Question

find each product. then determine which conclusion can be drawn based on the res
$\sqrt{2} \cdot \sqrt{2}$
$\sqrt{5} \cdot \sqrt{7}$
$\sqrt{2} \cdot \sqrt{18}$
$\sqrt{2} \cdot \sqrt{6}$
\bigcirc the product of two irrational numbers is always irrational.
\bigodot the product of two irrational numbers can be rational or irrational.
\bigcirc the product of two irrational numbers is always rational.
\bigcirc the product of two irrational numbers cannot be determined.

Explanation:

Step1: Calculate $\boldsymbol{\sqrt{2} \cdot \sqrt{2}}$

Using the property $\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}$, we get $\sqrt{2 \cdot 2} = \sqrt{4} = 2$ (rational).

Step2: Calculate $\boldsymbol{\sqrt{5} \cdot \sqrt{7}}$

Applying the same property, $\sqrt{5 \cdot 7} = \sqrt{35}$ (irrational).

Step3: Calculate $\boldsymbol{\sqrt{2} \cdot \sqrt{18}}$

Simplify: $\sqrt{2 \cdot 18} = \sqrt{36} = 6$ (rational).

Step4: Calculate $\boldsymbol{\sqrt{2} \cdot \sqrt{6}}$

Simplify: $\sqrt{2 \cdot 6} = \sqrt{12} = 2\sqrt{3}$ (irrational).
From these calculations, we see products can be rational (e.g., $2, 6$) or irrational (e.g., $\sqrt{35}, 2\sqrt{3}$).

Answer:

The product of two irrational numbers can be rational or irrational. (Option: The product of two irrational numbers can be rational or irrational.)