Question

Find the product of 3, square root of, 5, end square root3
5
​
and 2, square root of, 2, end square root2
2
​
in simplest form. Also, determine whether the result is rational or irrational and explain your answer.
Answer
Attempt 1 out of 5

Result:


The result is
because it
be written as the ratio of two integers and its decimal expansion
terminate or repeat.

Answer 1

To find the product of \( 3\sqrt{5} \) and \( 2\sqrt{2} \), we start by multiplying the two expressions:

\[
3\sqrt{5} \times 2\sqrt{2}
\]

We can multiply the coefficients (the numbers outside the square roots) and the square roots separately:

\[
(3 \times 2)(\sqrt{5} \times \sqrt{2})
\]

Calculating the coefficients:

\[
3 \times 2 = 6
\]

Now, calculating the square roots:

\[
\sqrt{5} \times \sqrt{2} = \sqrt{5 \times 2} = \sqrt{10}
\]

Putting it all together, we have:

\[
3\sqrt{5} \times 2\sqrt{2} = 6\sqrt{10}
\]

Next, we need to determine whether \( 6\sqrt{10} \) is rational or irrational.

1. **Identifying \( \sqrt{10} \)**: The number \( \sqrt{10} \) is irrational because 10 is not a perfect square, and the square root of a non-perfect square is always irrational.
2. **Product of Rational and Irrational**: The number 6 is rational, and when you multiply a rational number (6) by an irrational number (\(\sqrt{10}\)), the result is always irrational.

Thus, the result of the product, \( 6\sqrt{10} \), is irrational.

### Final Answer
The result is \( 6\sqrt{10} \) because it cannot be written as the ratio of two integers and its decimal expansion does not terminate or repeat.