If the roots of the equation x2-2ax+a2+a-3=0 are real and less than 3, then
a<2
2≤a≤3
3<a≤4
a>4
Given equation 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≤3
As roots are less than 3, hence f(x)>0
9-6a+a2+a-3>0⇒a2-5a+6>0
⇒(a-2)(a-3)>0⇒either a<2 or a>3
Hence a<2 satisfy all.