$\begin{array}{l}Ifthedifferencebetweentherootsof{x}^2+ax+b=0isequaltothedifferencebetweentheroots\\ of{x}^2+px+q=0then{a}^2-p^2=\end{array}$

$\begin{array}{l}Let\alpha ,\beta arerootsof{x}^2+ax+b=0\\ \alpha +\beta =-a,\alpha \beta =b\\ {(\alpha -\beta )}^2={(\alpha +\beta )}^2-4\alpha \beta =a^2-4b\\ and\gamma ,\delta arerootsof{x}^2+px+q=0\\ \gamma +\delta =-p,\gamma \delta =q\\ {(\gamma -\delta )}^2={(\gamma +\delta )}^2-4\gamma \delta =p^2-4q\\ Given|\alpha -\beta |=|\gamma -\delta |\\ \Rightarrow {(\alpha -\beta )}^2={(\gamma -\delta )}^2\\ \Rightarrow a^2-4b=p^2-4q\\ \Rightarrow a^2-p^2=4(b-q)\end{array}$