To rewrite the irrational square root \(\sqrt{63}\) as the product of an integer and another irrational square root, we first need to factor 63 into its prime factors:
\[ 63 = 9 \times 7 = 3^2 \times 7 \]
Now, we can separate the square root:
\[ \sqrt{63} = \sqrt{9 \times 7} = \sqrt{9} \times \sqrt{7} \]
Since \(\sqrt{9} = 3\), we can rewrite it as:
\[ \sqrt{63} = 3\sqrt{7} \]
Thus, \(\sqrt{63}\) rewritten as the product of an integer and another irrational square root in simplest form is:
\[ 3\sqrt{7} \]
1 answer