Which product yields an irrational number?

Question

A. Square root of 9 · 3 (The square root is only over the 9, not the · 3)
B. 7/8 · square root of 25
C. 2 square root of 64 (square root ends there and is only on the 64) · 7 square root of 81
D. 4 square root of 5 (Square root ends there and is only on the 5) · 2 square root of 49

Answer 1

To determine which product yields an irrational number, we need to analyze each option.

A. \( \sqrt{9} \cdot 3 \)

• \( \sqrt{9} = 3 \), so: \[ \sqrt{9} \cdot 3 = 3 \cdot 3 = 9 \]
• This is a rational number.

B. \( \frac{7}{8} \cdot \sqrt{25} \)

• \( \sqrt{25} = 5 \), so: \[ \frac{7}{8} \cdot \sqrt{25} = \frac{7}{8} \cdot 5 = \frac{35}{8} \]
• This is a rational number.

C. \( 2 \cdot \sqrt{64} \cdot 7 \cdot \sqrt{81} \)

• \( \sqrt{64} = 8 \) and \( \sqrt{81} = 9 \), so: \[ 2 \cdot \sqrt{64} \cdot 7 \cdot \sqrt{81} = 2 \cdot 8 \cdot 7 \cdot 9 = 1008 \]
• This is a rational number.

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

• \( \sqrt{49} = 7 \), so: \[ 4 \cdot \sqrt{5} \cdot 2 \cdot 7 = 8 \cdot 7 \cdot \sqrt{5} = 56 \cdot \sqrt{5} \]
• Since \( \sqrt{5} \) is irrational, \( 56 \cdot \sqrt{5} \) is also irrational.

Therefore, the product that yields an irrational number is:

D: \( 4 \cdot \sqrt{5} \cdot 2 \cdot \sqrt{49} \)