Simplify the following radical expression.
$\sqrt{108 s^2 t^{15}}$
We need to simplify the expression \(\sqrt{108 s^2 t^{15}}\). A radical expression \(\sqrt{X}\) is simplified by factoring out the largest perfect square from the radicand (the expression under the square root symbol). We treat the numerical coefficient, \(108\), and the variables \(s^2\) and \(t^{15}\) separately. We assume the variables \(s\) and \(t\) are such that the expression is defined (i.e., \(t \ge 0\)).
First, we find the prime factorization of \(108\):
\[108 = 2 \times 54 = 2^2 \times 27 = 2^2 \times 3^3 = (2^2 \times 3^2) \times 3 = 36 \times 3.\]
Since \(36\) is a perfect square (\(6^2\)), we can write \(\sqrt{108} = \sqrt{36 \times 3} = 6\sqrt{3}\).
Next, we simplify the variable part \(\sqrt{s^2 t^{15}}\).
For the variable \(s^2\), the square root is \(\sqrt{s^2} = |s|\).
For the variable \(t^{15}\), we can rewrite the exponent as an even number to extract a perfect square: \(t^{15} = t^{14} \times t\).
Then, \(\sqrt{t^{15}} = \sqrt{t^{14} \times t} = \sqrt{(t^7)^2 \times t} = t^7\sqrt{t}\).
Finally, we combine all the simplified components:
\[\sqrt{108 s^2 t^{15}} = \sqrt{36 \times 3 \times s^2 \times t^{14} \times t}\]
\[= \sqrt{36} \times \sqrt{s^2} \times \sqrt{t^{14}} \times \sqrt{3 \times t}\]
\[= 6 \times |s| \times t^7 \times \sqrt{3t}\]
\[= 6|s|t^7\sqrt{3t}.\]
Thus, the simplified form of the radical expression is \(6|s|t^7\sqrt{3t}\).
Simplifying radicals is a fundamental skill used in physics and engineering. For example, when calculating the time it takes for an object to fall a certain distance \(d\) under gravity \(g\), the formula involves a square root: \(t = \sqrt{2d/g}\). If you have complex expressions for distance, simplifying the radical allows you to isolate the variables, making it much easier to perform calculations or interpret the physical behavior of the system. Understanding these algebraic manipulations is key to mastering scientific formulas.