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

412−−√
4 start root 12 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 begin by simplifying \(\sqrt{48}\).

First, we notice that \(48\) can be factored into \(16\) and \(3\) (since \(48 = 16 \times 3\)). The square root of \(16\) is \(4\).

Now we can express \(\sqrt{48}\) as follows:

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

So, the correct response is: \(4\sqrt{3}\)