The values of ' a ' for which one root of the equation x^2-(a+1)...

x²-(a+1) x+a²+a-8=0 Since roots are different, therefore $\mathrm{D} \gt 0$$\begin{aligned}& \Rightarrow(a+1)^2-4 \left(a^2+a-8\right)>0 \\ & \Rightarrow(a-3)(3 a+1) < 0 \end{aligned}$$There are two cases arises. Case I.$a-3 \gt 0$and$3 a+1 \lt 0\Rightarrow a \gt 3 and a \lt -$\frac{11}{3} Hence, no solution in this case Case II :$a-3 \lt 0$and$3 a+11 \gt 0\Rightarrow a \lt 3 and a \gt -$\frac{11}{3}$\therefore \quad-\frac{11}{3} \lt a \lt 3 \Rightarrow-2 \lt a \lt 3