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=0 If 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(3)>0 9−6a+a2+a−3>0⇒a2−5a+6>0 ⇒(a−2)(a−3)>0⇒either a<2 or a>3Hence  a<2 satisfy all.