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

Question

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. $\sqrt{12}$
$$\sqrt{12}=\sqrt{4\cdot 3}=2\sqrt{3}$$ $$15\sqrt{3}\cdot 2\sqrt{3}=30\sqrt{3\cdot 3}=30\cdot 3=90\text{(rational)}$$

B. $\sqrt{9}$
$$\sqrt{9}=3$$ $$15\sqrt{3}\cdot 3=45\sqrt{3}\text{(irrational)}$$

C. $4\sqrt{2}$
$$15\sqrt{3}\cdot 4\sqrt{2}=60\sqrt{6}\text{(irrational)}$$

D. $\sqrt{15}$
$$15\sqrt{3}\cdot \sqrt{15}=15\sqrt{3\cdot 15}=15\sqrt{45}=15\cdot 3\sqrt{5}=45\sqrt{5}\text{(irrational)}$$

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.