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

Given equation is  x2−2ax+a2+a−3=0If roots are real then, D≥0⇒        4a2−4a2+a−3≥0⇒        −a+3≥0⇒        a−3≤0 ⇒a≤3As roots are less than 3, hence f(3)>0    9−6a+a2+a−3>0⇒    a2−5a+6>0⇒    (a−2)(a−3)>0⇒ Either a<2 or a>3.Hence, only a < 2 satisfy