Find the values of x if p + 7 = 0, q - 12 = 0 and x² + px + q = 0.

Since , p + 7 = 0, q - 12 = 0 it means p = -7 and q = 12.

Given,

$x^2 + px + q = 0 \\[0.5em] \Rightarrow x^2 + (-7)x + 12 = 0 \\[0.5em] \Rightarrow x^2 - 7x + 12 = 0 \\[0.5em] \Rightarrow x^2 - 4x - 3x + 12 = 0 \\[0.5em] \Rightarrow x(x - 4) - 3(x - 4) = 0 \\[0.5em] \Rightarrow (x - 3)(x - 4) = 0 \text{ (Factorising left side) } \\[0.5em] \Rightarrow x - 3 = 0 \text{ or } x - 4 = 0 \text{ (Zero-product rule) } \\[0.5em] x = 3 \text{ or } x = 4$

Hence, the values of x are 3 , 4.