The roots $x_1$ and $x_2$ of the equation $x^2 + px + 12 = 0$ are such that $x_2 - x_1 = 1$. Find $p$.
Given quadratic equation: $\;\;\;$ $x^2 + px + 12 = 0$
Relation between the roots $x_1$ and $x_2$ of the given quadratic equation is
$x_2 - x_1 = 1$ $\;\;\; \cdots \; (1)$
Sum of roots of the given quadratic equation: $\;$ $x_1 + x_2 = -p$ $\;\;\; \cdots \; (2)$
Adding equations $(1)$ and $(2)$ simultaneously gives
$2 x_2 = 1 - p$ $\implies$ $x_2 = \dfrac{1-p}{2}$
Substituting the value of $x_2$ in equation $(1)$ gives
$x_1 = x_2 - 1 = \dfrac{1 - p}{2} - 1 = - \left(\dfrac{1 + p}{2}\right)$
Product of roots of the given quadratic equation: $\;$ $x_1 \cdot x_2 = 12$ $\;\;\; \cdots \; (3)$
Substituting the values of $x_1$ and $x_2$ in equation $(3)$ gives
$- \left(\dfrac{1 - p}{2}\right) \left(\dfrac{1 + p}{2}\right) = 12$
i.e. $\;$ $1 - p^2 = - 48$
i.e. $p^2 = 49$
i.e. $\;$ $p = \pm 7$