Rewrite the irrational square root 48−−√

Question

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

43–√
4 start root 3 end root

224−−√
2 start root 24 end root

412−−√
4 start root 12 end root

212−−√
2 start root 12 end root

Answer 1

To rewrite the irrational square root \( \sqrt{48} \) as the product of an integer and another irrational square root, we can break down \( 48 \) into its prime factors:

\[ 48 = 16 \times 3 \]

Since \( 16 \) is a perfect square, we can take it out of the square root:

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

So the correct response is:

4 start root 3 end root (or \( 4\sqrt{3} \)).