Question

rewrite radical 45 in as a product of an integer and irrational square root

Answer 1

To rewrite \(\sqrt{45}\) as a product of an integer and an irrational square root, we start by factoring 45:

\[
45 = 9 \times 5
\]

Now, we can express the square root as follows:

\[
\sqrt{45} = \sqrt{9 \times 5}
\]

Using the property of square roots that \(\sqrt{a \times b} = \sqrt{a} \times \sqrt{b}\):

\[
\sqrt{45} = \sqrt{9} \times \sqrt{5}
\]

Since \(\sqrt{9} = 3\), we can simplify this further:

\[
\sqrt{45} = 3 \times \sqrt{5}
\]

Thus, \(\sqrt{45}\) is rewritten as a product of an integer and an irrational square root:

\[
\sqrt{45} = 3\sqrt{5}
\]