What is the irrational square root 63−−√

Question

rewritten as the product of an integer and another irrational square root, in simplest form?(1 point)

Answer 1

To rewrite \(\sqrt{63}\) as the product of an integer and another irrational square root, we can start by factorizing 63.

The prime factorization of 63 is: \[ 63 = 9 \times 7 = 3^2 \times 7 \]

Now we can take the square root: \[ \sqrt{63} = \sqrt{9 \times 7} = \sqrt{9} \times \sqrt{7} = 3\sqrt{7} \]

Thus, the irrational square root \(\sqrt{63}\) rewritten as the product of an integer and another irrational square root in simplest form is: \[ 3\sqrt{7} \]