Question

For what real numbers x is `x^2-6x+9` negative?

Answer 1

`x^2-6x+9 = (x-3)^2`

This is never negative using real numbers for `x`

(It is `0` for `x=3` and positive for every other real number value of `x`.)

Answer 2

The are none.

Method
To figure that out you let `x^2-6x+9<0`

factorize `x^2-6x+9`

By completing the square,
`x^2 -6x + (-6/2)^2-(-6/2)^2+9`
=`(x-3)^2-9+9`
=`(x-3)^2`

Hence,
`(x-3)^2<0`

Since the square of any real number is always positive there are really no values of `x` for which that expression is negative!