Answer:
[tex]\text{A.}\ \sqrt{2}\times\sqrt{18}\\\\\text{B.}\ \sqrt{2}\times\sqrt{6}[/tex]
Step-by-step explanation:
A: The root will be rational if the product of the numbers under the radicals is a perfect square. For this part, there are a couple of choices.
[tex]\text{1.}\ \sqrt{2}\times\sqrt{18}=\sqrt{36}=6\\\\\text{2.}\ \sqrt{6}\times\sqrt{24}=\sqrt{144}=12[/tex]
__
B: The root will be irrational if the product of the numbers under the radicals is not a perfect square. For this part, there are many choices.
[tex]\text{1.}\ \sqrt{2}\times\sqrt{6}=\sqrt{12}\\\\\text{2.}\ \sqrt{2}\times\sqrt{12}=\sqrt{24}\\\\\text{3.}\ \sqrt{2}\times\sqrt{24}=\sqrt{48}\\\\\text{4.}\ \sqrt{6}\times\sqrt{12}=\sqrt{72}\\\\\text{5.}\ \sqrt{6}\times\sqrt{18}=\sqrt{108}\\\\\text{6.}\ \sqrt{12}\times\sqrt{18}=\sqrt{216}\\\\\text{7.}\ \sqrt{12}\times\sqrt{24}=\sqrt{288}\\\\\text{8.}\ \sqrt{18}\times\sqrt{24}=\sqrt{432}[/tex]
