Silly question: Why is $\sqrt{(9x^2)} $ not $3x$?

$\sqrt{x^2}=x$ if and only if $x\ge 0$; if $x<0$, then $\sqrt{x^2}=-x$. The correct general formula is $\sqrt{x^2}=|x|$.

Hint: We always know that: $$\sqrt{a^2x^2}=|ax|$$

Remember that $\sqrt{x^2}$ is the positive number whose square is $x^2$ (this is the definition of the square root sign). If $x$ is positive, then that number is $x$, otherwise, it is $-x$. For example, $\sqrt{(-3)^2} = \sqrt{9} = 3$, not $-3$. You can always say that $\sqrt{x^2} = \left| x \right|$.