Question

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

Answer 1

To find a value to fill in the blank such that the product of \( 5 \sqrt{27} \) and that value yields a rational number, we first need to simplify \( 5 \sqrt{27} \).

We know:

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

So, we have:

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

Now, we want to find a number \( x \) such that:

\[
15 \sqrt{3} \cdot x \text{ is a rational number.}
\]

For this product to be rational, \( x \) must be a value that cancels out the \( \sqrt{3} \). Specifically, we can let:

\[
x = \frac{1}{\sqrt{3}} \quad \text{(which is the multiplicative inverse of } \sqrt{3} \text{)}
\]

Now we can check:

\[
15 \sqrt{3} \cdot \frac{1}{\sqrt{3}} = 15 \cdot \frac{\sqrt{3}}{\sqrt{3}} = 15 \cdot 1 = 15
\]

Thus, \( 15 \) is indeed a rational number.

Therefore, the blank can be filled with:

\[
\frac{1}{\sqrt{3}}
\]

Alternatively, you could also multiply by \( \sqrt{3} \) to result in:

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

So, another option is:

\[
\sqrt{3}
\]

To summarize, you can fill in the blank with either:

\[
\frac{1}{\sqrt{3}} \quad \text{or} \quad \sqrt{3}
\]