1. The problem asks to evaluate the square root of 9, written as $\sqrt{9}$.\n\n2. The square root of a number $x$ is a value that, when multiplied by itself, gives $x$. In other words, if $y = \sqrt{x}$, then $y^2 = x$.\n\n3. For perfect squares like 9, the square root is an integer. Since $3 \times 3 = 9$, we have $\sqrt{9} = 3$.\n\n4. Therefore, the value of $\sqrt{9}$ is 3.