The product of 5 square root 27 and blank would yield a rational number

1 answer

To make the product of \(5 \sqrt{27}\) and some number a rational number, you want to multiply it by a number that will remove the square root.

First, simplify \(5 \sqrt{27}\):

\[ \sqrt{27} = \sqrt{9 \times 3} = \sqrt{9} \times \sqrt{3} = 3 \sqrt{3} \]

So,

\[ 5 \sqrt{27} = 5 \times 3 \sqrt{3} = 15 \sqrt{3} \]

To make the product \(15 \sqrt{3} \cdot x\) rational, \(x\) must equal \(\sqrt{3}\) because when you multiply \(\sqrt{3}\) by itself, it yields a rational number:

\[ 15 \sqrt{3} \cdot \sqrt{3} = 15 \cdot 3 = 45 \]

Thus, the blank should be filled with \(\sqrt{3}\). Therefore, the complete statement is:

The product of \(5 \sqrt{27}\) and \(\sqrt{3}\) would yield a rational number.