The equation will have real and equal roots if its discriminant is zero. Given: \(x^2 + (p - 3)x + p = 0\) Discriminant = \((p - 3)^2 - 4 \cdot p \cdot 1\) So, according to the question: \[ (p - 3)^2 - 4p = 0 \Rightarrow p^2 - 6p + 9 - 4p = 0 \Rightarrow p^2 - 10p + 9 = 0 \Rightarrow p^2 - (9 + 1)p + 9 = 0 \Rightarrow p^2 - 9p - p + 9 = 0 \Rightarrow p(p - 9) - 1(p - 9) = 0 \Rightarrow (p - 9)(p - 1) = 0 \] \[ \therefore p = 9 \quad \text{or} \quad p = 1 \] Therefore, for \(p = 9\) or \(p = 1\), the equation \(x^2 + (p - 3)x + p = 0\) will have real and equal roots.