rewrite the irrational square root of √48 as the product of an integer and another irrational square root

1 answer

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

First, factor \( 48 \): \[ 48 = 16 \times 3 \]

Now, we can take the square root: \[ \sqrt{48} = \sqrt{16 \times 3} \]

Using the property of square roots that states \( \sqrt{a \times b} = \sqrt{a} \times \sqrt{b} \), we have: \[ \sqrt{48} = \sqrt{16} \times \sqrt{3} \]

Since \( \sqrt{16} = 4 \), we can write: \[ \sqrt{48} = 4\sqrt{3} \]

Therefore, the irrational square root of \( \sqrt{48} \) can be expressed as the product of an integer and another irrational square root: \[ \sqrt{48} = 4\sqrt{3} \]