Rewrite the irrational square root 48−−√

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

224−−√
2 start root 24 end root

43–√
4 start root 3 end root

212−−√
2 start root 12 end root

412−−√

1 answer

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

First, notice that:

\[ 48 = 16 \times 3 \]

Now we can rewrite the square root:

\[ \sqrt{48} = \sqrt{16 \times 3} = \sqrt{16} \times \sqrt{3} = 4\sqrt{3} \]

So, \( \sqrt{48} \) can be expressed as \( 4\sqrt{3} \).

Therefore, the correct response is:

4√3