Remember that the square root of a perfect square can be simplified using absolute value.

\[ \sqrt{x^2} = |x| \]

Therefore,

\[ \sqrt{p^2} = |p|. \]

Remember that the square root of a perfect square can be simplified using absolute value.

\[ \sqrt{x^2} = |x| \]

Also, we notice that $9$ and $a^2$ are both perfect squares.

First, we write the expression under the square root as a perfect square:

\begin{align*} \sqrt{9a^2} &= \sqrt{3^2\cdot a^2} \\[5pt] &=\sqrt{3\cdot 3\cdot a\cdot a}\\[5pt] &=\sqrt{3\cdot a\cdot 3\cdot a}\\[5pt] &=\sqrt{3a \cdot 3a} \\[5pt] &=\sqrt{(3a)^2} \end{align*}

Then, we take the absolute value of the squared quantity and simplify using the rules of absolute value:

\begin{align*} \sqrt{(3a)^2}&= |3a| \\[5pt] &=|3|\cdot |a| \\[5pt] &=3|a| \end{align*}

Remember that the square root of a perfect square can be simplified using absolute value.

\[ \sqrt{x^2} = |x| \]

Also, we notice that $p^2$ and $q^2$ are perfect squares.

First, we write the expression under the square root as a perfect square:

\begin{align*} \sqrt{p^2q^2} &=\sqrt{p\cdot p\cdot q\cdot q}\\[5pt] &=\sqrt{p\cdot q\cdot p\cdot q}\\[5pt] &=\sqrt{(pq)\cdot (pq)}\\[5pt] &=\sqrt{(pq)^2} \end{align*}

Then, we take the absolute value of the squared quantity:

\begin{align*} \sqrt{(pq)^2} &=|pq| \end{align*}

Remember that the square root of a perfect square can be simplified using absolute value.

\[ \sqrt{x^2} = |x| \]

Also, we notice that $81, a^2,$ and $b^2$ are perfect squares.

First, we write the expression under the square root as a perfect square:

\begin{align*} \sqrt{81a^2b^2} &=\sqrt{9^2\cdot a^2\cdot b^2}\\[5pt] &=\sqrt{9\cdot 9\cdot a\cdot a\cdot b\cdot b}\\[5pt] &=\sqrt{9\cdot a\cdot b\cdot 9\cdot a\cdot b}\\[5pt] &=\sqrt{(9ab)\cdot (9ab)}\\[5pt] &=\sqrt{(9ab)^2} \end{align*}

Then, we take the absolute value of the squared quantity and simplify using the rules of absolute value:

\begin{align*} \sqrt{(9ab)^2} &=|9ab|\\[5pt] &=|9|\cdot |ab| \\[5pt] &=9|ab| \end{align*}

Remember that the square root of a perfect square can be simplified using absolute value.

\[ \sqrt{x^2} = |x| \]

Also, we notice that $t^2$ and $16$ are both perfect squares.

First, we write the expression under the square root as a perfect square:

\begin{align*} \sqrt{\dfrac{t^2}{16}} &=\sqrt{\dfrac{t^2}{4^2}} \\[5pt] &=\sqrt{\dfrac{t\cdot t}{4\cdot 4}} \\[5pt] &=\sqrt{\dfrac{t}{4}\cdot \dfrac{t}{4}} \\[5pt] &=\sqrt{\left(\dfrac{t}{4}\right)^2} \\[5pt] \end{align*}

Then, we take the absolute value of the squared quantity and simplify using the rules of absolute value:

\begin{align*} \sqrt{\left(\dfrac{t}{4}\right)^2} &=\left|\dfrac{t}{4}\right| \\[5pt] &=\dfrac{|t|}{|4|}\\[5pt] &=\dfrac{|t|}{4} \end{align*}

Remember that the square root of a perfect square can be simplified using absolute value.

\[ \sqrt{x^2} = |x| \]

Also, we notice that $p^2, 81,$ and $q^2$ are all perfect squares.

First, we write the expression under the square root as a perfect square:

\begin{align*} \sqrt{\dfrac{p^2}{81q^2}} &=\sqrt{\dfrac{p^2}{9^2\cdot q^2}} \\[5pt] &=\sqrt{\dfrac{p\cdot p}{9\cdot 9\cdot q\cdot q}} \\[5pt] &=\sqrt{\dfrac{p\cdot p}{9\cdot q\cdot 9\cdot q}} \\[5pt] &=\sqrt{\dfrac{p}{9q}\cdot \dfrac{p}{9q}} \\[5pt] &=\sqrt{\left(\dfrac{p}{9q}\right)^2} \end{align*}

Then, we take the absolute value of the squared quantity and simplify using the rules of absolute value:

\begin{align*} \sqrt{\left(\dfrac{p}{9q}\right)^2} &=\left|\dfrac{p}{9q}\right| \\[5pt] &=\dfrac{|p|}{|9q|} \\[5pt] &=\dfrac{|p|}{|9| \cdot |q|} \\[5pt] &=\dfrac{|p|}{9|q|} \end{align*}