If roots of x2−(a−3)x+a=0 are such that at least one of them is greater than 2 , then

Given equation is x2−(a−3)x+a=0 Now discriminant (D)=(a−3)2−4a ∵D=b2−4ac=a2−10a+9=(a−1)(a−9)Case I:  Both the roots are greater than 2.D≥0,f(2)>0,−B2A>2⇒(a−1)(a−9)≥0,4−(a−3)2+a>0 and a−32>2⇒a∈(−∞,1]∪[9,∞),a<10 and a>7⇒a≥[9,10)…….(1)Case II:  One root is greater than 2 and the other root is less than or equal   to 2. Hence, f(2)≤0,D>0⇒4−(a−3)2+a≤0,(a−1)(a−9)>0⇒a≥10 and a∈(−∞,1)∪(9,∞)⇒a∈[10,∞)……..(2) From Eqs. (1) and (2), a∈[9,10)∪[10,∞)⇒a∈[9,∞)