Rewrite the irrational square root 48−−√ as the product of an integer and another irrational square root.(1 point)

Question

Responses

212−−√
2 start root 12 end root

43–√
4 start root 3 end root

224−−√
2 start root 24 end root

412−−√

Answer 1

To rewrite \( \sqrt{48} \) as the product of an integer and another square root, we start by factoring 48:

\[ 48 = 16 \times 3 \]

Since 16 is a perfect square, we can simplify \( \sqrt{48} \) as follows:

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

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

The correct response is:

4√3.