Question

the product of 3square root 12 and ___________ would yield a rational number
the product of 3square root 12 and ___________ would yield a irrational number

5square root 3
6square root 2

Answer 1

To determine the missing values that would yield a rational or irrational product with \(3\sqrt{12}\), let's first simplify \(3\sqrt{12}\).

We can simplify \(\sqrt{12}\):

\[
\sqrt{12} = \sqrt{4 \times 3} = \sqrt{4} \times \sqrt{3} = 2\sqrt{3}
\]

Thus,

\[
3\sqrt{12} = 3 \times 2\sqrt{3} = 6\sqrt{3}
\]

Now we have \(6\sqrt{3}\). We will analyze the products with the two values you mentioned: \(5\sqrt{3}\) and \(6\sqrt{2}\).

1. **For a rational product:**
The product of \(6\sqrt{3}\) and a number should equal a rational number.

So,

\[
6\sqrt{3} \cdot k \text{ is rational if } k \text{ is } \frac{1}{\sqrt{3}} \text{ or any rational multiple of } \sqrt{3}.
\]

However, using your provided options:

- \(6\sqrt{3} \cdot 5\sqrt{3} = 30 \cdot 3 = 90\) (a rational number)
- \(6\sqrt{3} \cdot 6\sqrt{2} = 36\sqrt{6}\) (this will be irrational because \(\sqrt{6}\) is irrational)

Therefore, **the blank which yields a rational number is \(5\sqrt{3}\)**.

2. **For an irrational product:**
The product would be irrational when multiplied with \(6\sqrt{2}\) because:

- \(6\sqrt{3} \cdot 6\sqrt{2} = 36\sqrt{6}\) (irrational product since \(\sqrt{6}\) is irrational)

Thus, **the blank which yields an irrational number is \(6\sqrt{2}\)**.

In summary:
- The product of \(3\sqrt{12}\) and \(5\sqrt{3}\) yields a rational number.
- The product of \(3\sqrt{12}\) and \(6\sqrt{2}\) yields an irrational number.