αt2=Fv+βx2 . Find the dimension formula for [α] and [β] (here t= time, F= force, v= velocity, x= distance )

Since, dimension of F v=[Fv]=MLT-2LT-1=ML2T-3So, βx2 should also be ML2 T-3[β]x2=ML2 T-3 [β]=ML4 T-3And Fv+βx2 will also have dimension ML2 T-3 so LHS should also have the same dimension ML2 T-3So, [α]t2=ML2 T-3 {α}=ML2 T-1