If the roots of the equation x2-2ax+a2+a-3=0 are real and less than 3, then

Given equation x2-2ax+a2+a-3=0If roots are real, then D≥0⇒4a2-4(a2+a-3)≥0⇒-a+3≥0⇒a-3≤0⇒a≤3As roots are less than 3, hence f(x)>09-6a+a2+a-3>0⇒a2-5a+6>0⇒(a-2)(a-3)>0⇒either a<2 or a>3Hence a<2 satisfy all.