Question

The product of 5 square root of 27 and BLANK would yield a rational number.
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:
\[
\sqrt{27} = \sqrt{9 \cdot 3} = \sqrt{9} \cdot \sqrt{3} = 3\sqrt{3}
\]
Thus,
\[
5\sqrt{27} = 5 \cdot 3\sqrt{3} = 15\sqrt{3}
\]

Now, we want to find which of the given options, when multiplied with \( 15\sqrt{3} \), results in a rational number. A product of two numbers is rational if the irrational parts (surds) cancel each other out.

We will analyze each option:

**A. \( \sqrt{12} \)**
\[
\sqrt{12} = \sqrt{4 \cdot 3} = 2\sqrt{3}
\]
\[
15\sqrt{3} \cdot 2\sqrt{3} = 30\sqrt{3 \cdot 3} = 30 \cdot 3 = 90 \quad \text{(rational)}
\]

**B. \( \sqrt{9} \)**
\[
\sqrt{9} = 3
\]
\[
15\sqrt{3} \cdot 3 = 45\sqrt{3} \quad \text{(irrational)}
\]

**C. \( 4\sqrt{2} \)**
\[
15\sqrt{3} \cdot 4\sqrt{2} = 60\sqrt{6} \quad \text{(irrational)}
\]

**D. \( \sqrt{15} \)**
\[
15\sqrt{3} \cdot \sqrt{15} = 15\sqrt{3 \cdot 15} = 15\sqrt{45} = 15 \cdot 3\sqrt{5} = 45\sqrt{5} \quad \text{(irrational)}
\]

Out of all options, only option A, \( \sqrt{12} \), produces a rational number when multiplied by \( 15\sqrt{3} \).

Thus, the correct answer is:
**A. Square root of 12**.