Verified Answer :
- Factor 27 as \(9 \times 3\) and simplify \(\sqrt{27}\) to \(3\sqrt{3}\), so \(3\sqrt{27}\) becomes \(9\sqrt{3}\).
- Factor 48 as \(16 \times 3\) and simplify \(\sqrt{48}\) to \(4\sqrt{3}\).
- Add the two simplified terms: \(9\sqrt{3} + 4\sqrt{3} = 13\sqrt{3}\).
- The final simplified form is \(\boxed{13\sqrt{3}}\).
Explanation
- Analysis of the Expression
We are tasked with simplifying the expression \(3\sqrt{27} + \sqrt{48}\). To simplify this expression, we need to express the radicands (27 and 48) in terms of their prime factors and identify any perfect square factors, as these can be pulled out of the square root radical. A square root of a product is the product of the square roots: \(\sqrt{x \cdot y} = \sqrt{x} \cdot \sqrt{y}\). For a square root of a perfect square \(k^2\), we have \(\sqrt{k^2} = k\).
- Simplifying the First Term
First, consider 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, incorporate the coefficient 3 from the original expression:
\[3\sqrt{27} = 3 \times (3\sqrt{3}) = 9\sqrt{3}.\]
- Simplifying the Second Term
Next, consider the term \(\sqrt{48}\). We can factor 48 by finding the largest perfect square factor. The factors of 48 include 1, 2, 3, 4, 6, 8, 12, 16, 24, 48. The largest perfect square factor is 16 (\(4^2\)). Thus, we write:
\[\sqrt{48} = \sqrt{16 \times 3} = \sqrt{16} \times \sqrt{3} = 4\sqrt{3}.\]
- Combining the Terms
Now, substitute the simplified terms back into the original expression:
\[3\sqrt{27} + \sqrt{48} = 9\sqrt{3} + 4\sqrt{3}.\]
Since both terms are multiples of \(\sqrt{3}\), we can add them like regular algebraic terms:
\[9\sqrt{3} + 4\sqrt{3} = (9 + 4)\sqrt{3} = 13\sqrt{3}.\]
The expression is now in its simplest radical form.
Examples
Simplifying radical expressions is a foundational skill in geometry and physics. For example, if you are calculating the diagonal of a rectangle with sides \(\sqrt{27}\) and \(\sqrt{48}\) units, you might use the Pythagorean theorem \(d = \sqrt{a^2 + b^2}\), but in other contexts, being able to add these lengths directly as \(13\sqrt{3}\) allows you to compare dimensions more easily, such as determining if a piece of material of length \(13\sqrt{3}\) will fit into a specific space. Mastering these simplifications helps in quick mental estimations and precise algebraic manipulation.