Question

The product of 5to the square root of 27 and ____ would yield a rational number.
A square root of 12
B square root of 9
C 4 to the square root of 2
D square root of 15

Answer 1

To determine which option, when multiplied with \(5 \sqrt{27}\), yields a rational number, we first simplify \(5 \sqrt{27}\):

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

Thus, we have:

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

Next, we look at each option and check if multiplying it by \(15 \sqrt{3}\) results in a rational number.

**Option A: \( \sqrt{12} \)**

\[
15 \sqrt{3} \cdot \sqrt{12} = 15 \sqrt{3} \cdot \sqrt{4 \cdot 3} = 15 \sqrt{3} \cdot 2 \sqrt{3} = 15 \cdot 2 \cdot 3 = 90
\]

This is rational.

---

**Option B: \( \sqrt{9} \)**

\[
15 \sqrt{3} \cdot \sqrt{9} = 15 \sqrt{3} \cdot 3 = 45 \sqrt{3}
\]

This is not rational.

---

**Option C: \( 4 \sqrt{2} \)**

\[
15 \sqrt{3} \cdot 4 \sqrt{2} = 60 \sqrt{6}
\]

This is not rational.

---

**Option D: \( \sqrt{15} \)**

\[
15 \sqrt{3} \cdot \sqrt{15} = 15 \sqrt{3 \cdot 15} = 15 \sqrt{45} = 15 \sqrt{9 \cdot 5} = 15 \cdot 3 \sqrt{5} = 45 \sqrt{5}
\]

This is not rational.

---

After examining all options, only **Option A: \(\sqrt{12}\)** results in a rational number when multiplied with \(15 \sqrt{3}\).

So the answer is:

**A. \(\sqrt{12}\)**