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

1 answer

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.