If (2+\sqrt{m}) is irrational and (m) is a positive integer, which (m) can be correct?

(18)

The key idea is that the square root of a positive integer is rational when the integer is a perfect square, and irrational when it is not a perfect square. Adding the rational number 2 to an irrational number keeps the result irrational. Therefore, we must find the option whose square root cannot be written as a fraction or whole number.

For 18,
\(\sqrt{18}=3\sqrt{2}\), which is irrational because 2 is not a perfect square. In contrast, \(\sqrt{16}=4\), \(\sqrt{25}=5\), and \(\sqrt{36}=6\) are rational. Hence \(2+\sqrt{18}\) is irrational, so option A is correct. The other choices produce rational sums.

This is a Class 10 Mathematics question. Chapter: Polynomials. Topic: Irrational numbers and real numbers.