Rewrite the irrational square root 48−−√

Question

as the product of an integer and another irrational square root.(1 point)
Responses

412−−√
4 start root 12 end root

43–√
4 start root 3 end root

224−−√
2 start root 24 end root

212−−√

Answer 1

To rewrite the irrational square root \(\sqrt{48}\) as the product of an integer and another irrational square root, we can factor \(48\) as follows:

1. Notice that \(48 = 16 \times 3\).
2. This allows us to rewrite \(\sqrt{48}\) as \(\sqrt{16 \times 3}\).
3. Using the property of square roots, we get \(\sqrt{48} = \sqrt{16} \cdot \sqrt{3} = 4\sqrt{3}\).

So the correct response is:

\[ 4\sqrt{3} \]