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:
-
Square root 12: \[ \sqrt{12} = \sqrt{4 \times 3} = 2 \sqrt{3} \] \[ 15 \sqrt{3} \times 2 \sqrt{3} = 30 \times 3 = 90 \quad (\text{rational number}) \]
-
Square root 9: \[ \sqrt{9} = 3 \] \[ 15 \sqrt{3} \times 3 = 45 \sqrt{3} \quad (\text{not rational}) \]
-
4 Square root 2: \[ 4 \sqrt{2} \] \[ 15 \sqrt{3} \times 4 \sqrt{2} = 60 \sqrt{6} \quad (\text{not rational}) \]
-
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).
1 answer