Simplify.
$3 \sqrt{27}+\sqrt{48}$
We are tasked to simplify the radical expression \(3\sqrt{27} + \sqrt{48}\). The goal is to express this in the form \(a\sqrt{b}\), where \(a\) is an integer and \(b\) is a square-free integer. To achieve this, we will simplify each radical term individually by extracting the largest possible perfect square factor from under the square root. Recall that \(\sqrt{x \cdot y} = \sqrt{x} \cdot \sqrt{y}\) for non-negative \(x\) and \(y\).
First, let's look at the term \(\sqrt{27}\). We can factor 27 as \(9 \times 3\). Since 9 is a perfect square (\(3^2\)), we have:
\[\sqrt{27} = \sqrt{9 \times 3} = \sqrt{9} \times \sqrt{3} = 3\sqrt{3}.\]
Now, substitute this back into the first part of our original expression:
\[3\sqrt{27} = 3 \times (3\sqrt{3}) = 9\sqrt{3}.\]
Next, let's look at the term \(\sqrt{48}\). We look for the largest perfect square factor of 48. Factoring 48 gives \(16 \times 3\). Since 16 is a perfect square (\(4^2\)), we have:
\[\sqrt{48} = \sqrt{16 \times 3} = \sqrt{16} \times \sqrt{3} = 4\sqrt{3}.\]
Finally, add the two simplified terms together:
\[3\sqrt{27} + \sqrt{48} = 9\sqrt{3} + 4\sqrt{3} = (9 + 4)\sqrt{3} = 13\sqrt{3}.\]
The expression is now in the simplest form, \(13\sqrt{3}\), where 13 is an integer and 3 is a square-free integer. The result of the simplification is \(\boxed{13\sqrt{3}}\).
Simplifying radical expressions is a foundational skill in geometry and physics. For instance, if you are calculating the length of a diagonal in a room with dimensions based on square roots, or working with vectors, you often end up with sums of radicals. Simplifying these into a common radical form, like \(13\sqrt{3}\) meters, makes it much easier to compare values or perform further calculations, such as estimating the numerical distance without needing a calculator for every step.