If the roots of the equation ${ x }^{ 2 }-8x+({ a }^{ 2 }-6a)=0$ are real, then

$x^{2}-8 x+\left(a^{2}-6 a\right)=0$

নিশ্চায়ক,

$\begin{array}{l} b^{2}-4 a c \\ =(-8)^{2}-4 \cdot\left(a^{2}-6 a\right) \\ =64-4 a^{2}+24 a\end{array}$

প্রশ্নানুসারে,

$\begin{array}{l}-4 a^{2}+24 a+64>0 \\ \Rightarrow 4 a^{2}-24 a-64<0 \quad[\text {(-)~দ্বারা~গুন }]\end{array}$

$\begin{array}{l}\Rightarrow a^{2}-6 a-16<0 \\ \Rightarrow a^{2}-8 a+2 a-16<0 \\ \Rightarrow a(a-8)+2(a-8)<0 \\ \therefore(a-8)(a+2)<0\end{array}$

$\begin{aligned} & a-8<0 \\ \Rightarrow & a<8\end{aligned}$

$\begin{array}{l}a+2<0 \\ a<-2\end{array}$

$\therefore$ শর্ত $-2<a<8$