Question

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

Square root 12
Square root 9
4 Square root 2
Square root 15

Answer 1

To find out which option combined with \(5 \sqrt{27}\) yields a rational number, we first need to determine the form of \(5 \sqrt{27}\) and simplify it.

First, simplify \(\sqrt{27}\):
\[
\sqrt{27} = \sqrt{9 \times 3} = \sqrt{9} \times \sqrt{3} = 3 \sqrt{3}
\]

Now, multiply it by 5:
\[
5 \sqrt{27} = 5 \times 3 \sqrt{3} = 15 \sqrt{3}
\]

Now, to yield a rational number, we need to find an option that, when multiplied by \(15 \sqrt{3}\), results in a cancellation of the \(\sqrt{3}\). This means we are looking for a term that contains \(\sqrt{3}\) such that when we multiply, the \(\sqrt{3}\) can be cancelled out.

Let's analyze the options:

1. **Square root 12**:
\[
\sqrt{12} = \sqrt{4 \times 3} = 2 \sqrt{3}
\]
\[
15 \sqrt{3} \times 2 \sqrt{3} = 30 \times 3 = 90 \quad (\text{rational number})
\]

2. **Square root 9**:
\[
\sqrt{9} = 3
\]
\[
15 \sqrt{3} \times 3 = 45 \sqrt{3} \quad (\text{not rational})
\]

3. **4 Square root 2**:
\[
4 \sqrt{2}
\]
\[
15 \sqrt{3} \times 4 \sqrt{2} = 60 \sqrt{6} \quad (\text{not rational})
\]

4. **Square root 15**:
\[
\sqrt{15}
\]
\[
15 \sqrt{3} \times \sqrt{15} = 15 \sqrt{45} = 15 \times 3 \sqrt{5} = 45 \sqrt{5} \quad (\text{not rational})
\]

The only option that gives a rational product when multiplied with \(15 \sqrt{3}\) is:

**Square root 12** (option 1).