Question

Which product yields an irrational number? ○ A. √√9.3 ○ B. 7 √√25 ○ C. 2√ √647√81 ○ D. 4√√5-2√ √ 49

Answer 1

Let's analyze each option to determine if they yield an irrational number:

A. \( \sqrt{\sqrt{9.3}} \)

To evaluate this, we can express it as \( \sqrt{(9.3)^{1/2}} = (9.3)^{1/4} \). Since \( 9.3 \) is not a perfect fourth power, this expression yields an irrational number.

B. \( 7 \cdot \sqrt{\sqrt{25}} \)

Here, \( \sqrt{25} = 5 \), so \( \sqrt{\sqrt{25}} = \sqrt{5} \). Thus, this yields \( 7 \cdot \sqrt{5} \), which is also irrational since \( \sqrt{5} \) is irrational.

C. \( 2\sqrt{\sqrt{647\sqrt{81}}} \)

First, \( \sqrt{81} = 9 \), so we have \( 2\sqrt{\sqrt{647 \cdot 9}} = 2\sqrt{\sqrt{5823}} \). Since \( 5823 \) is not a perfect square, \( \sqrt{5823} \) is irrational, and so is \( 2\sqrt{\sqrt{5823}} \).

D. \( 4\sqrt{\sqrt{5}} - 2\sqrt{\sqrt{49}} \)

Calculating \( \sqrt{49} = 7 \), we have \( 2\sqrt{\sqrt{49}} = 2 \cdot 7^{1/2} = 2 \cdot 7^{1/2} = 2 \cdot \sqrt{7} \). Now, we have \( 4\sqrt{\sqrt{5}} - 2\sqrt{7} \). The term \( 4\sqrt{\sqrt{5}} \) is irrational because \( \sqrt{5} \) is irrational. Since both terms \( 4\sqrt{\sqrt{5}} \) and \( 2\sqrt{7} \) are irrational, we cannot directly conclude that their difference is irrational, since irrational minus irrational can be rational.

In summary:
- Options A, B, and C yield irrational numbers.
- Option D does not guarantee an irrational result since it is the difference of two irrationals.

Since we're looking for a product (not the difference), each of options A, B, and C yield irrational numbers but their products or sums aren't being considered directly.

However, if we simplify:

- A: \( \sqrt{\sqrt{9.3}} \) is irrational.
- B: \( 7 \cdot \sqrt{5} \) is irrational.
- C: \( 2\sqrt{\sqrt{5823}} \) is irrational.
- D: The difference still remains uncertain.

Thus, based on the provided options, **the most straightforward answer for yielding an obvious irrational number is**:

**A. \( \sqrt{\sqrt{9.3}} \)**