$$(\sqrt3+\sqrt{11})^2$$​ is a/an:

Correct option is A

Given:

The expression is $$\left( \sqrt{3} + \sqrt{11} \right)^2.$$​​

Formula Used:

​$$(a + b)^2 = a^2 + 2ab + b^2$$​

Solution:

​$$\left( \sqrt{3} + \sqrt{11} \right)^2 = (\sqrt{3})^2 + 2 \times \sqrt{3} \times \sqrt{11} + (\sqrt{11})^2$$​​

​$$= 3 + 2 \times \sqrt{33} + 11 = 14 + 2\sqrt{33}$$​

Since $$\sqrt{33}$$​ is an irrational number, $$2\sqrt{33}$$​ is also irrational. Therefore, the sum $$14 + 2\sqrt{33}$$​ is an irrational number.

The expression $$\left( \sqrt{3} + \sqrt{11} \right)^2$$​ is an irrational number.

The correct answer is A: irrational number.