The values of p for which the difference between the roots of the equation$x^2+px+8=0$ is  2 are

Let  $\alpha , \beta$are the roots of$x^2+px+8=0$

$\alpha +\beta =-p , \alpha \beta =8$

$$\begin{gathered} NOW {\left(\alpha -\beta \right)}^2={\left(\alpha +\beta \right)}^2-4\alpha \beta \\ \Rightarrow {\left(\alpha -\beta \right)}^2=p^2-32 \\ \Rightarrow \left|\alpha -\beta \right|=\sqrt{p^2-32} \\ but given \left|\alpha -\beta \right|=2 \\ \Rightarrow \sqrt{p^2-32}=2 \\ \Rightarrow p^2-32=4 \\ \Rightarrow p^2=36 \\ \Rightarrow p=\pm 6 \end{gathered}$$