If 8 and 2 are the roots of $x^2+ax+\beta =0$and 3,3 are the roots of $x^2+\alpha x+b=0$then the roots of the equation $x^2+ax+b=0$  are

If 8 and 2 are the roots of $x^2+ax+\beta =0$

$\Rightarrow 8+2=-a \Rightarrow a=-10$

and 3,3 are the roots of $x^2+\alpha x+b=0$

$\Rightarrow 3\left(3\right)=b \Rightarrow b=9$

$$\begin{gathered} \therefore x^2+ax+b=0 \\ \Rightarrow x^2-10x+9=0 \\ \Rightarrow x^2-9x-x+9=0 \\ \Rightarrow x(x-9)-1(x-9)=0 \\ \Rightarrow (x-9)(x-1)=0 \\ \Rightarrow x=9,1 \end{gathered}$$