If \( p \in\{0 ; 1 ; 2 ; 3 ; 4\} \), determine the value of p such that \( \sqrt{\frac{16}{p-2}} \) 1.1 .1 is irrational 1.1 .2 is non real 1.1 .3 is undefined

To determine the value of \( p \) such that \( \sqrt{\frac{16}{p-2}} \) is irrational, non-real, or undefined, we need to analyze the expression under the square root. Given that \( p \in \{0, 1, 2, 3, 4\} \), we will substitute each value of \( p \) into the expression and determine the nature of the result. 1.1.1. Irrational: For the expression to be irrational, the denominator \( p-2 \) must be non-zero and the result under the square root must not be a perfect square. 1.1.2. Non-real: For the expression to be non-real, the denominator \( p-2 \) must be zero, which means the expression is undefined. 1.1.3. Undefined: For the expression to be undefined, the denominator \( p-2 \) must be zero, which means the expression is non-real. Let's substitute each value of \( p \) into the expression and determine the nature of the result. Simplify the expression by following steps: - step0: Solution: \(\sqrt{\frac{16}{\left(p-2\right)}}\) - step1: Remove the parentheses: \(\sqrt{\frac{16}{p-2}}\) - step2: Use the properties of radicals: \(\frac{\sqrt{16}}{\sqrt{p-2}}\) - step3: Simplify the expression: \(\frac{4}{\sqrt{p-2}}\) - step4: Simplify: \(\frac{4\sqrt{p-2}}{p-2}\) Substitute each value of \( p \) into the expression and determine the nature of the result. 1.1.1. Irrational: For \( p = 0 \): \( \sqrt{\frac{16}{0-2}} = \sqrt{\frac{16}{-2}} = \sqrt{-8} \) is irrational. 1.1.2. Non-real: For \( p = 1 \): \( \sqrt{\frac{16}{1-2}} = \sqrt{\frac{16}{-1}} = \sqrt{-16} \) is non-real. 1.1.3. Undefined: For \( p = 2 \): \( \sqrt{\frac{16}{2-2}} = \sqrt{\frac{16}{0}} \) is undefined. Therefore, the value of \( p \) such that \( \sqrt{\frac{16}{p-2}} \) is irrational is \( p = 0 \), non-real is \( p = 1 \), and undefined is \( p = 2 \).