Find the product of 2, square root of, 4, end square root2 4 ​ and 5, square root of, 13, end square root5 13 ​ in simplest form.

Answer 1

To find the product of \( 2\sqrt{4} \) and \( 5\sqrt{13} \), we will first simplify each term:

1. Simplifying \( 2\sqrt{4} \): \[ \sqrt{4} = 2 \] Therefore, \[ 2\sqrt{4} = 2 \times 2 = 4 \]

2. The expression \( 5\sqrt{13} \) is already in its simplest form.

Now, we will multiply the two simplified terms:

\[ 4 \times 5\sqrt{13} = 20\sqrt{13} \]

Thus, the product of \( 2\sqrt{4} \) and \( 5\sqrt{13} \) in simplest form is:

\[ \boxed{20\sqrt{13}} \]

Answer 2

The result is irrational because it CANNOT be written as the ratio of two integers and its decimal expansion does NOT terminate or repeat. correct?

Answer 3

Yes, that is correct. The result \( 20\sqrt{13} \) is indeed irrational.

Explanation:

1. Irrational Numbers: A number is irrational if it cannot be expressed as a fraction (the ratio of two integers), and its decimal expansion does not terminate or repeat.

2. Square Roots of Non-Perfect Squares: Since \( 13 \) is not a perfect square, \( \sqrt{13} \) is irrational. Consequently, any non-zero multiple of an irrational number is also irrational.

3. Decimal Expansion: The decimal expansion of \( \sqrt{13} \) is approximately \( 3.60555... \), which continues infinitely without repeating. Therefore, \( 20\sqrt{13} \) will also have a non-terminating, non-repeating decimal expansion.

4. Conclusion: So, you are indeed correct in stating that \( 20\sqrt{13} \) is irrational.