^2-4x+7` is in the form `ax^2+bx+c`, with `a=1`, `b=-4` and `c=7`.

So its zeros occur for:

`x = (-b+-sqrt(b^2-4ac))/(2a) = (-(-4)+-sqrt((-4)^2-(4xx1xx7)))/(2xx1)`

`=(4+-sqrt(16-28))/2 = (4+-sqrt(-12))/2 = (4+-i sqrt(12))/2`

`=(4+-i*2sqrt(3))/2 = 2+-i sqrt(3)`

Alternatively, you can complete the square as follows:

`0 = x^2-4x+7 = x^2-4x+4+3 = (x-2)^2+3`

Subtract `3` from both ends to get:

`(x-2)^2 = -3`

So:

`x-2 = +-sqrt(-3) = +-i sqrt(3)`

Add `2` to both sides to get:

`x = 2 +-i sqrt(3)`