Question

What is the sum of the following expression? 2 108−−−√+48−−√+5 27−−√ (1 point)

Answer 1

To find the sum of the expression \(2\sqrt{108} + \sqrt{48} + 5\sqrt{27}\), we will simplify each square root first.

1. **Simplifying \(\sqrt{108}\)**:
\[
\sqrt{108} = \sqrt{36 \times 3} = \sqrt{36} \times \sqrt{3} = 6\sqrt{3}
\]
Thus, \(2\sqrt{108} = 2 \times 6\sqrt{3} = 12\sqrt{3}\).

2. **Simplifying \(\sqrt{48}\)**:
\[
\sqrt{48} = \sqrt{16 \times 3} = \sqrt{16} \times \sqrt{3} = 4\sqrt{3}
\]

3. **Simplifying \(\sqrt{27}\)**:
\[
\sqrt{27} = \sqrt{9 \times 3} = \sqrt{9} \times \sqrt{3} = 3\sqrt{3}
\]
Thus, \(5\sqrt{27} = 5 \times 3\sqrt{3} = 15\sqrt{3}\).

Now, substituting back into the original expression:

\[
2\sqrt{108} + \sqrt{48} + 5\sqrt{27} = 12\sqrt{3} + 4\sqrt{3} + 15\sqrt{3}
\]

We can combine these terms:
\[
12\sqrt{3} + 4\sqrt{3} + 15\sqrt{3} = (12 + 4 + 15)\sqrt{3} = 31\sqrt{3}
\]

Thus, the sum of the expression is:
\[
\boxed{31\sqrt{3}}
\]