Question

The product 5 square root of 27 and _ would yield another rational number
Fill in the blank
A. Square root of 12
B. Square root of 9
C. 4 square root of 2
D. Square root of 15

Answer 1

To determine which option would yield a rational number when multiplied by \( 5\sqrt{27} \), we first simplify \( 5\sqrt{27} \).

We know that:

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

So:

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

Now we need to find which option, when multiplied with \( 15\sqrt{3} \), results in a rational number. A product will be rational if the square root factors cancel out.

Let’s analyze each option:

A. \( \sqrt{12} \)

\[
15\sqrt{3} \cdot \sqrt{12} = 15\sqrt{3} \cdot \sqrt{4 \cdot 3} = 15\sqrt{3} \cdot (2\sqrt{3}) = 15 \cdot 2 \cdot (\sqrt{3} \cdot \sqrt{3}) = 30 \cdot 3 = 90 \quad (\text{Rational})
\]

B. \( \sqrt{9} \)

\[
15\sqrt{3} \cdot \sqrt{9} = 15\sqrt{3} \cdot 3 = 45\sqrt{3} \quad (\text{Not Rational})
\]

C. \( 4\sqrt{2} \)

\[
15\sqrt{3} \cdot 4\sqrt{2} = 60\sqrt{6} \quad (\text{Not Rational})
\]

D. \( \sqrt{15} \)

\[
15\sqrt{3} \cdot \sqrt{15} = 15\sqrt{45} = 15\cdot 3\sqrt{5} = 45\sqrt{5} \quad (\text{Not Rational})
\]

From these findings, only Option A \( \sqrt{12} \) yields a rational number when multiplied by \( 15\sqrt{3} \).

Thus, the answer is:

**A. Square root of 12**