$$\begin{gathered} Ifonerootoftheequation{x}^2+px+12=0is4,whiletheequation \\ {x}^2+px+q=0hasequalroots,thenthevalueof\text{'}q\text{'}is \end{gathered}$$

$\begin{array}{l}If4isrootofx+px+12=0\\ \Rightarrow 4^2+p\left(4\right)+12=0\\ \Rightarrow 4p=-28\\ \Rightarrow p=-7\\ TheEquation{x}^2+px+q=0hasequalroots\\ then{p}^2-4q=0\\ \Rightarrow 49-4q=0\\ \Rightarrow 4q=49\\ \Rightarrow q=\frac{49}{4}\end{array}$