Quadratic Inequality of perfect square trinomial

Question

Need to solve the following and write the answer in interval notation.

$x^2-6x+9<0$

I know that factors to $(x-3)^2<0$. I presume I need to take the sqrt of both sides but I'm stuck from that point.

Answer 1

Do case analysis for two different cases (I am going to use the example you have given).

Case 1: $(x-3)$ is positive. Then $x-3 \lt 0$ (Divide both sides by $x-3$). But wait! $x-3$ is said to be positive, which means $x-3 \gt 0$. Then $x-3$ can't be positive. That means no solution for case 1.

Case 2: $(x-3)$ is negative. Then $x-3 \gt 0$ (Divide both sides by $x-3$). But $x-3$ is said to be negative, so no solution for this case too. $x-3 = 0$ also won't work since $0$ is not less than 0.

This happens to be a example where there is no solution. In this particular case, this can be achieved faster: set $x-3 = u$. Then $u^2 \lt 0$. Since squaring a real number can never be negative, there's no solution.

The general approach to these kinds of problem is as follows: separate the problem into cases and analyze those case by case.