If a∈R and both the roots of x2−6ax+9a2+2a−2=0 exceed 3, then a lies in the interval

Given equation can be written as (x−3a)2=2(1−a)We must have 1−a≥0 or a≤1Also, x=3a±21−a.Both the roots will exceed 3, if smaller of the two roots viz3a−21−a exceeds 3, that is, 3a−21−a>3⇒ 3(a−1)>21−a                                            (1) For a=1 this is not possible. For a<1we can write (1)as                      −3(1−a)2>21−awhich is no possibleThus, there is not real value of a.