1.1 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 0,

To determine the values of \( p \) that make \( \sqrt{\frac{16}{p-2}} \) irrational, non-real, or undefined, let's break it down: For \( \sqrt{\frac{16}{p-2}} \) to be undefined, the denominator must equal zero. Therefore, when \( p - 2 = 0 \) or \( p = 2 \), the expression becomes undefined. To find when it's non-real, we look for when \( p - 2 < 0\). If \( p \) is any value less than 2, \( \sqrt{\frac{16}{p-2}} \) would yield a negative radicand (since 16 is non-negative and can't balance out a negative denominator), thus it’s non-real. This applies for \( p = 0, 1 \). For it to be irrational, the denominator \( p - 2 \) must not be a perfect square. When \( p = 3 \), \( 3 - 2 = 1 \) (a perfect square), which makes it rational. For \( p = 4 \), \( 4 - 2 = 2 \) (not a perfect square). Hence, the expression is irrational when \( p = 4\) and defined. In summary: - Undefined: \( p = 2 \) - Non-real: \( p = 0, 1 \) - Irrational: \( p = 4 \)